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In "Principles of mathematical analysis" Walter Rudin gives the definition of

$$\lim_{x \rightarrow p} f(x)=q$$

for $f: X \supset E \rightarrow Y$ with $X,Y$ metric spaces, in this way:

for every $\epsilon>0$ there exists a $\delta >0$ such that

$$d_Y(f(x),q)<\epsilon$$

for all points $x \in E$ for which

$$0<d_X(x,p)<\delta$$

where $d_X$ and $d_Y$ are the distances in $X$ and $Y$. After few pages he gives the definition of continuity:

$f$ is said to be continuous at $p$ if for every $\epsilon>0$ there exists a $\delta >0$ such that

$$d_Y(f(x),f(p))<\epsilon$$

for all points $x \in E$ for which

$$d_X(x,p)<\delta$$

My question is, why did he drop the $0<\dots$ in $d_X(x,p)<\delta$? I though that continuity in $p$ means that the limit of the function in $p$ exists and is equal to $f(p)$, so one should simply replace $q$ with $f(p)$ in the above definition of limit, why changing the condition on $x$?

Doing so a function like

$$f: \{1\} \rightarrow \mathbb{R}$$ which maps 1 to 5, for example, would be continous in 1, since $x=p$ is allowed, but what is the limit of $f$ in 1? It doesn't exist, right?

Is there a deep reason for such a counterintuitive definition?

EDIT:

Maybe I should better clarify my question. I know that this definition works, I am asking why considering a function continuos at isolated points (which is an immediate consequence of this defintion ---> It turned out I was wrong about this).

The only reason I can think of is that this definition agrees with the topological one for continuity, but it doesn't seem to me a good reason since topology came after (and would have agreed no matter the convention).

Emmet
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  • Consider the statement: $\lim_{x\to p} f(x) = q$ to mean "what could we set $f(p)$ to, to make $f$ continuous at $p$? In that case, $\lim_{x\to p} f(x)$ usually has only one value, but when $p$ is "isolated," the definition of limit allows any $q$ on the right. – Thomas Andrews Feb 26 '13 at 16:08
  • @ThomasAndrews is the limit multi-valued or undefined? I don't understand. – Emmet Feb 26 '13 at 16:09
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    We write: $\lim_{x\to p} f(x) = q$ but the definition does not require $q$ to be distinct, so the equals sign is lazy language. The only case where $\lim$ is multi-valued is when $p$ is isolated, however, so most of the time we are safe. – Thomas Andrews Feb 26 '13 at 16:12
  • To be honnest, I would do exactly the contrary... I would omit the $0<$ for the definition of limit, and include it in the definition of continuity.. Continuity only makes sense to me when you analyze what happens in the surroundings of a point, knowing the value at the point. Existence of a limit simply means that a value can be chosen arbitrarily close to another value, and this doesn't exclude taking the target value itself in my opinion. – Jonathan H Feb 26 '13 at 16:19
  • @Sh3ljohn Then if $f(p)=1$ and $f(x)=0$ if $x\neq p$ would have $\lim_{x\to p} f(x)$ undefined, when you want it to be $0$. – Thomas Andrews Feb 26 '13 at 16:21
  • @ThomasAndrews I was looking for something to say, but I think my logic is wrong and you are right. So "continuity" and "limit" are more closely tied than I thought. :/ The only thing that bothers me is that in the case you describe, do we, indeed, want it to be 0? – Jonathan H Feb 26 '13 at 16:27

4 Answers4

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The real question is, why is the $0<d_X(x,p)$ condition in the definition of the limit?

Basically, it doesn't hurt to allow $x=p$ in the continuity example, because (1) we know that $p\in E$, and (2), when $x=p$, $d_Y(f(x),f(p))=0<\epsilon$, so there is no reason to leave it out.

Essentially, in the continuity case, the case $x=p$ is trivially true.

On the other hand, when $p\in E$ in the limit definition, we don't want the limit to depend on $f(p)$, but only on the values $F(x)$ when $x\neq p$.

For example, when $f(x)=0$ for $x\neq p$ and $f(p)=1$, we want $\lim_{x\to p} f(x) = 0$, but that would not be true if we didn't have the condition $0<d_X(x,p)$ - without that condition, the limit is undefined.

Isolated points:

We consider isolated points to be points of continuity because we want in general that if:

$f:E\to Y$ is continuous at $p$ and $p\in E'\subset E$ then $f_{|E'}:E'\to Y$ to also be continuous at $p$.

However, note that the continuity definition could have said $0<D_X(x,p)$. That doesn't affect the continuity at $p$ one bit, so if you wanted to define all isolated points as points of discontinuity, you'd want some other definition completely.

