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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


for all points $x \in E$ for which


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


for all points $x \in E$ for which


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?


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).

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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
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. – Sh3ljohn 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
up vote 3 down vote accepted

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 of $0<d_X(x,p)$ was not a condition - 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.

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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
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

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, then, 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 aproach points at which they are undefined.

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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

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.

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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

"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.

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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

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