Wait, this is continuous??

[@glennberry4829]

near 4:45, the lower blue line should have been extended to the right and dropped down again where the second intersection is. That might be closer to a than the left side drop. Delta has to be in the narrower of the two sides. If the upper blue line had intersections those also would need to be dropped, and delta would need to be in the narrowest of all drops...

[@yf-n7710]

7:55 Why? Wouldn't it end once f(x) exited N? That still happens, it just happens from the same direction it entered N.

[@nodrogj1]

Came here to say this, +1

[@RexxSchneider]

That assumes that for some value of x > a, we get f(x) ∉ N. And, of course nobody told us how f(x) behaves when x >> a. It could decrease to -∞ or turn upwards +∞, in either case there would be a point where f(x) is no longer in N, as you say. I suspect that Michael Penn had it in his mind that as x increased, f(x) became asymptotic to some value that was within N, in which case the open neighbourhood of a would extend to ∞, as he suggested.

[@jaddaj5881]

I think this is just an error. If you want to imagine that f has values inside of N outside of his drawing then as its also conceivable this happens on the the left why didn’t he extend infinity to the left.

[@RexxSchneider]

@@jaddaj5881 I concede that it's just as likely to be an error, but I prefer to at least consider a charitable interpretation. Michael Penn is hardly renowned for the accuracy of his drawings, but he did draw an intercept so it's not possible for the function to remain within N to the left.

[@yf-n7710]

@@RexxSchneider if it goes back into N, wouldn't it just be union of two open sets in that case? It still wouldn't include the part where it is outside N

[@maryamer8774]

4:30: we can't go as far as we want to the right, you have to extend the line from f(a)-eps until it intersects with the curve again.

[@landsgevaer]

Yeah, that was a bit weird. Also the delta can be chosen much narrower, as long as the image of the delta-neighborhood falls in the epsilon-neighborhood.

[@shanathered5910]

"continuous in what sense?" :)

[@debtanaysarkar9744]

It's continuous at every point except when x is an integer

[@normanstevens4924]

@@debtanaysarkar9744 And right continuous everywhere.

[@writerightmathnation9481]

I’m not done watching the video, but his opening remarks suggest to me that he’ll introduce a different topology on the real line that makes the floor function continuous. There are many such topologies, I’m watching and waiting to see what topology he chooses to describe.

[@shanathered5910]

I would imagine he means continuous on the set of non-integers

[@debtanaysarkar9744]

@@shanathered5910 That's what I meant LoL

[@hxc7273]

It would be great if you could do a topology course on your other channel

[@kenhill2125]

This is why we sometimes call these sets "clopen". This "continuity" of the floor function is a peculiarity of using what is a non-intuitive concept of an open set. Basically, what is true in most of Mathematics, he who controls the definitions controls the conclusions 😀

Great! I was struggling with this question and found very few sources which describe this interaction between topology and calculus. You nailed it!

[@kumoyuki]

This is the first time I've heard of lower-limit topology, but it's really important for various kinds of numerical computation. Specifically, if you're trying to differentiate a function which is specified by an algorithm, you have to deal with neighbourhood boundary conditions quite a lot. It's actually a point where I got stuck in my research program, and I'd love to hear what you have to say about differentiation/integration in a lower-limit topology. I ended up working in a synthetic differential geometry framework to avoid having to think too much about the limits, but SDG also comes with a number of baked in assumptions that introduce their own complications. (BTW, a side effect of learning SDG was to become convinced that its a much better way to think/learn about calculus than coming in from limits theory. Fight me bro. But seriously, I think there's an interesting discussion to be had about a lot of these "non-standard" approaches to various topics in calculus and beyond...)

[@TheEternalVortex42]

My favorite example of this is that if we consider only computable functions (ie on computable numbers) then every function is continuous (or another way to say it is that discontinuous maps aren't functions because they aren't well-defined).

[@omargoodman2999]

There are even physical corollaries when you consider the Planck limits and the Uncertainty Principle. There's a certain minimum threshold of even physical space, below which if there *is* a concept of "measurable length", it doesn't matter because it isn't well defined, so all lengths might as well round down to no less accurate than the Planck Length.

[@robmac8590]

What a mess. The inverse image of the set N, i.e., f^{-1}(N) is the FINITE region between the two points that map to f(a) - epsilon. The other mistake is that if x_1 and x_2 are these two points that map to f(a) - epsilon, i.e., the points to the left and to the right, respectively, then the delta needs to be the minimum of a - x_1 and x_2 - a when you are using the second definition of continuity. You can't assume that a - x_1 is less than x_2 - a from the picture.

[@TranquilSeaOfMath]

Very nice presentation and discourse.

[@dominik.sauer1]

well the negborhood does not extend inifinitely to the right, its bounded by second intersection with the graph.

[@jplikesmaths]

7:51 why is the green line extending infinitely to the right, can’t I do the same to the same to the left, provided that I also change the neighbourhood of f(a)?

