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In highschool, I learned continuity of a function $f: \mathbb{R} \to \mathbb{R}$ at a point $a$ to be defined as the following condition holding:

$$ \lim_{x \to a} f(x) = f(a) \tag{1}$$

However in undergraduate analysis book, the continuity is defined as so. A function $f: \mathbb{R} \to \mathbb{R}$ at a point is continous, if for every $\epsilon>0$, there exists a $\delta$ such that:

$$ |x-a | < \delta \implies |f(x) - f(a) | < \epsilon \tag{2}$$

In this video by Michael Penn , it's noted that somehow the undergraduate definition is more general although it is not explained precisely how. I think that they are actually the same. I think so clearly (1) implies (2) [take definition of limit], but I am having difficulty understanding why (2) doesn't imply (1). Could someone explain abstractly what the issue is? ( Examples are helpful but please don't make the entire explanation based on that, thanks)

  • $1)\iff 2)$ . Even in any metric spaces $1) \iff 2) $ – Sourav Ghosh Sep 05 '22 at 14:04
  • $\epsilon-\delta$ continuous iff sequentially continuous. $1) \iff 2)$ can be extended to a function $f:(X, \tau) \to (Y, \tau') $ where $(Y, \tau)$ is Hausdoff. – Sourav Ghosh Sep 05 '22 at 14:08
  • $f:E\subset (X,\tau)\to (Y, \tau')$ and $a\in E'$ then $\lim_{x\to a} f(x) =f(a) $ iff $\forall V\in\tau'$ and $f(a) \in V$ there exists $U\in \tau$ containing $a$ such that $f(E\cap U\setminus {a})\subset V$. – Sourav Ghosh Sep 05 '22 at 14:13

3 Answers3

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Whether $(1)$ and $(2)$ are actually different depends on how you define $\lim_{x\to a}f(x)$.

There are two typical ways to define it, one is so called $\varepsilon-\delta$ definition and the other uses sequences.

Def. 1. Let $L$ such that $$(\forall\varepsilon>0)(\exists \delta > 0 )\quad 0<|x-a|<\delta \implies |f(x) - L| < \varepsilon.$$ Then we write $L = \lim_{x\to a}f(x)$.

Def. 2. Let $L$ such that for every sequence $(x_n)$ ($x_n\neq a$) that converges to $a$ we have that sequence $(f(x_n))$ converges to $L$. Then we write $L = \lim_{x\to a}f(x)$.

When we write $\lim_{x\to a} f(x) = f(a)$, if we use def. 1., we say that $f$ is continuous at point $a$. If we use def. 2., we say that $f$ is sequentially continuous at point $a$. It is a well known theorem that for real functions of a real variable these two definitions coincide, so definitions of continuity $(1)$ and $(2)$ are actually equivalent in this case.

However, there are general types of spaces where limits and continuity make sense called topological spaces. In such spaces every continuous function is sequentially continuous, but sequentially continuous function doesn't need to be continuous. The reason for this is that topological spaces can be complicated and sequences alone might not be enough to describe behaviour around a single point. In that case, instead of sequences we need generalized sequences called nets or hypersequences. The main distinction is that sequences are indexed by a countable set, while nets can be indexed by uncountable sets.

Ennar
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If we're talking about a real-valued function defined on $\Bbb R$ the two are the same. Maybe point in that video is that (1) makes sense in the context of topological spaces, where there's no such thing as $|x-a|$.

In fact, in the context of topological spaces, (1) is equivalent to the standard definition of continuity. I've been asked why.

Say $X$ and $Y$ are topological spaces and $f:X\to Y$.

Def. Given $a\in X$ and $b\in Y$ we say that $\lim_{x\to a}f(x)=b$ if for every neighborhood $V$ of $b$ there exists a neighborhood $U$ of $a$ with $f(U)\subset V$.

Def. $f$ is continuous if the inverse image of any open set is open.

Prop. $f$ is continuous if and only if (1) holds for every $a\in X$.

Proof. Suppose first that $f$ is continuous and $a\in X$. Let $V$ be a nbd of $f(a)$. Now $U=f^{-1}(V)$ is a nbd of $a$ with $f(U)\subset V$.

Otoh suppose (1) holds for every $a\in X$ and let $V\subset Y$ be open. Let $U=f^{-1}(V)$; we need to show that $V$ is open. For each $a\in U$ there exists an open set $U_a$ with $a\in U_a$ and $f(U_a)\subset V$. Hence $U_a\subset U$. So $U=\bigcup_{a\in U}U_a$, hence $U$ is open.

