Can A Function Be Continuous At Only One Point?

[@ExplosiveBrohoof]

My favorite example of a function with degenerate continuity properties is this: If x is irrational, let f(x) = 0. If x is rational and nonzero, write x = p/q, where p and q are coprime and q is positive. Then set f(x) = 1/q. Finally, set f(0)=1. This function ends up being continuous at every irrational point, and discontinuous at every rational point.

[@atreidesson]

Wow that's very clever! I love it too

[@52flyingbicycles]

So a function can be continuous at any number of finite points, countably infinite points, or uncountably infinite points. Very nice.

[@dolorsitametconsectur]

52flyingbicycles moment

[@p07a]

Continuous at Uncountably infinite points? Which function is that?

[@52flyingbicycles]

@@p07a most continuous functions lol

[@albertonullstein3631]

The identity, exponential, all polynomials etc. ​@@p07a

[@TayaTerumi]

@p07a y=x²

[@bjornfeuerbacher5514]

As far as I know, it's usual to say that f(x) = 1/x is neither continuous nor discontinuous at x = 0, since it only makes sense to talk about continuity for values of x where a function actually is defined.

[@johnk6190]

Yes, exactly in fact f(x)=1/x is continuous for all x≠0, in other words f is a continuous function. If you define f:IR-->IR with f(x)=1/x for all x≠0 and f(0)=a for some real number a, then f is not continuous at 0 (because the limit of f at 0 does not exist). In an intuitive scene, the property "you can draw the graph without lifting your pen" only holds if f is continuous in a single interval (as one can see for example with f(x)=1/x with x>0 or x

[@koenth2359]

No, if a function f(x) is undefined for a certain x, it is not continuous at that x. A function is continuous in a point if the value at that point equals the limit in its direct vicinity, otherwise it is discontinuous. If the value, the limit, or both don't exist, one can never claim equality. Furthermore, a function is continous on a domain interval if it is continuous on every value of that interval. Would you say that 1/x is continuous on the interval [-1,1]?

[@idjles]

If it’s undefined it is definitely discontinuous

[@johnk6190]

@@koenth2359 It really depends on your definition of continuity. If we use the one presented in the video then you need to have f(x_0) so f needs to be defined at x_0. Wikipedia actually has the same example with 1/x. You can find it here: en.wikipedia.org/wiki/Continuous_function. With that definition in the points that f is not defined then we cannot talk continuity. So 1/x is neither continuous nor discontinuous on the interval [-1,1] as it is not defined at 0. Furthermore, I disagree with this: "A function is continuous in a point if the value at that point equals the limit in its direct vicinity, otherwise it is discontinuous." because that means that a function is discontinuous at its isolated points if it has any. For example take sqrt(x^4-x^2) at 0.

[@jamesfullwood7788]

A point not in the domain of a function does not satisfy the definition of the function being continuous at that point. Therefore, a function is not continuous at every point not in its domain.

[@oli_dev]

What a brilliant solution. It seems so obvious in hindsight too!

[@yakov9ify]

Nice video as always. Many interesting generalizations of this question exist, one of my favourites is that the set of continuity points of any function into R is always a G_delta set, a countable intersection of open sets.

[@alejrandom6592]

I saw the thumbmail, kept scrolling, and then had to come back cuz the doubt consumed me. "Continuous everywhere function?" I said, and then I watched the entire video, mesmerized by the elegant explanation. My doubts vanished, replaced by a deeper understanding and appreciation for the beauty of mathematics. Reminds me of my college days, when my professor, a kind elderly man, took extra time to explain a tricky concept. His patience ignited a passion within me, and I went on to pursue a career in math. This video brought back those fond memories. Thank you for creating content that not only educates but inspires. You're keeping the spark of curiosity alive, one video at a time. ❤

[@DrBarker]

Thank you, glad you enjoyed it!

[@mikenorman2525]

Having studied analysis as an undergraduate (many decades ago) with the formal delta/epsilon definition, my gut feeling was no of course you can't have a function that is only continuous at a single point. But then the fact that the question was even posed in the first place led me to think the answer must in fact somehow be yes which is what I went for. And so I was right! However if you asked me to define such a function I would have failed. Bah! Maths eh?

[@ZeanIkLaurie]

I feel like our definition of continuous is kinda random, incomplete, and doesn't actually match the stuff we wanted it originally to describe

[@alexwarner3803]

​@@ZeanIkLaurie the definition of continuity, I assume you mean strictly "continuous" isn't random, is completely by design and led by sound principles with a goal in mind, and is quite robust. It matches the "stuff (then and currently) well enough to get to where we are in differential geometry, topology, and more. Not sure what you mean by "incomplete (sparing talk of Gödel Incompleteness)." Strictly defined depending on what type of continuous you are talking about, really. Continuously differentiable ⊂ Lipschitz continuous ⊂ α-Hölder continuous ⊂ uniformly continuous = continuous A fun pathological example is the Weierstrass function. Continuous everywhere, but nowhere differentiable 🤯 did challenge notions at the time it was published. Stating the concept of continuity being random has no bearing tbh.

[@bobbun9630]

Similar background here. I did figure out where he was going before he got there, but I also agree a little bit with the other poster in this thread. Not that the definition is "random", but that the notion of "continuous at an isolated point" represents a case where the delta/epsilon definition leads to a conclusion that is at odds with our intuitive understanding of what continuity means. Of course, there are a lot of mathematical results that are counterintuitive.

