# motivation of limit points

Lets use Wikipedia's definition of a limit point and let $\lim(A)$ denote the set of limit points of $A$. $a\in \lim (A) \leftrightarrow a\in\operatorname{cl}(A\setminus\{a\})$, $\lim (A)\cup A = \operatorname{cl}(A)$. Is there a situation where lim suits us and cl does not? lim is not a generalization of a limit of sequence, because a constant sequence has the limit in any topological space and no limit points in any $T_1$-space. It seems that $a\in\operatorname{cl}(A)$ is an appropriate formalization of the intuitions “$A$ converges to $a$” and “$a$ is infinitely close to $A$”.

Maybe lim is helpful to define a closed set via open balls, like in Rudin's “Principles of Mathematical Analysis” in Definition 2.18.d? Here it is:

A point $p$ is a limit point of the set $E$ if every neighborhood of $p$ contains a point $q\neq p$ such that $q\in E$.

(Author's “neighborhood of $p$” is my “open ball of $p$”, check Definition 2.18.a.) But a slightly modified and simpler definition “every open ball of $a$ meets $A$” is equivalent to $a\in\operatorname{cl}(A)$. No limit points are needed.

(There is a similar question “Limit points and interior points”. On the contrary, I'm not asking for understanding.)

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I changed \operatorname{lim} to \lim. It's already a standard operator name. When used in "display" format as opposed to "inline" format, this operatorname puts subscripts directly below "lim", thus: $\displaystyle\lim_{n\to\infty}$. – Michael Hardy Jul 9 '12 at 17:29

## 2 Answers

Both of the sets $\operatorname{cl}A$ and $\lim A$ are useful, but for different purposes. Cameron Buie has already given a nice example of a setting in which $\lim A$ is important and $\operatorname{cl}A$ is not, and the reverse situation is very common, so I’ll try to address the question of motivation in more general terms.

The points of $A\setminus\lim A$ are in $\operatorname{cl}A$ almost by accident, simply by virtue of being in $A$; they aren’t structurally tied to the rest of $A$. Points of $\lim A$, on the other hand, whether they’re in $A$ itself or not, are structurally tied to $A$: you can’t surround one of them with an open nbhd without picking up another point of $A$. In fact, if $X$ is $T_1$, a weak property that amounts to saying that finite subsets of $X$ are closed, you can’t surround a point of $\lim A$ with an open nbhd without picking up infinitely many points of $A$. Thus, $\lim A$ is likely to be the set of interest when you’re dealing with questions of convergence, continuity (as in Cameron’s answer), or anything else that depends as much on what happens near a point as it does on what happens at the point.

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Let me give you an instance when $\lim(A)$ can come in handy and we don't care about $\mathrm{cl}(A).$ First, a few definitions.

We say a topological space $Y$ is Hausdorff if and only if for every $x,y\in Y$ we have either (i) $x=y$ or (ii) there exist neighborhoods--I'll abbreviate by "nbhds"--$U,V$ of $x,y$ (respectively) such that $U\cap V=\emptyset$. (Put another way, distinct points of Hausdorff spaces can be "separated by" open sets.)

A punctured neighborhood--I'll abbreviate by pnbhd--of a point $x$ in a topological space $X$ is a set $U\smallsetminus\{x\}$, where $U$ is a nbhd of $x$ in $X$.

Suppose $X,Y$ topological spaces, $Y$ Hausdorff, $f:A\to Y$ with $A\subseteq X$, $y\in Y$, and $a\in\lim(A)$. Then we say that $$y=\lim_{x\to a}f(x)$$ if and only if for every nbhd $V$ of $y$ in $Y$, there is a pnbhd $P$ of $a$ in $X$ such that $f(x)\in V$ whenever $x\in A\cap P$.

Now, suppose $X,Y$ are topological spaces, $Y$ Hausdorff, $f:A\to Y$ with $A\subseteq X$, and $a\in A$. Then (one can prove that) $f$ is continuous if and only if either (i) $a$ is isolated in $A$ or (ii) $a\in\lim(A)$ and $f(a)=\lim_{x\to a}f(x).$ Note that this generalizes the "limit definition" of continuity that one generally encounters in real analysis to an arbitrary Hausdorff space $Y$ and an arbitrary subset $A$ of an arbitrary topological space $X$. Note that we do not need to know everything about $\mathrm{cl}(A)=A\cup\lim(A)$. In case (i), we're looking at $A\smallsetminus\lim(A)$, and in case (ii), we're looking at $A\cap\lim(A)$. Unless $A$ is closed, there will be limit points of $A$ not lying in $A$, so in general, we're using less information, here.

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"$\mathrm{cl}(A)=A\cap\lim(A)$" Is this a mistake? – beroal Jul 9 '12 at 21:50
Yes, it is! '\cap' should be '\cup'.... Thank you, beroal! – Cameron Buie Jul 10 '12 at 0:05
Now I think that requirement $a\in\lim(A)$ implies that $\lim_{x\to a}f(x)$ is unique. If $a\not\in\lim(A)$, then there is a punctured neighborhood $P$ of $a$ such that $P\cap A\subseteq \varnothing$, therefore $Y\subseteq \lim_{x\to a}f(x)$, these limits are not interesting. – beroal Jul 10 '12 at 3:39