[@bgbd182]

yes. that's a mistake

[@morgengabe1]

Also continuous in the sense that within a von neumann based number system, each step corresponds to the value of its height which is a set that contains some set that is an element of its predecessors. It's like a filtration.

[@drdca8263]

Why’d you say you could go as far as you want to the right? The lower limit matters on both sides.. And then you say the pre-image extends infinitely to the right? That’s not true for the function and interval drawn though?

[@gp-ht7ug]

Why is the floor function your favorite function?

[@zapazap]

​@@MyOneFiftiethOfADollar Why do you assume that he cares, rather than being, simply, idly curious?

[@archimidis]

1:41 Isn't (2) just the definition of limit (1) ?

[@TheEternalVortex42]

Typically in initial calculus courses limit is defined in a simpler more "intuitive" way. Something like this: "lim f(x) = L is defined as when x gets closer to a, f(x) gets closer and stays close to L". Only in later courses (or real analysis) do you get into the full epsilon/delta definition.

[@metalsonic8859]

In the definition of {limit f(x) as x goes to a} deleted neighborhood of limit point {a} is used whereas in the definition of continuity neighborhood is not deleted. In other words in definitoon of limit {x} can approach point {a} at points near {a} excluding it and in definition of continuity f(x) can be evaluated at point {a}

[@JayTemple]

I think that's how I learned it.

[@matheusjahnke8643]

Cumulative distributions functions(say... given a random variable X... F(x) = the chance of X being less than or equal to x) are always continuous with respect to the Rl topology; If we X is a continuous random variable then F(x) is continuous on both senses, but X also allows for "points"(which may behave like a discrete variable) with non zero chance(or some k * delta(x) density)... But not that for the jump... the value at the discontinuity is the right limit, not the left one(so if we limited the scope and scaled the floor function... we would have a valid CDF... corresponding to a discrete random variable over the a specific range of integers)

[@frobeniusfg]

So it's probably true that continuous functions wrt lower limit topology correspond to right-continuous functions wrt "standard" topology

[@marcushletko8258]

This is the weirdest thing, but the first time I saw this video, I ignored it because it wasn’t a good time for math then. I’m also self studying real analysis through baby rudin, so as of an hour ago I just did chapter 4.8, which is pretty much exactly this topic. I’m weirdly creeped out and thankful because I didn’t understand it.

[@julianbushelli1331]

I really don't recommend baby Rudin as a primary text, unless you're already very comfortable with analysis. His language is sparse, and the book is not really written to be approachable. I'd suggest instead (or as a supplement) something like Reed's "fundamentals of real analysis" which will walk through the (still very theoretical) material in a more verbose and grounded way, and not assume prior familiarity. That said, if baby Rudin is working for you, great. It's at least a very nice reference.

[@joshuascholar3220]

I would have liked some context of what ""continuousness" implies in each topology. And I would have liked examples of what those definitions were defined to exclude.

[@EebstertheGreat]

It's interesting how the continuity of a function depends on its codomain, not just its domain. I've tried and failed to explain this to people before. This can happen even if the codomain is a subset of R with the subset topology. For instance, consider Y = R\((-1,0)U(0,1)) and the function f:R→Y defined by f(x) = x for x∈Y and f(x) = 0 otherwise. This function is actually continuous on Y. The preimage of every open set in Y is open in R (with the usual topology). The annoying result is that if you use the set-theoretic definition of a function which does not specify the codomain, then a function can't actually just be "continuous." It can only be continuous _on_ a particular codomain. So this function is continuous on Y (its range) but not on R. Another way of looking at it is that a continuous function from one topological space (M,Tₘ) to another topological space (N,Tₙ) corresponds to a function Tₙ→ Tₘ between their topologies, in the opposite direction. Whether this function exists clearly depends on Tₙ. So it seems like essential properties of a function depend on its codomain.

[@jellyfrancis]

How about topology course in math major ❤?

[@gene51231356]

@michaelpennmath, could you perhaps cover how to calculate the 0-to-1 integral of the power tower (x^x^x^x^...)dx?

[@DaveSalwinski]

This only converges for x>1/e, so you can't start the integral at 0

[@dontthrow6064]

2:10 isn't 2 simply the definition of a limit?

[@dawkinsfan660]

This makes me think about the exponential function: it is invertible if and only if you define it as a function from all real numbers to the strictly positive real numbers. If you define the exponential function from all real numbers to all real numbers it is not invertible.

[@General12th]

Hi Dr. Penn! Very cool!

[@snnwstt]

So by any of these definitions of continuity a circle is continuous at its most (maximum) horizontal point. And since a circle owns symmetry, we conclude that a circle is nowhere continuous!? (A circle exists independently of the system of coordinates used to describe it.) Even if, in parametric form, x=r*cos t ; y = r*sin t; cos and sin are continuous versus the parameter t, in the reals.

[@i_amscarface_the_legend9744]

Open/closed sets are not things written in some holy book. It is just a question of choice, and each topology gives some new stuff and even rearrange the real numbers in some ways.