[@jonahansen]

@@bobbun9630 I'm in the same boat as you guys; took analysis as an undergrad years ago. I was pretty sure it was going to turn out no, but see what happened. The setup is contrived but clever, and it seems sort of pathological and counterintuitive. I spent some time trying to figure out a definition that wouldn't allow it, but in the end decided I just didn't have the right intuition in the first place and that that was what I needed to change.

[@ZeanIkLaurie]

@@alexwarner3803 the "stuff" is what should guide our definitions. It's also mental abstractions. In any case what he talks about is a function R to R but that should be distinguished as a projection or 1D version, based on a specific parametrization of what is actually continuity for most. And in this sense now you should see what I mean by incomplete.

[@koenth2359]

Dr Barker keeps impressing with great insights and new information. Really great channel!

[@DrBarker]

Thank you so much!

[@alexwarner3803]

Inb4 Continuously differentiable ⊂ Lipschitz continuous ⊂ α-Hölder continuous ⊂ uniformly continuous = continuous A fun pathological example is the Weierstrass function. Continuous everywhere, but nowhere differentiable 🤯 did challenge notions at the time it was published.

[@isomeme]

I love a good paradox, and the concept of a function being continuous only at a single point is a *great* paradox. I'll be turning this one over in my head for quite a while. Thanks for this video!

[@MichaelRothwell1]

I enjoyed this video, but I'm afraid I have a small nit to pick. Right near the start, I didn't feel very comfortable about the choice of f(x)=1/x as an example of a discontinuous function. Whilst (as you said) it's not continuous everywhere, it is continuous on its domain. Later at 3:10, when you attempted to illustrate the discontinuity of f(x)=1/x at a point on its domain, I really cringed! It would have been better (in my opinion) to give a step function as a simple example of a discontinuous function.

[@davidcarter8269]

Is it really worthy of being cringe-inducing or is it just a simple mistake that warrants a small and simple correction?

[@jesusthroughmary]

Are you really afraid, though? It seems like you're rather chuffed at the opportunity to pick a small nit.

[@DrBarker]

@@MichaelRothwell1 Fair point, I wanted to avoid step functions in the beginning, just to make the video more accessible, but yes the example f(x) = 1/x certainly doesn't work with the more rigourous definition of continuity!

[@ValentineRogue]

If you really want to nitpick, it's easy to define a function continuous at only one point by expressing it as a function in a topology with only 1 point to begin with. Any topology on the set results in the function being continuous. All discussions in the video seem to have implied the idea that we're looking at continuity on the set of all real number under the usual metric, a reasonable thing to have as an unspoken assumption, since this is a youtube video and not an academic paper. It's true that f(x):R/{0}->R=1/x is continuous everywhere as you have pointed out, but f(x):R->R=1/x is discontinuous at x=0, which is the point he was making graphically in the video.

[@moonshine7753]

​@@ValentineRoguesorry to interrupt but f(x):R ---> R = 1/x is not well defined. You'd need to choose a point for f(0)

[@j.r.8176]

Such an underrated channel. Lovely.

[@arthurkassis]

I entered in the video already knowing the answer, but the explanation was really good.

[@Dr.Cassio_Esteves]

Great explanation! Every thing was so clear! I would love if you did a topology playlist.

[@flockofwingeddoors]

Amazing watch at usual. You do incredibly clear explanations, and you always pick such interesting topics to talk about. Bravo!

[@DrBarker]

Thank you!

[@jakobr_]

My favorite way to think about continuity is to say that a function is continuous at an input if that output is what we “expect” it to be by looking at nearby outputs. This works really well as a plain english version of “continuous means the output at an input and the limit of outputs as the inputs approach that input are equal”

[@chonkycat123]

Always a pleasure watching your videos. More analysis !!

[@theimmux3034]

short answer: yes. F.ex. f: R -> R x -> x if x is rational and otherwise x -> -x is only continuous at x = 0.

[@XahhaTheCrimson]

Even the answer is pretty trivial, I haven't thought about that and I think it's really worth it.

[@landsgevaer]

So, for f(x) = { 0 if x is rational { cos(1/x) if x is irrational we even have *infinitely many* continuous points in any open interval that contains zero, yet not continuity in *any* open interval as a whole. Funny...

[@MihaiNicaMath]

The hardest bonus problem in my 1st analysis class was about proving that not all sets are possible as the set of discontinuity points of a function. It turns out the Baire category theorem limits what is possible! For example you can't make a function that is only continuous on the irrational numbers.

[@oniondesu9633]

This brings up an almost philosophical question to me of what continuity is. Don't quite know how to describe this well, so stay with me. But I presume the way we arrived at this rigorous definition of continuity is that we already had a fuzzy concept of a "continuous function" at first, then thought about examples of continuous and discontinuous functions and looked for a property that distinguished between them. By looking at it this way, you could argue that this example you've shown is a counterexample which shows the definition isn't perfect and has edge cases. I mean the sentence "continuous at a single point" is almost an oxymoron in itself. But ultimately I think it doesn't matter, this definition has served us very well and this is just a philosophical argument over what the purpose of the term "continuous" is.

