What really is a function?

[@MichaelPennMath]

Two things: 1. New studio space is still under construction. 2. not every transition can be super smooth and cinematic...

[@tomkerruish2982]

Your sound level is low. I'm certain it's something you're working on; I just thought the feedback would be useful.

[@srki22]

I can hear you well but if you are in nature, I would like to hear some birds, crickets and other nature sounds :)

[@laurenzhartmann9433]

Actually, many define a function f: X->Y as the triple (X,G,Y), where G ist the Graph of the function - the Graph is what you defined as the function here. But two functions can have the same graph and be different in the sense that they have different domains and codomains. For example: f: R->R>=0, x->x^2 and g:R->R, x->x^2 have the same graph und would therefore be the same using your definition. But there is a good reason to motivate a distinction between f and g as functions, for example when one wants to pose the question of surjectivity. f is surjective while g is not surjective. So in conclusion, there is a good reason to disagree with the definition of a function presented in this video.

[@martinnyberg9295]

😂 Who cares!? We love your videos in whatever form they come. Particularly this one - both a slideshow and an outdoor lecture in the same video. 😊👍🏼 I say keep doing outdoor lectures year round. Just get chalk holders you can use with mittens. 😁

[@taufiqutomo]

One underrated thing he did: The slides are friendly to screens of even the smallest of phones. Edit: even in 144p

[@tomkerruish2982]

7:16. I'm confused. The way h is written, it can't contain (1,-1) since x can't be negative.

[@debdami]

I agree. As defined here, h is the square-root function.

[@owenlam3693]

I think he meant to write x^2 in [0, infinity) instead of x, in which case h would not be a function, using the same reasoning from the video.

[@tomkerruish2982]

@@owenlam3693 That makes sense.

[@loicdelzenne7684]

@Owen Lam : we shouldn't wonder what he was thinking. The conclusion based on what is put on the blackboard is not correct ^^

[@martinnyberg9295]

@@loicdelzenne7684 Have you NEVER made a mistake on the board during a lecture? 😂 Most of us are thankful when students notice those mistakes and point them out.

[@caldersheagren]

7:40 Isn't x restricted to be positive? That would mean (1,-1) ot\in h ...

[@jojanen9144]

Yes, x is strictly non-negative real number.

[@graikoyt]

Maybe he meant the domain is non-negative, so it should be x squared that's non-negative?

[@lucakoch3432]

Awesome video! Also love that you‘re experimenting with different locations and camera shots, looks great! Love your content. Greetings from Germany ^^

[@goodplacetostop2973]

3:43 Now this is production value 9:43 Good Place To Stop

[@EpicMathTime]

I don't think your leading definition is quite correct as displayed. I think you should say "such that (a,b) \in f" at the end. Otherwise, your statement about a and b do not actually involve the function.

[@hopelessdove]

7:36 (1,-1) isn't in h

[@Chess-ks8lk]

I would like to buy a full academic course from you. Have u considered making such courses?

[@oida10000]

Isn't h={(x^2, x) | x in [0, infty)} still a function since (1, -1) would not be in there as -1 not in [0, infty)?

[@linggamusroji227]

We can rewrite h as {(t^2,t)| x€[0,~)}. If x=t^2, then t=+-√x. So, y=t=+-√x. h is not a function

[@subpopulations]

No it's a blackboard mistake Michael meant to restrict x squared but wrote a restriction on x. As the restriction is on x the h function as written is a function For it to not be a function x needs to be unrestricted.

[@goodplacetostart9099]

Good Place To Camp at 3:51

[@michaelsteinle]

why is -1 in [0,inf) ?

[@MaxG628]

At the bottom of the first slide, I think you meant “is an element of f” not “is an element of a cross b”.

[@chaost11]

No, I'm pretty sure it's correct. A x B is the the image "drawn" by f, made up of ordered pairs, meaning (a, b) ∈ A x B | a ∈ A, b ∈ B. The value y=f(x) would only be an element of B => y=f(x) ∈ B | x ∈ A. I might be mistaken, too. We were recently taught this at my uni so I can't say with 100% confidence.