[@erikcarp9359]

What about piecewise continuity? I don’t remember what it means just that it came up in class this year but I never really got it

[@kkanden]

i think piecewise continuity means that on certain intervals the function is continuous but not necessarily on all R. with the floor funciton as an example, it is continuous on all intervals [k, k+1) for integer k but it's not continuous on all R.

[@schweinmachtbree1013]

f : ( D ⊆ *R* ) → *R* is piecewise continuous on two pieces if the domain can be decomposed as D = A ∪ B and the restrictions f|_A : A → *R* and f|_B : B → *R* are continuous - for example the sign function sgn(x) = {1 if x>0, -1 if x

[@evankalis]

There are finite subintervals in the interval that are continuous

[@schweinmachtbree1013]

@@evankalis that is very imprecise

[@erikcarp9359]

@@schweinmachtbree1013 though it is much easier to understand and seems to be generally correct(?)

[@dadoo6912]

2:00 am I seeing things or you just rewrote the definition of continuity in epsilon delta terms? it's not like they are equivalent, those two "definitions" are identically the same. lim notation is just a shorthand for Cauchy's definition of a limit of a function. and what calculus class doesn't teach you the definition of a limit and yet introduces you to it? how is it brave to teach proper calculus?

[@marc-andredesrosiers523]

Would have been nice to consider the p-adic topology. 😀

[@pierreabbat6157]

I agree. Though the p-adic topology isn't defined on the reals, but is defined on the rationals. e is not a p-adic number for any p.

[@demenion3521]

the floor function is continuous on R\Z which i always find weird

[@ethandavis7310]

Thanks for the head-scratcher

[@schweinmachtbree1013]

Any function will become continuous if you restrict it to the points where it is continuous! One can even do this with a simple formula: given a function f : ( D ⊆ *R* ) → *R* , its "continuization", say cont(f), is defined by (cont(f))( _a_ ) = lim_(x → _a_ ) f(x). cont(f) is only defined where the limit exists, so its domain can be a subset of D (possibly all of D, possibly empty). For the floor function, ⌊ ⌋ : *R* → *R* becomes cont(⌊ ⌋) : *R* - *Z* → *R* . Actually this continuization operation doesn't just restrict f to the largest subset of its domain on which it is continuous (although it does in the example of the floor function); it can also change the value of f, since the limit lim_(x → _a_ ) f(x) can exist without equaling f( _a_ ) - for example the function I_{0} : *R* → *R* defined by I_{0}(0) = 1 and otherwise I_{0}(x) = 0 will become the constant function 0, so cont(I_{0})(0) = 0. In fact if lim_(x → _a_ ) f(x) exists but _a_ is not in the domain of f then _a_ _will_ be in the domain of cont(f). If you know what removable and non-removable discontinuities are, then what cont(f) is doing is repairing the removable discontinuities of f (which can add points to the domain) and deleting the non-removable discontinuities (which removes points from the domain).

[@landsgevaer]

@@schweinmachtbree1013 Some functions may not have any domain left though.

[@schweinmachtbree1013]

@@landsgevaer Indeed, but it would be mean to say the empty function isn't a function - it's a Strong Independent continuous injection, and a homeomorphism onto its image (no matter the topology!) :p

[@landsgevaer]

@@schweinmachtbree1013 Yes, no cancel-culture among mathematicians: every function counts, no matter how empty. 🙂

[@eddietime1811]

Isn’t the sequential definition of continuity (limf(x)=f(a)) equivalent to the epsilon delta, and thus it is silly to call it the “real” definition just because it looks more complicated?

[@zapazap]

Be careful. There are sequential characterizations of some properties that are equal equivalent in some topologies but not in others.

[@RadicalCaveman]

You forgot to explain what the lower limit topology is.

[@dondisco6399]

Hi Michael, could you make a video explaining why if f:[a,b] -> R is a continuous function then its derivative function f':(a,b) -> R is definided, if it exists, inside [a, b], i.e. (a, b). Thank you.

[@schweinmachtbree1013]

This is just because the most common meaning of derivative is "two-sided derivative", i.e. the derivative using the two-sided limit f'(a) := lim_(x → a) (f(x) - f(a))/(x-a). One can also define left derivatives and right derivatives by f'(a^-) := lim_(x → a^-) (f(x) - f(a))/(x-a) and f'(a^+) := lim_(x → a^+) (f(x) - f(a))/(x-a). Then the derivative of a function f : [a, b] → *R* is also a function from [a, b] to *R* (if it exists, i.e. if the usual limit of (f(x) - f(a))/(x-a) exists for x inside [a, b], and if the left and right limits of (f(x) - f(a))/(x-a) exist for x on the boundary of [a, b].) It should be noted that the meaning of "derivative" and "limit" being interpreted as two-sided, i.e. only considering points _a_ of the domain _D_ which are contained in some open interval inside _D_ (such points _a_ are called "interior points" of _D_ ) is only a convention, usually used for simplicity. When one gets to more advanced analysis or topology one starts considering limits in general, and here instead of requiring "all points near _a_ " to be inside the domain _D_ , we only need there to be points arbitrarily close to _a_ inside _D_ (such points _a_ are called "cluster points" or, appropriately, "limit points"). In this more general setting, the derivative of f : [a, b] → *R* is just f' : [a, b] → *R* with left and right derivatives taken at the endpoints (if it exists).