[@tombratcher6938]

Saw the picture, thought, "huh, nice example". I assume you explained it well too :-)

[@scollyer.tuition]

1/x is usually described as continuous since it's continuous at every point in domain, x= 0 not being in its domain. I guess this may depend on your precise definition of "continuous" however.

[@df-163]

Generalisation: if f is a function from R to R, then the set of continuity points of f is a countable intersection of open sets. In descriptive set theory, this is called a G-delta set So in this case, a singleton set {x} is the countable intersection of (x-1/n, x+1/n) open intervals as n goes to infinity. But the set of rational numbers Q is NOT a G-delta set. As such, it is impossible for a function to only be continuous at rational numbers! However, the set of irrational numbers is a G-delta set. Exercise for the reader to show that Thomae's function satisfies this!

[@MercuriusCh]

Simple question for 1y of Uni. Just x * D(x) is continuos at x=0 Where D(x) is 1 if x in Q, 0 if x in R\Q

[@jellybabiesarecool4657]

Can you please do more analysis videos analysing non-elementary functions?

[@jcsahnwaldt]

The sine example function is continuous at countably many points, but discontinuous in the neighbourhood of each such point. Can this be extended to uncountably many points? Is there a function that is continuous at uncountably many points, but discontinuous in the neighbourhood of each such point? Question 457893 on math stackexchange provides a similar function based on the Cantor set. Are there others, maybe based on less "pathological" sets / functions?

[@spineo2387]

I was trying to disprove it could be done, but this is what i found. Had to search around a bit at some points cause it's been awhile: You can argue that the points of continuity these kinds of functions yield will always be closed. Therefore, the set of points of discontinuity (call it U) must be open. We are also assuming that U must be dense, else there exists some interval containing no discontinuities. Since U is open and dense, it's complement (our set of continuities) must be nowhere dense. So sounding pretty cantor-set-y at that point no matter what. The example I could find was using the function f(x) = d(x,A) / [d(x,A)+d(x,b)] With A some cantor set (or similar) and b a point outside of A. Which like I never would have come up with, but is surprisingly simple and similar to the video's concept. It still feels like one of those things a continuous function shouldn't be able to swing to me, but I guess it can!

[@jcsahnwaldt]

@@spineo2387 Great stuff! Quick question: I don't know what d(x,A) and d(x,b) means. I guess it's a distance, but I'm not sure. Could you explain? Thanks!

[@spineo2387]

@@jcsahnwaldt yeah, exactly it's the distance function where d(x,y) is the distance between two points. You can also extend such functions to define the distance between two sets, which we would need to define the "distance to tge cantor set A" in that function

[@jcsahnwaldt]

@@spineo2387 If I understand correctly, d(x, A) is the distance from x to the nearest point in A, right?

[@spineo2387]

@@jcsahnwaldt in this case yes, that can be our definition. In general, defining the distance to a set can be just slightly more subtle (what if there is no single nearest point? For example, what should d(0, R+), the distance from zero to the set of positive numbers, be?). But in this particular case, that concern isn't relevant to a closed set A in the real numbers. There will always be nearest point(s) in A that can give us our distance to A.

[@archangecamilien1879]

My first reflex was that it can't be possible, lol, was trying to find a demonstration, not sure I could...now, as soon as I clicked the link to comment, spoiler alert, it seems to be possible...wanted to try to figure this one out, lol, the type of question I find interesting...

[@honeybee9455]

Same reaction, this video keeps reminding me that the formal definition is the first step to gaining new intuition.

[@archangecamilien1879]

I have to say I didn't watch the video yet...

[@cparks1000000]

If you have any bounded function f, then g defined by g(x)=(x-a)f(x) is continuous at a. You can also get a function that is discontinuous at every rational point in [0, 1].

[@77Chester77]

I always like when you backwards and disappear out of the screen. It reminds me of the Homer Simpson Meme where he disappears in the bushes 🙂

[@DrBarker]

@@77Chester77 I'd never made that connection, you're right! 😂

[@YoungPhysicistsClub1729]

it's questions like this that push math to the edge, the very essence of calculus and limits is based off of a very fuzzy definition of a point

[@wqltr1822]

I have a real analysis book by jay cummings which has some nice examples of 'unintuitive' (when it comes to continuity) functions, one of them is similar to this

[@Meghana_Nallamilli]

Same!

[@MC5677]

1/x is continuous, but you need to wrap around infinity. that's not really possible for us to do, but the line does continue like any other continuous function.

[@JacksonBockus]

Pick a value of δ that satisfies the definition of continuity at x = 0 for ε = 0.1 in the function f(x) = 1/x If one exists, the function might be continuous. If it doesn’t, the function is discontinuous at x = 0.

[@fernandojackson7207]

To add a nerdish point. The set of points of continuity is a G_ {delta} set, meaning the countable intersection of open sets.

[@JulienCote-mx6bf]

Fun question! Very interesting.

[@CatholicSatan]

Not being any sort of mathematician, the property of continuity occurring at just the one minuscule point seems to be a bit of a misnomer (given by what I'd understand as "continuous"). What is "continuous" at a single point (or at multiple discontinuous(?) points as mentioned below) then telling me?