Thomas Andrews
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  • How can you take the limit at isolated points without considering the point itself? This is what I don't understand – Emmet Feb 26 '13 at 16:28
  • @Emmet. In general, consider a statement like: "Forall $x$, if $P(x)$ then $Q(x)$." Now, what if $P(x)$ is never true? Mathematicians then consider the statement" "Forall $x$, $P(x)\implies Q(x)$" to be trivially true. Int this case $P(x)$ is $0<d_X(x,p)<\delta$ where $\delta$ is chosen so that only $p$ is within $\delta$ of $p$. – Thomas Andrews Feb 26 '13 at 16:36
  • In the definition of limit $x=p$ is not allowed, so $\delta=0$ is not allowed. If I cannot consider $x=p$ then the limit of an isolated point is not defined because for all punctured Balls centered on $p$ and small enough there would be no points to consider. So the limit is not defined there. Why would $f$ still be continuos with $0<\delta$? Maybe I am confused by the definition of continuity and the fact that a function is continuos in $p$ when it's limit in $p$ equals $f(p)$. – Emmet Feb 26 '13 at 16:37
  • Right, but when $p$ is an isolated point, there is a $\delta>0$ such that there are no points $x$ that satisfy $0<d_X(x,p)<\delta$ – Thomas Andrews Feb 26 '13 at 16:38
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    Your comment about $P(x) \Rightarrow Q(x)$ being trivially true when $P(x)$ is false is very interesting. Maybe that is the point that I am missing. – Emmet Feb 26 '13 at 16:40
  • Let me see if I understood, you are saying that requiring $x \neq p$ in the definition of continuity (and so simply replacing $q$ with $f(p)$ in the definition for limit) would still make a function continuos at isolated points. My example is a bad choice because my point is open and closed since it's the only one, but if choose ${0} \cup [2,3]$ as my domain, no matter $\epsilon>0$ I choose, every $\delta$ works for $f(0)$? – Emmet Feb 26 '13 at 16:52
  • It takes a while getting used to, but the idea is that: $\forall x\in\emptyset: Q(x)$ is true, no matter what $Q$ is. This is one of those really useful logical definitions. It is similar to why we want empty sums to be defined as zero. In general, you want: $\forall x\in X_1\cup X_2: Q(x)$ to be equivalent to $\forall x\in X_1: Q(x) \land \forall x\in X_2: Q(x)$. If $X_1$ is empty, you still want $\forall x\in X_1:\dots$ to be true. – Thomas Andrews Feb 26 '13 at 16:53
  • All isolated points are opened an closed. That is their nature. All points in a metric space are closed, and isolation means that the there is a ball that is equal to ${p}$, so it is open, too. – Thomas Andrews Feb 26 '13 at 16:55
  • Opennes shouldn't depend on the topology? You are saying that points are open in the usual topology for $\mathbb{R}$ (the distance one)? – Emmet Feb 26 '13 at 16:59
  • The topology is defined by the metric. Once you have a metric space, the topology is defined by the basis of open balls under the metric. – Thomas Andrews Feb 26 '13 at 17:00
  • Anyway +1 for kindness. – Emmet Feb 26 '13 at 17:01
  • They are open when they are isolated. So in $\mathbb R$ there are no isolated points, but in set $\mathbb Z\subset \mathbb R$, all points are isolated, and any function $\mathbb Z\to Y$ are continuous. In general, it is not true that if $p\in E'\subset E$ and $f:E\to Y$ is such that $f_{|E'}$ is continuous at $p$ then $f$ is continous at $p$. For example, if $E=\mathbb R$ and $E'=\mathbb Q$, then $f(x)=1$ if $x$ is rational and $f(x)=0$ otherwise is not continous at any point of $E$, but is continuous when restricted to $E'$. – Thomas Andrews Feb 26 '13 at 17:02
  • There is still something I can't really catch about your argument but you helped me to understand I need some more thoughs on those topics. Thank you very much. – Emmet Feb 26 '13 at 17:05
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It's fine to let $x=p$ in the definition of continuity, since then $f(x)=f(p)$, and so $$d_Y(f(x),f(p))=0<\epsilon.$$ It just isn't very interesting. When simply discussing limits of functions, though, we may want to examine behavior near but not at a point (such as when taking derivatives), so we require $x\neq p$ in such cases.

Note: It's possible that $p$ is isolated in $X$, so that we can't talk about the limit as $x$ approaches $p$ at all. A function $f:X\to Y$ is continuous at $p\in X$ if and only if either (i) $p$ is isolated in $X$ or (ii) $p$ is a limit point of $X$ and $f(p)=\lim\limits_{x\to p}f(x)$.

Edit: Even with your proposed alteration to the definition, a function is necessarily continuous at isolated points of its domain. Suppose that $f:E\to Y$ with $E\subseteq X$ and $p$ an isolated point of $E$. In other words, there is some $\delta>0$ such that for any $x\in E$, we have $d_X(x,p)<\delta$ if and only if $x=p.$ In particular, there is no $x\in E$ such that $0<d_X(x,p)<\delta$.

Now, take any $\epsilon>0$. Then for all points $x$ in the (empty) set $$\{x\in E: 0<d_X(x,p)<\delta\},$$ we have that $$d_Y\bigl(f(x),f(p)\bigr)<\epsilon,$$ vacuously. Since $\epsilon>0$ was arbitrary, then $f$ is continuous at $p$, under your proposed alteration to the definition. In fact, your altered version is mathematically equivalent to Rudin's version.