[@nestorv7627]

@@chaost11 No. He says that (a,b) is an element of A×B. Yet, (a,b) may not be an element of f. So it's better to say "an element of f" so that by definition of subsets, it's an element of A×B

[@EpicMathTime]

@@chaost11 A × B is the set of _all_ ordered pairs (a,b) such that a is in A, b is in B, by definition. Saying that (a,b) is in A × B doesn't even involve the function f.

[@MaxG628]

The essence of a function is to associate EVERY value in the domain with EXACTLY ONE value in the codomain. The “such that” part of the slide is trying to say that in the language of sets, but does so incorrectly.

[@chaost11]

Awesome, thanks for clearing that up 😄

[@shivam5105]

Question: at 7:35, you took the set (1, -1) and (1,1) however, the first one does not satisfy the condition (x^2, x), actually they would be (-1,1) and(1,1) respectively, which is a set since no element in the domain is assigned to 2 different elements in the co domain

[@MrRyanroberson1]

try that again. for x = -1, (1, -1) is a valid (x^2,x) pair. the mistake was: x could never have been -1 in the first place

[@nathangreene3]

Why do we require f:A -> B map to all of B at 1:25? I would have thought a subset of B would suffice. For example, f:R -> R defined by f(x) = x^2 doesn't map to anything negative. Do we really need to specify f is defined over the reals onto the positive reals instead or am I misunderstanding something? To site something, W.A. Kosmala's Friendly Introduction to Analysis defines the range or image of a function to be a subset of the codomain.

[@GhostyOcean]

I'm not sure what you're trying to do at the given time. There was no requirement that f maps to all on B.

[@EpicMathTime]

He is talking about b (the element of A), not B. As an element of the domain, yes, b has to map to something.

[@nathangreene3]

@@EpicMathTime I thought b was an element of B as in his definition. I see now there's no conflict with his definition and the one I cited.

[@roberttelarket4934]

A definition for a function will vary with certain mathematicians even though the core meaning is the same. All you need are two sets(non-empty preferred). If W is to be a function all you need is again two non-empty subsets one called the pre-image and the other the image such that if p is in the pre-image it is "mapped" to only one element from the image. Otherwise it is called a multi-valued function.

[@heinerkaiser8673]

Or better a relation.

[@gaborendredi8161]

I am really confused or I missed something in the last but one example. I see there a condition that x >= 0, therefore for me h(1)=-1 is not a counterexample. What did I miss?

[@gaborendredi8161]

OK. I see, I am not the only one.

[@AM-hj2bo]

Love the outdoor blackboard. Do you use Hagaromo chalk?

[@shanmukeshr1696]

Is that hagaromo?

[@samdean1966]

Your definition should say (a,b) is in f, not in AxB

[@rekingooo]

Mistake occured at 0:00

[@CodingDragon04]

What always doesn't quite work for me with this definition of a function, is that a function f:A -> B would be identical to a function g:A -> C, x |-> f(x), where B is a subset of C. I thought that functions with different codomains should be considered different.

[@ConManAU]

I think you could make an argument that at the very least there is an equivalence between those two functions (something like f ~ g if the sets of ordered pairs are identical regardless of the codomain). And while sometimes the difference will be important, a lot of the time it won’t and you can just say you’re working in the equivalence classes instead. It definitely makes a difference if f and g have different domains, but that’s captured in this definition already.

[@akashpoudel]

hey Michael Penn, just letting you know you are amazing ❤, Your videos really help me understand the concepts so I'd like to encourage you to make more videos on the topics of mathematics. Your videos have helped me when I didn't understand the concepts in calculus and real analysis. Thank you so much!

[@iabervon]

The first one is a function, but not one whose domain is the integers like it was presumably supposed to be. We can guess what the domain ought to be from the form of the definition, but there's nothing in the set of pairs to let us know that 1 is supposed to have a value.

[@ConManAU]

I think it’s important to make the distinction that f is definitely not a function, even though its elements are given by a relation that could define a function elsewhere. Which I suppose is the point - it reinforces the idea that functions aren’t necessarily just rules.