[@jellybabiesarecool4657]

A function being continuous as a point means that any "small" change in input from that point gives a "small" change in the function's output

[@kappasphere]

Continuity at a point doesn't describe a property of only that point, instead it relates the point to its neighborhood of an arbitrary size. Because it has to work for any size, you can always still see it at some macroscopic scale that is still larger than the infinitesimal.

[@CatholicSatan]

@@jellybabiesarecool4657 "Small" in quotes. But in these cases, small implies _any_ shift from that point causes a (large) jump, not a small change in output. Hence my query about it being a misnomer and wondering what "continuous" at a point means.

[@russellsharpe288]

@@CatholicSatan Why do you say that "any shift from that point [zero here] causes a (large) jump, not a small change in output". In the example presented, any shift from zero produced a change in output no greater than that shift itself (zero for irrationals, the shift itself for rationals). The change in output is bounded by the change in input. But you are right to say that this example is very odd intuitively. Perhaps the best way to think about it is to say that the formal definition captures what we want for "normal" functions, and we just accept somewhat counterintuitive implications (like continuity at a single point) for bizarre ones for the sake of having a nice consistent and homogenous theory. After all bizarre functions like f(x) = x for x in Q, f(x) = 0 for x not in Q are quite counterintuitive already. (I believe it took a long time for mathematicians to become comfortable even accepting such things as functions at all)

[@fahrenheit2101]

At a point vs on an interval The latter is more intuitive

[@sergodobro2569]

Let the function be x when it is rational x, and -x when irrational numbers f(x) = x, x is rational; -x, x is irrrational It will be continuous only in 0 Yay! I wrote the example when I haven't watched the video and now he introduced almost the same idea!

[@sergodobro2569]

Yay! I wrote the example when I haven't watched the video and now he introduced almost the same idea

[@allvods1385]

You can also build a function that is continuous over a closed interval, by multiplying the Q indicator function by a C infinity function that vanishes over [-1,1] and then goes to 1 gradually on each side. But is the set of continuity always closed?

[@spineo2387]

I think with this style of construction it should be: I started writing out a longer proof working directly from using limit points of the set of continuity, but all we really need is two things: 1) such functions will be continuous at their zeros (and no where else) 2) the preimage of a closed set is closed Since {0} is closed, that should do it!

[@deinauge7894]

it is not. say for example f(x)=0 for all irrational x, and for all x with 0

[@spineo2387]

@@deinauge7894 that's distinct from the videos example in the sense that that original f(x) was discontinuous everywhere on that interval. Clearly it's not true that every function in general is continuous on a closed set. But *if you're starting out with a continuous function* then the set of continuity for that function times our indicator function will be closed by the above argument.

[@rtg_onefourtwoeightfiveseven]

Mostly good video, but I do have to raise a point out the gate. The function 1/x is (at least to my understanding) continuous, because it's continuous at every point where it's defined. The intuitive idea of "you can draw it without lifting your pen" does not work in this example; standard definitions of continuity such as the epsilon-delta definition have that the function is continuous.

[@tfae]

Indeed, if you extend it with a point at infinity, then it's continuous at x = 0 as well.

[@atussentinel]

Direchlet function bounded by y=|x|, really easy to construct and prove.

[@TsuzuraYuuki]

I haven’t seen the solution yet, my idea is a function defined as follow: If rational, y=x If irrational, y=-x Limit doesn’t not exist everywhere except at 0, and just so happen y(0)=0 Edit: just saw the solution, have to say this is very fun maths snack, thanks for the video

[@toddshoemaker4285]

OK, I went on and on about this question with copilot by initially asking the simple question: Is there any logical point for stating that a function is continuous at only one point? His answer was "yes, because it is useful in understanding the behavior of functions that are not uniformly continuous across their entire domain." I then followed up with the question: "You stated that there is a logical point for stating that a function is continous at only one point by stating that it is useful in understanding the behavior of functions that are not uniformly continuous across their entire domains. This turns out to be a circular argument because the user has already defined the discontinuous domains, and thus it is assumed to be understood" His answer: "I see your point"

[@VidNudistKid]

Can we create a function f(x) that is continuous where x is rational, and discontinuous everywhere else?

[@dlevi67]

I don't think so - between any 2 rational points there are infinite irrational points. So you could build a function that is discontinuous there, but not continuous.

[@GhostyOcean]

It seems quite simple to construct a function using the fact that an isolated point is automatically continuous and the function f(x) = { 0 if x=0; 1 if x is irrational and |x|>1; 0 if x is rational and |x|≥1

[@asdfghyter]

could you also have a function only be continuous on an open or closed interval? I think a piecewise function could give the answer yes to one of them at least. what about only continuous in countably infinitely many points between 0 and 1?

[@johnpaterson6112]

This suggests to me that the formal definition of continuity is in need of improvement.

[@anandarunakumar6819]

Great video, perhaps gives a good extension to spectral theory and Hilbert space concepts.

[@notmymain2256]

Fun fact, you can even have a function differentiable in one point which is only continuous in that point (es. x^2*dirichlet)

[@דניאלהאיש]

For a function to be continues it limits should be equal it value so for x+ epsilon the value should be near the point for every delta so the function also continues for x+epsilon because you can choose smaller epsilon to reduce this epsilon

[@andylenk959]

Can use rounding to draw on more than one point as with this formula: floor(6*floor((n^2+23)/6)/(n^2+23))n

[@kotnihal]

Too me down memory lane. Super!