[@dondisco6399]

@@schweinmachtbree1013 First of all thank you for your comment. Theorems such as Rolle or Mean Value, assume f is continuous in [a,b] and derivable in (a,b) at least ... There are also a theoreme that claims that if a function f is continuous in (a,b) i has an unique relative extreme in that interval, then this relative extrem is also an absolute extreme. I wonder if that result won't be true if I change (a,b) for [a,b]. When I try to find out the absolute extrems of a function f defined in [a,b] I creat a table of values in this points of domain: x=a, x=b, and in all points of (a,b) where f'(x) = 0 or f'(x) doesn't exist (critical points). Again, when I derivate I only work in (a,b). Is that correct? Some authors take x=a i x=b as critical points too because they consider that f' only exists in interior points of [a,b], so f' doesn't exist in x=a i x=b and so, by definition, these 2 points are also critical points. And the last case: when you work with a piecewise function. Imagine a function f:[a,b] -> R continuous defined by several analytical expressions from a to a1, by f_1(x), from a1 to a2 by f_2(x), from a2 to a3 by f3(x) ... and from a_(n-1) to a_n=b by f_n(x). How to proceed to find relative extremes and absolute extremes?

[@schweinmachtbree1013]

​@@dondisco6399 The other theorem you are thinking of, the Extreme Value theorem, applies to closed intervals [a, b] - for an example of the extreme value theorem failing on an open interval, consider f(x) = 1/x on (0, ∞) or tan(x) on (-pi/2, pi/2): no extremum exists. The fact that a unique extremum c is an absolute ("global") extremum is immediate: for example if c is a unique maximum on [a, b] then we know that f(c) > f(x) for all x near c, but since it is unique there are no other maximums, so it must also be the case that f(c) > f(x) for all x not near c too (otherwise there would be another maximum). The usual derivative rules apply just as well to left derivatives and right derivatives, so deriving f : [a, b] → *R* to give f' : [a, b] → *R* will tell you the right and left derivatives f'(a^+) and f'(b^-). These values do give you some information: if f'( _a_ ^+) > 0 then _a_ is a minimum, if f'( _a_ ^+) < 0 then _a_ is a maximum, if f'( _b_ ^-) > 0 then _b_ is a maximum, and if f'( _b_ ^-) < 0 then a is a minimum - if f'( _a_ ^+) = 0 then we can't tell, so you have to look at x = _a_ separately, and similarly if f'( _b_ ^-) = 0 or if f'(a^+) or f'(b^-) doesn't exist. For your last paragraph, you can probably figure that out yourself: the extremums of f are just the extremums on (a=a_0, a_1), (a_1, a_2), ..., (a_(n-1), a_n=b), plus what you get from looking at the points x = a_0, a_1, ..., a_(n-1), a_n separately. The absolute ("global") extremums are of course then just the biggest maximum and the smallest minimum.

[@dondisco6399]

@@schweinmachtbree1013 Thank you very very much. I still thinking that the topology of opened and closed intervals talking about continuity and derivability is very important to be clear but as Michael has shown about that a funtion is continuous if the antiimage of an opened is an opened. Of course this version only makes reference to two-sided of concept of limit.

[@schweinmachtbree1013]

​@@dondisco6399 Yes, the topology of open an closed intervals is indeed relevant - for example, the Extreme Value theorem, which says that any continuous function f : [a, b] → *R* has extremums and attains these extremums, also generalises to functions f : C → *R* where C = {z ϵ *C* : |z| = 1} is the unit circle, and to functions f : S^n → *R* where S^n = {(x_0, x_1, ..., x_n) ϵ *R* ^(n+1) : x_0^2 + x_1^2 + ... + x_n^2 = 1} is the n-sphere - this generalises to the topological theorem that any function f : X → *R* where X is a _compact topological space_ has extremums and attains these extremums: closed intervals [a, b] are compact, the unit circle is compact, and the n-sphere S^n is compact. We can also see from a topological perspective that open versus closed intervals do _not_ matter for the Intermediate Value theorem. (An equivalent statement to) the intermediate value theorem is that any continuous function f : [0, 1] with f(0) < 0 < f(1) has a zero between 0 and 1 - this follows from the topological theorem "the continuous image of a connected space is connected": the topological space [0, 1] is connected so it's continuous image f([0, 1]) is a connected subspace of *R* . since f(0) ϵ f([0, 1]), f(1) ϵ f([0, 1]), and f(0) < 0 < f(1), this means that 0 ϵ f([0, 1]), which is to say that there exists some _c_ ϵ [0, 1] such that f( _c_ ) = 0. However the open interval (0, 1) is a connected space too, so there is also a version of the intermediate value theorem for open intervals - it would say: "for any continuous function f : (0, 1) → *R* (this can easily be generalized to f : (a, b) → *R* ) and any _d_ between lim_(x → 0^+) f(x) and lim_(x → 1^-) f(x), there exists _c_ ϵ (0, 1) such that f( _c_ ) = _d_ ". Note that these limits could be infinite. For example with f = tan: (-pi/2, pi/2) → *R* we have -∞ = lim_(x → -pi/2^+) f(x) < 2 < lim_(x → pi/2^-) f(x) = ∞, so by the IVT for continuous functions f: (-pi/2, pi/2) → *R* there exists _c_ ϵ (-pi/2, pi/2) such that tan(c) = 2. You can see that the statement of IVT for functions f: (a, b) → *R* is more complicated than IVT for functions f: [a, b] → *R* , which is why IVT is always stated for functions defined on closed intervals, but the point is that IVT is still true for functions defined on open intervals, unlike, for example, the Extreme Value theorem - this is because [a, b] and (a, b) are both connected topological spaces, but [a, b] is a compact space and (a, b) isn't.