  • But topological spaces need pre image of open is open but how is that captured by (1). Non obviosu to me – tryst with freedom Sep 05 '22 at 12:24
  • It's easy to show that (1) is equivalent to the condition about inverse images – David C. Ullrich Sep 05 '22 at 12:26
  • When I expand out the condition of 1 using definition of a limit, I get that it is same as 2 (mentioned in post). Maybe I miss something, Could you tell me how you would do it? – tryst with freedom Sep 05 '22 at 12:28
  • @TrystwithFreedom I have no idea what you're asking. If you're still talking about $f:\Bbb R\to\Bbb R$ then (1) is the same as (2). So what's the problem? (How would I do what???) – David C. Ullrich Sep 05 '22 at 12:41
  • Oh sorry, my brain died when I wrote that comment I think. I meant, how can one show that (1) is equivalent to the inverse image definition from topology? – tryst with freedom Sep 05 '22 at 12:42
  • @TrystwithFreedom See edit. – David C. Ullrich Sep 05 '22 at 12:59
  • Don't you need first-countable or something for one implication? – Lukas Heger Sep 05 '22 at 13:07
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    @LukasHeger I don't see why - I suspect you're thinking of a similar result (or more or less the same result with a different definition of limit?) – David C. Ullrich Sep 05 '22 at 13:17
  • Yes, I was thinking of sequential continuity, but your definition is different – Lukas Heger Sep 05 '22 at 13:22
  • Reiterating my comment to Arthur; the limit you’re defining is an undeleted limit. While Arthur was talking about a deleted limit. – Vivaan Daga Sep 05 '22 at 13:42
  • @Shinrin-Yoku If we're talking about limits in general then yes it's very important that the definition says $0<|x-a|<\delta$ and not $|x-a|<\delta$. But here, when we're just talking about whether $\lim_{x\to a}f(x)=f(a)$, it makes no difference, because the conclusion $|f(x)-f(a)|<\epsilon$ is automatically true when $x=a$. – David C. Ullrich Sep 05 '22 at 13:45
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I can't read the mind of Michael Penn, but here is one case where the undergraduate definition is needed: A domain with isolated points.

Say, for instance, a function defined on the integers. An expression like $\lim_{x\to a}f(x)$ (more commonly written as $\lim_{n\to a}f(n)$) makes no sense for finite $a$, so the highschool definition can't be used to even ask the question about continuity. But the undergraduate definition still lets you probe whether the function is continuous (and in this particular case you will find that all functions are continuous).

It is true that for functions with, say, some Euclidean space $\Bbb R^n$ as domain (and standard metric / topology), the two notions are equivalent. And even more generally, whenever both notions of continuity make sense, they will agree on the question of continuity. But for general metric spaces (with a trivial rewrite $|x-a|\mapsto d(x, a)$), the undergraduate definition is the only one of the two that makes sense everywhere.

(Do note that it goes the other way too: there are domains where limits make sense but distances don't, so there are times where the highschool definition works but the undergraduate definition doesn't. So I wouldn't say that either is more general than the other.)

Arthur
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  • Someone had written this in the comments of the video but it still doesn't resolve my query :/ – tryst with freedom Sep 05 '22 at 12:23
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    Limits make perfect sense in the integers, it's just that all convergent sequences are eventually constant, which gives the same conclusion that every function on the integers is continuous. – Lukas Heger Sep 05 '22 at 12:26
  • How does that not resolve your query? If they both make sense, they agree on whether a function is continuous, but the undergraduate one makes sense in places where the highschool one doesn't. – Arthur Sep 05 '22 at 12:26
  • @LukasHeger That's a different kind of limit from the one we are talking about here. You mean the limit of a sequence, while we are interested in the limit of a function as the variable approaches (but is never equal to) some point in the domain. A limit like $\lim_{n\to 0}3^n$ does not make sense in the integers, and that's the type of limit that the highschool definition of limits above talks about. – Arthur Sep 05 '22 at 12:29
  • @LukasHeger That doesn't work. You need to specify that the sequence is never equal to $a$. The whole point of limits is that they never, ever actually touch the value in question (in this case $a$). And that makes it not work to probe continuity at isolated points. Also, sequential continuity is not the same as limit continuity, so you can't read one and then just blindly assume it means the other. – Arthur Sep 05 '22 at 12:40
  • "$\lim_{x\to a}f(x)=f(a)$" is an entirely different thing from "$\lim_{n\to\infty}f(x_n)=f(a)$ whenever $\lim_{n\to\infty}x_n= a$". – Arthur Sep 05 '22 at 12:45
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    I think this eventually boils to deleted limits vs non deleted limits for instance if you give the def of a limit in general top. spaces like this: https://math.stackexchange.com/questions/3151413/how-to-define-limit-operations-in-general-topological-spaces-are-nets-able-to-d then there is no problem about talking about limits on functions in the integers. – Vivaan Daga Sep 05 '22 at 13:01
  • @Shinrin-Yoku I think that that definition only works for functions that are already established to be continuous. Take, for instance, the function $f:\Bbb R\to \Bbb R$ given by $$f(x)=\cases{1& if $x=0$\0& otherwise}$$Any reasonable definition of limits would say that that function has a limit at $0$, and that limit is $0$. But your linked definition fails to yield a limit, precisely because the function is discontinuous. And limit definitions that require continuity cannot be used to define continuity. – Arthur Sep 05 '22 at 13:07
  • (One could presumably declare the existence of that limit as a definition of continuity. But that's not really one of the options in question here.) – Arthur Sep 05 '22 at 13:08
  • I don’t disagree with you; I don’t do much topology but as far as I know the notion of a undeleted limit is rarely useful; if it were me I would define it like this: https://en.wikipedia.org/wiki/Limit_of_a_function#Functions_on_topological_spaces. And then this answer would be completely correct. Though to be honest it’s written so vaguely that I can sort of understand where the downvotes are coming from. (But I don’t feel they are deserved). – Vivaan Daga Sep 05 '22 at 13:41