[@chinchao]

Brilliant!

nice video! but there is a small error at 1:58, because you say, that "this is not connected to each other" but it is because Q is dense in R. between two numbers (from R or from Q) there is always a number from Q, therefore Q is connected.

[@peterruf1462]

No the density works both ways. Between two irrational numbers is a rational one and between two rational ones is a irrational.

[@poisonedward7104]

This is probably not the point of the video, but if you have a function f : [0,1] U {2} -> R, then take f(x) = 0 if x is irrational, 1 if x is rational(both inside [0,1]) and let f(2) = 2 say, this function is continuous at 2 only.

[@mathunt1130]

Classic example that every undergraduate encounters on a second course in analysis...

[@GabriTell]

It certainly depends on the topology we are working on. For example, if we take "τ = {ℝ, ∅, {1}}" then the function "f : ℝ-->ℝ" such that "f(x) = 1" for all "x∈ℝ" is only continuous at "x=1" on the topological space "(ℝ, τ)".

[@blizzard8958]

The definition of continuous i learned in my clac class that it is continuous at f(x) iff f(x) exists, lim as c approaches x of f(c) exists, and that those two equal each other, is this the same or a different definition?

[@kristenmork4316]

I think your definition of continuous needs to be less than or equal to epsilon for the delta = epsilon example to work because f(delta) = delta = epsilon for rational delta.

[@worldnotworld]

Hmmm... the possibility of such an answer actually makes me dissatisfied with the epsilon-delta definition of continuity. It becomes somehow meaningless; there seems to be nothing special about such points despite their meeting the definition! Or is there anything special about them?

[@GabriTell]

🤷

[@viktorsmets29]

One small note, the sine function example doesnt work with f(x) = sin(x) for rational x, as the zeros of this function are integer multiples of pi, and thus not rational (besides 0). f(x) = sin(pi * x) for rational x and 0 for irrational x would work however.

[@Papierkorb2292]

I think it would still be continues whenever x is a multiple of pi though, because you can pick arbitrarily close rational numbers

[@MichaelRothwell1]

Well spotted! f(x)=sin(x/π) when x∈ℚ would work a lot better!

[@garyzan6803]

With only rational numbers, you can only get arbitrarily close to zero, but you cannot get zero itself You could however choose f(x) to be sin(π*x), and you would have infinite zeroes Edit: I just saw that the original comment suggested the same function >.

[@Papierkorb2292]

@@garyzan6803 The function from the video is defined to be zero for every irrational number x, so f(π) would be zero. Then, for every ε>0, you can pick a range around x, such that the absolute value of the sine of every rational number in that range is smaller than ε (and for irrational numbers the difference is 0 of course).

[@m.guypirate6900]

no, his example does work because the rational sines approach zero, the irrational numbers approach zero, and the exact value of the function is also zero

[@rescyy2235]

Can we see an integral from -inf to inf of a discontinuous function, and continuous at some point? Like the first one you came up with

[@moonshine7753]

Well, it turns out that it doesn't change much, and with this type of functions it's actually surprisingly simple. Intuitively, since the rational numbers are numerable, they are a smaller infinity than the irrational numbers, so they count pretty much nothing. This can be formalised in measure theory and it turns out that the measure of a numerable set is 0. Now, if you integrate over a set of measure 0, you get 0 no matter which function you have. So for the integral of that function you can separate it in a sum of two integrals, one over the rationals and one over the irrationals. But the first one is 0, and the second one is also 0 because the function is 0 at the irrationals. So in this case it's just 0

[@rescyy2235]

@@moonshine7753 Fascinating. I would love to see a proof of this. Can you reference me one?

[@moonshine7753]

@@rescyy2235 well... it's knowledge I got from my course on measure theory, so I don't think there is just a single proof. If you want, my professor used Rudin's "Real and complex analysis" as a base for her lessons (you can probably find a pdf online), but it's a terrible book, very dense. If you want I can give some explanation by not really proving everything. The idea is that you assign a measure to (almost) every set, which represents how big it is. In the real numbers the measure of an interval for example is its length. This is because to create this measure we take as a basis Riemann's integral. So first of all we define an integral with respect to a measure (the Lebesgue integral), then for every function that we can integrate with Riemann we want a measure that makes both integrals coincide. If you have a very simple function, for example one that is 1 in some set and 0 everywhere else, then the Lebesgue integral of that function is just the measure of that set. You can then extend this to every other function basically by doing some limits and using simple functions. The point is that the measure of a single point is 0, because you can see that point like an interval of length 0. Another good property is that if you have a numerable amount of disjoint sets, the measure of their union is just the sum of their measures. So now you have that Q is just a numerable union of single points of measure 0, so the measure of Q is still 0. I'm... really not good at explaining things, but we had to define and prove a lot of things to get to these results. There are a lot of intricacies in how you define rigorously these things. I think if you search measure theory you might find some useful sources, but I honestly don't know.

[@Fine_Mouche]

in projective world, do 1/x keep his asymptote or it became continue ? (in this same world x² give an elipse)

[@JacksonBockus]

1/x is a hyperbola, and since all conic sections are closed loops in the projective plane 1/x will be too, as far as I know.