[@flavioxy]

i learned that definition of continuity around 2 min at calculus one (if you will, at university anyway)

[@MyOneFiftiethOfADollar]

Limit as x approaches 1 from right = 1. Limit as x approaches 1 from left = 0 Function did discontinuous at 1 with regard to usual open set topology of real numbers.

[@bot24032]

but your definition of pre-image just doesn't work in (3)? on this open interval to infinity not everything maps to N? and what about when there is more than one point with the value equal to f(a)? even if the function would normally be declared as continuous it wouldn't under your definition because the pre-image wouldn't be an interval

[@bot24032]

also your example for lower limit topology is bad because the (1/2; 3/2) actually isn't an open set (even though that doesn't change anything)

[@sieni221]

just take a point set topology course and u will come across induced topology

[@hyjn4884]

Yeah i think he marked the preimage wrong but the definition still works and is equivalent to the other ones. A neighbourhood doesn't have to be an interval, topologically a neighbourhood of a point p is any set containing an open subset which contains p. So an open neighbourhood is just an open set with p in it, which again doesn't have to be an interval

[@bot24032]

@@hyjn4884 oh ok sorry

[@bot24032]

@@sieni221 im in 9th grade:\ hearing about this for the first time and just assuming stuff

[@spicymickfool]

Can a contraposition proof prove continuity? Standard def: f is continuous at c iff for all ep>0, exists delt>0 so that |x-c|0 , |x-c|\epsilon implies f is not continuous at c. Let f(x)=[x] and c=1. Let delt=1/N, N>1. f(c-\delta)=0, f(c+delta)=1. f(c)=1. So take epsilon=half and we have a proof by modification of standard def that [x] is not continuous at 1.

[@schweinmachtbree1013]

One wouldn't call this the "contrapositive" of continuity, but rather the _negation_ of continuity - then quite clearly it can't prove that a function is continuous but it can prove that a function is discontinuous. Also you're forgetting a quantifier: in the epsilon-delta definition the "x" is universally quantified, and in its negation it is existential quantified: f is continuous at c: ∀ε>0 ∃δ>0 ∀xϵdom(f) : (|x-c| < δ ⇒ |f(x)-f(c)| < ε). f is not continuous at c: ∃ε>0 ∀δ>0 ∃xϵdom(f) : (|x-c| < δ & |f(x)-f(c)| ≥ ε). Since delta is universally quantified in the negation of continuity, you can't say "let delta = 1/N" - you can however argue by cases on delta: if delta < 2 then f(1) = 1 and f(1 - delta/2) = 0, so for epsilon = 1/2 we can take x = 1 - delta/2 since we have |x-1| = delta/2 < delta and |f(x)-f(1)| = |0 - 1| = 1 ≥ 1/2. If instead delta ≥ 2 then we are looking in a very big interval around c=1, namely (c-delta, c+delta) ⊇ (c-2, c+2), so it's easy to find an x in this interval with |f(x)-f(1)| ≥ 1/2, e.g. x=2.

[@nimennacnamme6328]

Complete the following sentence: The floor function is l… a) lava b) love

[@tomlavelle8340]

I’ve always thought it was continuous.

[@Bodyknock]

8:15 I don’t think the preimage definition is quite right here because it’s possible the preimage is a union of sets that contains a locally open neighborhood but isn’t itself a single neighborhood. For instance, consider cos(0) and pick the neighborhood of cos() to be (0.5, 1.5). Then the preimage of that interval is the union of all open intervals of the form (π/4 +2n π, 3 π/4 +2n π) for all integers n. In other words that preimage is an infinite union of open neighborhoods, not a single neighborhood. Instead what you want is to say that for any given open neighborhood B in the range of f() there exists an open neighborhood A in the domain such that the image f(A) is a subset of B. P.S. This is only for the first section of the video where he is talking about if a function f(x) is continuous at a chosen point a. The remainder of the video looks correct because there he’s talking about if the function as a whole is continuous across its entire domain. In that case, for instance, sin(x) produces unions of open sets for any open neighborhood in its range so is continuous across its entire domain. But something like the Floor function can be continuous at a specific point, such as x=0.5, but not continuous as a whole, which is where you need to modify that topological definition slightly.