[@iabervon]

@@ConManAU I disagree; there are times when you want to use techniques that apply to functions on objects that you know don't have any duplicate values, but where you only know a superset of the domain. For example, finding poles in complex analysis involves knowing that something is a function over a subset of the complex numbers, and finding what subset that is. There are other times when you state up front what the domain has to be (like when you're working with functions over the reals), and this is where some relations will fail to be functions due to not being defined everywhere, but you need to include this in the question, especially if you're asking about a bunch of relations with clearly different domains. Note that set theory doesn't give you any way to find out that f and g were defined with elements from ZxZ; f might have been defined with 3ZxZ, and g might have been defined with Rx2Z, and they would be the same objects.

[@natepolidoro4565]

Hello I think you're videos are great and very interesting.

[@kingplunger1]

you're ?

[@martinnyberg9295]

@@kingplunger1 Autocorrect has crappy context awareness sometimes. 😏

[@natepolidoro4565]

great vid

[@fernandoanchaluisa2378]

So a funtion is a subset or a rule? and if it is both why ?

[@subpopulations]

Formally a function is a subset of the set of all ordered pairs between two sets. In practice we would define this set with a rule.

[@fernandoanchaluisa2378]

@@subpopulations Thank u so much!!! Regards from Ecuador🇪🇨🇪🇨🇪🇨

[@nahuelcaruso]

I don't know if the definition is ok, I think it would end with (a,b) in f

[@aaronhunt4302]

Functions are my favorite!

[@mescellaneous]

i'm struggling to see the significance of bringing up the power set in the fourth example. I feel like that was covered with the set like notation in the first half.

[@EpicMathTime]

The function reports the cardinality of any subset of A, so the domain is the power set of A.

[@ConManAU]

I think it shows a useful example of a function whose domain isn’t numeric and which isn’t just arbitrarily defined like the set example. It then motivates the idea that many useful operations we use in mathematics can be expressed as functions (e.g. taking the derivative is a function that operates across sets of functions, even proving a theorem can be treated as a function from the set of theorems to the set of proofs).

[@alumi9818]

8:15

[@quant-mind9539]

function is determistic ?

[@rajeevgopeesingh5314]

Surely in the first slide the last thing should be f not AXB

[@nathanisbored]

At the end, theta is not a function because the empty set maps to 0 and, as we all know and unanimously agree, the natural numbers do not include 0 😏

[@goodplacetostop2973]

Some man just wants to watch the world burn 😛

[@fernandoanchaluisa2378]

Thank you professor I've waited this video for a long time, in my language there isn't a lot information about this kind of details😁😁😁 Could I answer u something please? A relation could be defined something like R: { (x,y) | p(x,y)=truth ^ (x€A ^ y€B)) A ordered pair such as the predicate of two variabmes will be truth I am your fan from Ecuador 🇪🇨🇪🇨🇪🇨🇪🇨

[@alxjones]

Yeah, a relation is just any subset of A x B, without the restriction that a R b and a R c implies b = c.

[@stephendavis4239]

Where's Church's lambda when you need it?

[@marciocastro7101]

c -> 1...

[@nestorv7627]

I think you could make the definition more precise. A function is a special kind of relation. But you need to be precise on your definitions of domain and codomain. A Relation is a subset of the set A×B. Meaning that the Domain of a relation R is the subset X⊆A, where X={a∈A| aRb for some b∈B}. The Codomain (range) of R is the subset Y⊆B, where Y={b∈B| aRb for some a∈A}. Thus, a function f: X -> Y is a subset of A×B where Y⊆B and X={ a∈A | afb for exactly one b∈Y}. Hence, the domain subset X may not be the mother set A, and similarly, the codomain subset Y may not be equal to the mother set B; i.e. a function f:X->Y may not be a function f:A->B on the mother sets. That's why it's important to make such a distinction. For example the relation R={(2,5), (3,6)} is a function on X={2,3} Y={5,6}, but not a function on A={-2,2,3} B={5,6} even though X⊆A and Y⊆B. This is a flaw in your 3rd example. You set your domain and codomain to be both [0,∞), but for some reason you tested out inputs not in your proper domain and claimed that it wasnt a function. In your 1st example you do not make a distinction between the domain/codomain subsets X,Y and the mother sets A,B. It was a bit confusing at first to understand your judgement.

[@mescellaneous]

wow, i did not catch that for the third example. perhaps he meant x^2 ∈ [0,∞)