[@alexblokhuis]

Great video! 💛

[@Tata-ps4gy]

Okay, I did this function before watching the video. It's domain are rational numbers. First we have to create the function LD(q) which returns the last digit of q. This is ofc discontinuous everywhere. F(q) = LD(q) • q This is discontinuous everywhere except in q=0 😎

[@stQZuO]

Nice!

[@ТимурГубаев-ж8ы]

Virgin: Actually, you can‘t talk about discontinuity of f(x) = 1/x at x = 0 😧 How could you make that silly mistake? 😠 My day is ruined now thanks to you 😫 Chad: As always, great video, Dr Barker 👍 Keep them coming 💪

[@DrBarker]

@@ТимурГубаев-ж8ы 😂😂😂

[@sil1235]

Well done

[@retrogamingfun4thelife]

Question related to the function where rationals output 1 and other numbers output 0. Since Q is countable and R is not countable, i expect to have continuity on the 0es. Why is this not the case?

[@bjornfeuerbacher5514]

How do you get from "the number of points where the value of the function is 0 is not countable" to "the function is continuous at these points"? The second statement does not follow from the first one, why should it?

[@russellsharpe288]

Actually the restriction of the function to the rationals - that is, the function considered only as acting on rationals - IS continuous, since it is everywhere 1. (This has nothing to do with countability though: the restriction of the function to the irrationals is also continuous, since it is everywhere 0). But the function as defined is from the reals, and since every real number r (rational or irrational) has both rational and irrational numbers arbitrarily close to it, there will be (given any delta you care to choose, however small) some f(x) = 0 and some f(y) = 1 with both x,y within delta of r. That makes f discontinuous at r. And since r was arbitrary, this shows that it is discontinuous everywhere

[@jotch_7627]

​@@bjornfeuerbacher5514these definitions dont work on vibes. you cant just point at two sentences and decide they mean different things because they use different words. in this case the two really are unrelated, but that must be proven, not assumed

[@jotch_7627]

the epsilon-delta definition of continuity requires that given any finite non-zero vertical range (epsilon), there must exist a finite non-zero horizontal range (delta) where *all* inputs map to outputs within that vertical range. in this case, you can never pick a delta small enough to exclude the nearby rational points, so the function cant be continuous by that definition

[@retrogamingfun4thelife]

@@jotch_7627 thanks. But this “suggest” to me that R and Q have the same cardinality because any element in R-Q is “surrounded” by an element in Q. I’m perfectly conscious that this is just a wrong deduction, but i can’t see where is the mistake. Another doubt i have: does a definiton like the one choosed for the above function, needs a proof to show that the function actually exists?

[@dean532]

Can you try applying this notion of continuous at “one point” with imaginary numbers (probably needn’t any Complex Analysis) just the imaginary Im{}

[@david-komi8]

I first thought about sqrt(-abs(x)), what about that?

[@ivanovtv9817]

thumbnail kinda spoiled the answer. after seeing two dotted lines intersecting I instantly thought of x*Dirichlet function

[@ReaganStoleMyDick]

Can you make a separate Real Analysis ASMR channel?

[@sheikchilli8670]

what happens if we fourier transform the rational/irrational function?

[@wydadiyoun]

basically if it were proven (it was not) the direct conclusion to the result of this exercise is that the cardinal of Q is equal to cardinal of R-Q

[@gileadedetogni9054]

But the cardinalities of Q and R-Q are different, Q is countable and R-1 is uncountable

[@wydadiyoun]

@@gileadedetogni9054 but he just did has a direct consequence that they have the same cardinaly as each irrational number is surrounded only by two rational and vice versa! if there is an irrational surrounded by another irrational (or vice versa) then there is a continuity there and thus this video is garbage

[@gileadedetogni9054]

@wydadiyoun ah, got your point, but that not the thing. This property is called being dense, and two subsets of a set can be dense in the bigger set without having the same cardinality

[@jotch_7627]

​@@wydadiyounthe real numbers are not discrete. no number is "surrounded" by two other numbers. *any* range that includes more than one number includes an uncountably infinite amount. in the case of rationals, it is only countably infinite, but infinite nonetheless.

[@davidgillies620]

It seems counterintuitive that the cardinality of the reals is strictly greater than the cardinality of the rationals, but the rationals are dense in the reals. I bet this is an area of confusion for those just beginning measure theory or transfinite numbers.

[@GabriTell]

It stops looking counterintuitive when you realise that all rationals can be expressed as "a/b". And, since there are as many ordered pairs "(a,b)∈ℤ²" as intergers (you can even check it intuitively by putting all "(a,b)" pairs on a Cartesian plane and matching them all in a spiral shape) and since the set of all irreducible pairs is a subset of "ℤ²", there must be as many rationals as intergers. This is not a rigorous argument, but an intuitive way to think of it.

[@jotch_7627]

​@@GabriTellthis can explain why the set of rationals is countably infinite, but it doesnt explain density. after all, the integers are *not* dense in the reals even though they can map 1:1 to rationals

[@MikeRosoftJH]

Yes: f(x)=x for x is rational, f(x)=-x for x is irrational. This function is only continuous at 0.

[@ronbannon]

1/x is continuous on its domain

[@fbrtnrsthf]

Just think of f(x)=x on the rationals and constantly zero elsewhere…. Only continuous at the origin.