[@schweinmachtbree1013]

Actually it's fine because (π/4 + 2nπ, 3 π/4 + 2nπ) is an open neighborhood of 0. The definition of an open neighborhood of _a_ is an open set containing _a_ , and infinite unions of open sets are open. "Open set containing _a_ " and "open neighborhood containing _a_ " meaning the same thing admittedly makes the term "open neighborhood" not very useful, but one can more generally consider just "neighborhoods", a neighborhood of _a_ being defined as a superset of an open set containing _a_ . The topological definition of continuity at _a_ given in the video is then equivalent to "the preimage of any neighborhood of f( _a_ ) is a neighborhood of _a_ ", and the topological definition of continuity given in the video is then equivalent to "f is continuous at _a_ for all _a_ in X", i.e. "for any _a_ in X and any neighborhood of f( _a_ ) in Y, the preimage is a neighborhood of _a_ in X".

[@Bodyknock]

@@schweinmachtbree1013 Except that in the part I bookmarked he was talking about the function being continuous at a specific point, and it’s possible for functions to be locally continuous at a given point but discontinuous elsewhere. (See my endnote above). If the question is whether a function is continuous at a specific point a then you can’t just look at the preimage around f(a) because that preimage might not be a union of open sets even though it contains an open set that contains a. Using the floor function, for instance, it’s the difference between the function being continuous over its entire domain (which it isn’t as demonstrated in the video) being continuous at specific points like x= 1/2 (which it is continuous at that point even though the preimage of Floor(1/2) is not an open neighborhood.)

[@schweinmachtbree1013]

​@@Bodyknock -The preimage of floor(1/2) -_-is-_- an open neighborhood of 1/2.- The phrase "union of open neighborhoods, not a single neighborhood" doesn't make sense because any union of neighborhoods is a neighborhood - likewise "locally open" and "locally continuous" are not terms used in topology or analysis. What you seem to have a problem with is connectedness, but connectedness has nothing to do with openness/continuity.

[@Bodyknock]

@@schweinmachtbree1013 That’s incorrct. Floor(1/2) = 0, and the preimage of 0 is [0,1) which is not an open interval or open neighborhood.

[@schweinmachtbree1013]

​@@Bodyknock Sorry yes, my mistake - the preimage of floor(1/2) is a neighborhood of 1/2 but not an open neighborhood of 1/2. The reason we run into trouble here is because the singleton set {floor(1/2)} is not a neighborhood of floor(1/2) (a neighborhood of _a_ is any superset of an open set containing _a_ , not just any set containing _a_ ). when you say "If the question is whether a function is continuous at a specific point _a_ then you can’t just look at the preimage around f( _a_ )" this is incorrect - you can just look at the preimage around f( _a_ ); what you can't do is just look at the preimage *at* f( _a_ ). For instance it _is_ valid to consider the preimage of the neighborhood (-1/4, 1/4) of floor(1/2)=0 which in this case gives the same set [0, 1), which is indeed a neighborhood of 1/2. Sorry if my mistake has confused matters, but everything else I have said stands. As I said, your main issue seems to be with connectedness, which has no bearing on continuity - open sets are closed under unions, whether finite or infinite and whether intersecting or disjoint, as are neighborhoods (and open neighborhoods).

[@Anonymous-zp4hb]

Aren't 1 and 2 the same "level" ? The epsilon delta thing just breaks down the meaning of the "lim" part of the 1st level, no?

[@julianbushelli1331]

They are equivalent, yes. What the first definition is saying is "a function f is continuous if whenever you have a sequence x_n that converges to x, the sequence f(x_n) converges to x."

[@Datamining101]

For us non-experts it would have been nice to have some sort of motivation for this. Right now it's just "not continuous...? just change your definition." In what context is this important/what does it help us do that we otherwise couldn't?

[@kumoyuki]

Among other things, this is super important in digital signal processing. In DSP, we have to construct approximations to continuous-time functions out of digital-friendly functions. This usually starts with the Dirac delta function, which you can integrate to the Heaviside operator, which then integrates to the ramp function; and differentiates to the doublet function. You can construct the floor function as a summation of Heaviside operators, so having a reasonable topology for these kinds of functions is really useful.