This sort of vacuous truth is bothersome to many, and permitting $x=p$ as in Rudin's definition of continuity allows us to avoid such things. As I said, allowing $x=p$ isn't a problem, but it isn't very interesting, either. The primary reason that we have to have $0<...$ in the limit definition, but not necessarily in the continuity definition, is so that we can examine limits of functions with (for example) removable discontinuities at a point, or limits of functions as we approach points at which they are undefined.

Cameron Buie
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  • It's a bit confusing, but he really seems to mean $f:E\to Y$, with $E\subseteq X$. So all you really need is that $p$ is an isolated point of $E$. Also, if $p$ is an isolated point, with $0<d_X(x,p)$ you get continuity at $p$ is trivially true. – Thomas Andrews Feb 26 '13 at 16:04
  • @ThomasAndrews Yes, sorry, should I edit the question to make it clearer? – Emmet Feb 26 '13 at 16:07
  • @Cameron_Buie Ok, so, why not choosing a definition that avoids this distinction and call a function discontinuos at isolated points? But maybe I am missing something here... – Emmet Feb 26 '13 at 16:27
  • @Emmet: Even if you altered the definition of continuity as you suggest, a function will be continuous at isolated points. See my updated answer. – Cameron Buie Feb 26 '13 at 17:00
  • @CameronBuie thank you, I was busy talking with Thomas and didn't notice it. – Emmet Feb 26 '13 at 17:06
  • @CameronBuie Your answer is very nice. What answer should I choose? Your answer is clearer to me, but Thomas took a lot of time and patience to clarify all my doubts. What do you suggest? – Emmet Feb 26 '13 at 17:11
  • @Emmet: I'd upvote both, regardless. I'm not sure which answer I'd accept in your place, and for precisely the reasons you state! Sorry! – Cameron Buie Feb 26 '13 at 17:16
  • I already upvoted both :) Anyway the vacuos thing was very helpful, it was already mentioned by Thomas but not so clearly (at least for my weak mind). I'll choose him but you can consider yourself the moral winner (I hope everybody has his reward this way). – Emmet Feb 26 '13 at 17:22
  • @Emmet: Perfectly fine. I do love moral victories. ^_^ – Cameron Buie Feb 26 '13 at 17:24
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In the situation where $p$ is an isolated point, then continuity at $p$ is guaranteed no matter what and it doesn't make sense to talk about the limit of $f(x)$ as $x \to p$, so let's suppose that $p$ is not isolated.

A function which is continuous at $p$ in particular has to be defined at $p$. If you read the definition of limit of a function as the input approaches $p$ carefully, you'll notice that it doesn't require the function to be defined in $p$ but only on a set which has $p$ as a limit point.

kahen
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  • I see that It works when $f$ is defined in $p$, but I don't understand why doing it that way. It works even with a punctured neighbourhood of $p$. – Emmet Feb 26 '13 at 16:00
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"I though that continuity in p means that the limit of the function in p exists and is equal to f(p), so one should simply replace q with f(p) in the above definition of limit"

Doing so, the definition says for all $\epsilon>0$ there is $\delta>0$ such that if $0<d_X(x,p)<\delta$ then $d_Y(f(x),f(p))<\epsilon$. But note that if $d_X(x,p)=0$ then $x=p$ so $d_Y(f(x),f(p))=0$, so one may trivially drop the "$0<$".

Rudin stipulates $d_X(x,p)>0$ in the definition of limit just so he may say things like $$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}=\text{stuff},$$ without having to worry that the fraction isn't defined at $h=0$.

A possibly clearer definition of $\lim_{x\to t} f(x)$ is the common value of $\lim f(x_n)$ for all sequences $x_n$ in $X\backslash\{t\}$ such that $x_n\to t$, provided that these limits exist and are equal.

Sean Eberhard
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  • In the case of the derivative limit, that's just a case where $E=\mathbb R\setminus {0}$ so we don't need the condition $d_x(h,0)>0$, since $h\notin E$. You need the condition when $f(0)$ is defined - then you want to ignore that value when determining the limit – Thomas Andrews Feb 26 '13 at 16:16
  • But is $x_n=q,q,q,\dots$ (where $q$ is the limit considered) allowed? – Emmet Feb 26 '13 at 16:18
  • Where is $x_n$ coming from, @Emmet. We are not talking about sequences at all here. And $q\in Y$. – Thomas Andrews Feb 26 '13 at 16:19
  • @ThomasAndrews Yes, you are right, I meant $x_n=p$ when considering the limit of $x \rightarrow p$. Sean talked about sequences. – Emmet Feb 26 '13 at 16:24
  • @Emmet No, it's not. We consider sequences $(x_n)$ in $X\backslash{p}$ such that $x_n\to p$, so no $x_n$ is allowed to be $p$. – Sean Eberhard Feb 27 '13 at 10:11