[@romywilliamson4981]

Yes. The domain could be a single point. Then any function on it is continuous (in a boring, vacuous sense).

[@JacksonBockus]

Nope, can’t be continuous in the rigorous sense if it is only defined at one point. f(x) = 0 if x = 0; undefined if x ≠ 0. What is the limit as x approaches 0 of f(x)?

[@milasudril]

If you took f(x) = x^2 for x rational, would it then be differentiable at zero?

[@nobody-sq3nq]

assuming that f(x) = 0 for x ∉ ℚ then yes, here's why: for differentiability at zero we need to look at the limit as x approaches zero of (f(x) - f(0)) / (x-0) and since f(0)=0 that's just f(x) / x now let's plug in the definition of f(x): for x ∈ ℚ we get x^2 / x = x and for x ∉ ℚ we get 0 / x = 0 which is exactly the first example given in the video as shown in the video that function is continuous at zero and it has a value of zero, therefore the limit exists and is zero, meaning f is differentiable at zero and f'(0)=0

[@milasudril]

@@nobody-sq3nq And if we take x^n n-> infinity it is infinitely differentiable only in one point. Or is it the entire interval ]-1, 1[

[@nobody-sq3nq]

@@milasudril depends a bit on what you mean with n -> infty, i haven't checked thoroughly, but I'm pretty sure for a fixed n the function would be n times differentiable at 0 as we are dealing with a sequence of functions, we can look at pointwise and uniform convergence for taking the limit, for |x| > 1 the sequence diverges, so let's only look at [-1, 1] for pointwise convergence we look at the limit for a fixed x, which leads to the function f(1) = f(-1) = 1 and f(x) = 0 for x ∈ (-1,1); this function is constant and therefore infinitely differentiable except for x = 1 and x = -1 where the function is not continuous the sequence x^n does not converge uniformly if x can get arbitrarily close to 1 or -1, so you have to choose a smaller interval than (-1,1); and if you do, the resulting limit is the constant zero function, which again is infinitely differentiable

[@hlvaneeden]

It is my understanding that there are countably many rational numbers, but uncountably many irrational numbers. So there are many (infinitely many) more irrational numbers than rational numbers. That implies that the function f(x) = 1 when x is irrational and f(x) = 0 when x is rational should be continuous in infinitely many points, since there is not a rational number between every irrational number?

[@nobody-sq3nq]

while there uncountably many irrational numbers and only countably many rationals, there actually exists a rational number between any two irrational numbers, which works because infinities are weird and break intuition specifically the rationals are a dense subset of the reals

[@schwingedeshaehers]

there is a prove, that you can get a rational between 2 real numbers. i think the easiest was to understand is, if a real number has a fixed length, and then is only followed by 0, it is rational. if you have 2 real numbers, they have to be different at some point (and assume that they are bigger then 0), then you can use the bigger ine, and replace anything behind the difference with 0. it is lower then the bigger value, because we cut some places, and it's higher then the other one, because it has a higher number at a position before it ends

[@hlvaneeden]

@@nobody-sq3nq So there is a rational number between any two irrationals, but there are infinitely many irrationals between any two rational numbers?

[@jotch_7627]

​@@hlvaneedenthere are infinitely many rationals between any two numbers, just as there are infinitely many irrationals. both are infinite, but one is countably infinite and the other is uncountable. countability is not what matters here, *density* is

[@midas-holysmoke7642]

So in between 2 rational numbers, chosen arbitrarily close to each other (whatever you can define this), there is no continuous set of rational numbers. So rational and irrational numbers keep switching to each other for arbitrarily small steps? Sounds very counterintuitive for me... I wonder how we can prove that

[@jotch_7627]

yep, and the same holds for irrationals. the property is called "density", if you want to look up various proofs for it

[@midas-holysmoke7642]

@@jotch_7627 I find this amazing

[@VeteranVandal]

So, f(x)={x, if x is in Q; 0 otherwise} is an example? It's continuous in 0 only. Ah, it's exactly the example you gave.

[@jcsahnwaldt]

Is there a function that is continuous at uncountably many points but also discontinuous at uncountably many points?

[@ajfernandez216]

f(x)=x when x 0 and is irrational f(x)=1 when x > 0 and is rational The above function is continuous for positive negative values of x and discontinuous for nonnegative values.

[@jcsahnwaldt]

@@ajfernandez216 Oh, you're right. That's pretty simple. I was hoping for something more interesting. 🙂 I'll reword my question.

[@LaTortuePGM]

before i watch the vid, i'm gonna say smth like 1_Q(x)*x would work. that means it's x if x is rational, 0 otherwise.

[@LaTortuePGM]

let's fucking gooooooo

[@Joffrerap]

i guess now we can ask if a function can be discontinuous at every point except one where it's differentiable. edit: just thought about it, it's not that hard, f(x) = x² over the rationnals and 0 everywhere else is differentiable at 0 new challenge : can we find a function discontinuous everywhere but convex?

[@nightowl9512]

ehem... 1/x is continuous everywhere where its defined. It's not even defined for x=0 so asking whether its continuous at x=0 is kinda meaningless...