[@legitjimmyjaylight8409]

y = floor(x)

[@writerightmathnation9481]

I’m concerned that at @1:24, you state that the definition one first learns is the limit definition but then later call the other definition the “real analysis” definition, when all the best calculus 1 textbooks include both of those statements, which, in the case of any separable metric space, are equivalent. You’re right, to some extent; my concerns aren’t with how correct you are, but how you seem to accept this very poor state of affairs, in which the notion of an analysis type of definition is being postponed for mathematics majors because non-majors don’t want to learn serious mathematics and their preferences are now governing the way mathematics is taught to mathematics majors, delaying their progress in rigor. I don’t recall as a student covering the limit definition of continuity as a definition, but I expect that it was covered as a theorem then. Either statement can be used as a definition of continuity in the calculus 1 context, as I remarked above, so I disagree with your claim that the seconds one is “more precise”. As long as the corresponding rigorous - and precise - definition of a limit was provided prior to using it, it can be rigorously to provide a precise definition of continuity. Claiming otherwise is mathematically and logically misleading. The separability requirement can be removed by introducing the notion of a net or a Moore-Smith sequence, which can be indexed by a set of any cardinality. In a reasonable real analysis course, this should be done, In addition to the topological (neighborhood) definition.

[@schweinmachtbree1013]

You seem to have the misconception of many math majors (and even mathematicians) that rigour and logical purity are all-important. It is _perfectly fine_ to use a non-rigorous intuitive definition of limit in a Calc 1 course before giving a rigorous definition later in an Advanced Calc or Analysis course, not because "non-math-majors are governing how math majors are taught" or because anyone "doesn't want to learn serious mathematics", but simply because _it leads to better outcomes_ . When _learning_ mathematics, pedagogy is more important than logical purity - once you have learnt the material effectively, using ample amounts of intuition if beneficial, _then_ you can go back and justify everything rigorously, now that you understand it. Learning is not a straight line from A to B; it is a journey from A with lots of winding paths where you will probably get stuck or run into a dead end and have to retrace your steps, until you understand the landscape well enough to make it to B - once you have a path from A to B you can then refine your understanding and improve your path, and once you have a very good understanding of the landscape between A and B you can find the most efficient rigorous route from A to B ("the" if there is only one; often there are several). Like it or not, intuition is more valuable than rigour in actual mathematics: yes you of course need rigour to prove theorems, but by being a logical purist one is going to be much worse at _discovering_ theorems to prove in the first place. Practising mathematics is not only a logical process but also a _creative_ one, and the creative part is what comes first.

[@writerightmathnation9481]

@@schweinmachtbree1013 You seem to have not understood my concerns.

[@boblewis5762]

Very nice

[@RM-xq7gf]

good thumbnail/title

[@chimetimepaprika]

My teachers use this function to calculate when my grade is 69.99 and I still get a D

[@PleegWat]

I don't like that pre-image definition of continuity. It may give the 'normal' result when checking for continuity everywhere, but I'm pretty sure it breaks down when checking for continuity in a point.

[@BrooksMoses]

I don't think that definition of continuity has a concept of "continuity at a point". Possibly you could extend it to talk about continuity at a point on the domain of the function (the y-axis), but it doesn't have a concept of continuity at a point on the range (the x-axis). It's not so much that it breaks down as it doesn't define it in the first place.

[@joeltimonen8268]

The standard definition of continuity absolutely includes a concept of being continuous at a point. First it's defined what being continuous at a point means. Then continuity is just being continuous at every point of the domain. Uniform continuity doesn't care about single points of the domain, it only deals with the distances between them. The pre-image definition of continuity is sometimes very handy as some results' proofs are much more elegant if you use that definition as opposed to the standard epsilon-delta one. But the best thing about these definitions is that they're all equivalent. Whichever definition you take to be THE definition of continuity, you get all the other propoperties for continuous functions "for free".

[@PleegWat]

@@BrooksMoses Continuity in a point means for a specific x∈X, for all ε>0, there is a δ>0 such that |y-x|

[@thedanishelf]

Wouldn't (1) be the definition of a function being smooth and (2) the definition of continuity. like f(x)=|x| is continuous but not smooth as it is not differentiable at 0 I remember learning that continuous basically means that you have 1 value for each x and you can draw the function without lifting the pen. (crude simplification)

[@nothayley]

|x| is still continuous under (1) Smooth means that it can be differentiated infinitely many times (so all its derivatives are continuous). |x| fails this test, since its derivative is noncontinuous at 0.

[@joeltimonen8268]

All functions have just one value for each x, that's part of a function's definition. Continuity indeed intuitively means that you can draw the graph of the function without lifting the pen.

[@TheZoltan-42]

We were taught the definition 2) in high-school. The definition 1) only works if limes for both x->a+0 and x->a-0 converge to f(a), assuming f(a) makes sense in the first place.

[@Cloud88Skywalker]

WAIT. Isn't the conclusion wrong? The thing about open sets in standard and lower topology doesn't really matter because the previous condition is never met. You don't really get to find an open set in the codomain, because *the codomain is only integer numbers.* (1/2, 3/2) is not part of the codomain because there are infinite points in that interval (actually all of them except the number 1) which are never the image of a point in the domain regardless of whether we consider standard or lower topology.

[@MuffinsAPlenty]

This is a matter of codomain vs range. The codomain are elements in the "output set" based on how the function is defined. Here, the codomain is R. Range, on the other hand, are elements in the codomain which are actually hit by the function. Here, the range is the set of integers. This might feel weird because codomain might feel a bit artificial, whereas the range might feel a bit more substantive. The range is the sort of "minimal possible codomain". But, from the perspective of defining a function in the first place, we must specify a codomain in order to define a function at all. The floor function from R to R is technically considered to be a different function than the floor function from R to Z because they have different definitions (specifically, the codomain is different). The definition of continuity from topology is based on the topologies of the codomain and domain.

[@Cloud88Skywalker]

@@MuffinsAPlenty I perfectly understand what you say. And I concur with you, that notion of "codomain" is completely artificial and irrelevant. It affects the function in exactly nothing. The codomain, as you defined it, is completely useless. Notice how there's no analogous structure for the input. The domain truly is the set of all possible inputs and not just anything that includes all possible inputs and whatever else. The fact many times functions are defined R->R (for example) is nothing but laziness and usually only relevant to know what kind of structures are being used as input and output (are they vectors? matrices? natural numbers?) but don't really mean anything else. If I define the function f: R->R, f(x)=1/x, is it wrong because I didn't used the correct domain (R\{0}) but it isn't wrong for the very same reason with respect to codomain because that is, for some reason, allowed to bey nearly anything? It's very obvious to me, that what matters is what you call range (I wouldn't have distinguished between the two), hence my point stands.

[@MuffinsAPlenty]

@@Cloud88Skywalker Part of the reason it seems artificial is that we are taught most of our lives in dealing with "functions" from R to R, or functions with some type of formula. But when we move into the realm of using functions as a tool to study the relationships between structures/mathematical objects, codomain is incredibly natural. As an example, how is the set of rational numbers related to the set of real numbers? One is a subset of the other, so we can define the inclusion function ι : Q → R by ι(x) = x. This function codifies the subset relation there. If we were to only care about the range, then we would have the identity function ι : Q → Q by ι(x) = x. This removes the codification of the relationship between Q and R by making it entirely about Q itself. This is particularly useful in subjects like abstract algebra where we may have two groups G and H and want to know how they're related by considering all group homomorphisms from G to H. A common way that a group G can be related to another group H is if a quotient of G is isomorphic to a subgroup of H. If we eliminated the concept of codomain and only used range, it would be impossible for functions to be used to detect this relationship. Even something as simple as G being a subgroup of H would be completely undetectable using functions. In this video, the idea of a continuous function describes a relationship between two topological spaces: R with the standard topology and R with the lower limit topology. If we ignore the codomain and stick with the range only, then we lose the information the step function gives us about the relationship between these two topologies.

[@MuffinsAPlenty]

@@Cloud88Skywalker Also, as a follow-up (I was reminded of the post recently), there is a concept of a "partial function" f from a set X to a set Y. A partial function need not be defined on all of X, and the natural domain may be a proper subset S of X. This setting is similar to the codomain/range situation. It's way less common to deal with partial functions, but I have seen it come up in some situations. For example, rational maps between projective spaces in algebraic geometry are often partial functions, rather than total functions.

[@victorespeto]

Clickbait

[@pyropulseIXXI]

There is only one correct answer, since math is deductive from axioms; you just literally defined it to be continuous by saying [a,b) is open. I can define my face to be continuous as well, and this would be correct, since I just defined it to be so

[@MuffinsAPlenty]

Is 1/2 + 1/4 + 1/8 + ... = 1 or is it undefined?

[@nHans]

I see. If you have a discontinuous function, but you'd like it to be continuous, you can redefine the notion of continuity itself and get what you want. Is that only in math, or can you do it in the real world as well? Like, when the judge tells me that something I did was illegal according to some law, can I redefine that law so that my action is no longer illegal? Anyways, I'm now gonna redefine zero (or division, or both) so that y'all can start dividing by zero without problems. You're welcome!

[@BrooksMoses]

It's been done. There are extensions to the real set where division of a non-zero number by zero is defined as infinity (and both positive and negative infinity are considered to be the same number). Likewise, dividing a finite number by infinity produces zero in this extended set. As to your rhetorical question, yes this works in the real world too -- because what is being redefined here is just the words that we are calling things. Choosing to use the word "continuity" to mean "continuous on lower-limit topology" rather than "continuous on standard topology" did not change anything about the floor function or its behavior. Similarly, you can choose to use the word "legal" to refer to your actions, but the judge is still going to have you involuntarily placed in a concrete room with a locked door that you may or may not choose to call "jail". The main difference is just that mathematicians tend to react to other people saying "I'm going to use this alternate definition for this word" with "oh, cool, what interesting things do you get by doing that?" whereas most people's reaction will be "no, I'm not playing that game."

[@zapazap]

​@@BrooksMoses I like your snark repellant sir.

[@kumoyuki]

@@BrooksMoses Bravo! Bravo, bravissimo!