[@schwingedeshaehers]

(x-2)/(x-2) is not defined at 2, but still continues (as it is 1 everywhere except at 2)

[@nightowl9512]

@@schwingedeshaehers … which is a fundamentally different type of function than 1/x (with a removable discontinuity)

[@schwingedeshaehers]

@@nightowl9512 that's true, but it still shows that it doesn't make sense to say, that just because it isn't defined, that it can't be interesting if it is continuous at a point

[@nightowl9512]

one has to be careful when comparing singularities, and my original comment was for 1/x especially, ie with an essential singularity (which I could of course have made more clear). Removable singularities constitute a whole different type of singularities ​@@schwingedeshaehers

[@hoggoe7623]

Would an example of a function continuous at only one point be:- f(x)={x, xEQ {-x, xE/Q Where it's only continuous at 0?

[@AhmedHan]

y = x ^ (-|x|) Is this a continuous function anywhere?

[@nobody-sq3nq]

it's continuous for positive x and for zero if you define 0^0 as 1 I don't know a definition that works for (negative number)^(non-integer) but if you restrict the domain to only allow negative integers then it's still continuous as isolated points are always continuous

[@minerscale]

If you set delta = epsilon won't it |f(delta)| = epsilon when delta js rational? But the requirement was thst delta was strictly less than epsilon. Unless you meant less than or equal to? Or maybe I'm missing something.

[@jotch_7627]

delta isnt bounded by epsilon at all, it just has to exist and be non-zero for any possible epsilon value. if you can prove that the value of epsilon itself would work for delta, then you meet that requirement, but it isnt the only way to do it.

[@minerscale]

@@jotch_7627 Ah of course thank you. Epsilon delta is such a simple concept but I get lost in the weeds of notation all the time, I should sit down and do a few problems.

[@RGP_Maths]

Don't forget that the stipulation was to apply to the OPEN interval (x0 - d, x0 + d), so x is always < delta away from x0, at which f(x) is therefore < epsilon away from f(x0). So delta = epsilon was legit.

[@minerscale]

@@RGP_Maths ah I see. Got it cheers :)

[@hellbowe.]

if there is x_0 that is continuous compared to x, so x must be continuous compared to x_0, so there are at least two points, am I wrong?

[@ninwu]

Wait so what is the slope at that point?

[@zunaidparker]

It doesn't have a defined slope. The function is continuous at 0, but it isn't differentiable there. Now if he has picked x^2 as the value for X rational, then the derivative at zero would exist and would be equal to 0.

[@dneary]

Isn't this a pretty trivial question? f(x) = 0 (x in Q) or x (x !in Q) - I seem to recall proving that this function is continuous only at x = 0 in an undergraduate course (possibly Analysis 1)

[@obansrinathan]

I mean yeah it’s analysis 1 so trivial for trained mathematicians, but even general science doesn’t do analysis, let alone general KZbin which is the audience.

[@tukan1652]

the next question would be could a function be differentiable at only 1 point?

[@edmundwoolliams1240]

Do you teach in schools?

[@marcosmerino9369]

i am not a mathematician, so maybe i have somethinf missing, but this relies in the fact that irrational numbers and rational numbers intercalate each other. But i am not sure that is true. Say you have some irrational number Q and you add an infinitesimally small irrational number q, then wouldn't the sum be irrational and have no rational numbers in between this making this false. Hope i made myself clear :)

[@spineo2387]

The issue would be that there is no such thing as an "infinitely small number" in the rea number system. The closest we have would be zero, which is rational. In discussing calculus we sometimes appeal to the idea of an infinitely small number to get intuitive arguments more simply stated, but that is best understood as (usually) just an intuition, not a rigorous, literal statement *or* (much, much less often) as us working in a more subtle system than the real numbers that includes objects that can be interpreted that way. In either case, for the real numbers, the construction you describe can't actually be achieved

[@marcosmerino9369]

@@spineo2387 hmm ok, but what I'm trying to convey is that the assumption that rational and irrational numbers intercalate doesn't need to be true, if in any case at all the function has two points with irrational numbers then it'd be continuous also there

[@spineo2387]

@@marcosmerino9369 let me know where i might clarify better, but the point i was making above is that that conclusion you're reaching is faulty because there is no such thing as an infinitely small number. "Intercalate" isn't necessarily the best way to visualize things because there is no such thing as the "next" real number after some specific x. However, if we look at some interval around any x, *no matter how small* there will always be infinitely many irrational and rational numbers. Therefore what you describe is simply impossible within the real number system.

[@marcosmerino9369]

@@spineo2387 ok I’m following, thanks for the clarification. I will do a bit more research on this but I am starting to understand, thanks!

[@jotch_7627]

also good to keep in mind that two numbers being irrational does not suggest, let alone guarantee that their sum is also irrational

[@sleekweasel]

I always thought continuity had something to do with differentiability - this feels cheaty - like engineering vs physics. Not saying it's wrong, just doesn't feel very elegant.

[@JacksonBockus]

Differentiability implies continuity but continuity does not imply differentiability. For example, f(x) = |x| is not differentiable at x = 0 but it is continuous at x = 0 because anywhere a function is pointy it isn’t differentiable. Using fractals you can define a continuous function that is pointy everywhere and therefore differentiable nowhere.

[@doraemon402]

Odd concept, nice.

[@oniondeluxe9942]

He talks about drawers. What drawers? Where he puts his pens?