2
$\begingroup$

I'm reading through Functional Analysis by Bachman.

He defines a limit point as follows:

The point $x$ is said to be a limit point of $A \subset X$ iff for every $r$, $S_r(x) \cap A$ contains infinitely many points of $A$.

Then, later in the book, he gives the following example:

Suppose $A$ is a non-countably compact set with some infinite subset $B$ with no limit point in $A$. Let $M=\{x_1,x_2,...\}$ be a denumerable set of distinct points from $B$. Let $E_1, E_2,....$ be a sequence of $\epsilon_n - $ neighbourhoods of $x_1,x_2,...$ respectively, where the $\epsilon_n$ have been chosen to guarantee that $$E_n \cap M = \{x_n\}.$$ If it were not possible to construct such neighborhoods for every point, this would mean that for some $x_n$, every neighbourhood of it would contain some other point of $M$ distinct from $x_n$. Thus $x_n$ would be a limit point of the set $M$.

But didn't we require that each neighbourhood contain infinitely many points of the set? Here each neighbourhood may contain even only one of ther point of $M$, so wouldn't it be a point of closure, rather than a limit point?

EDIT: note that the definitions refer to metric spaces

$\endgroup$
3
  • 1
    $\begingroup$ I am assuming you have a metric space. Pick one of the neighbourhoods of this $x_n$ say $S_r(x)$ if $S_r(x)\cap M=\{x_{a_j}\}$ is finite let $a=\min\{d(x_n,x_{a_j})\}$ then by taking $S_{\frac a 2}(x_n)$ you get a neighbourhood with the required property. $\endgroup$
    – Spotty
    Jul 8, 2016 at 12:03
  • $\begingroup$ Yes, that's right, I should have mentioned the context is a metric space. I'll edit the question in case anyone stumbles upon it $\endgroup$ Jul 8, 2016 at 12:34
  • 1
    $\begingroup$ What Bachman calls a limit point is what other authors call an accumulation point. $\endgroup$ Jul 31, 2016 at 17:19

3 Answers 3

6
$\begingroup$

By definition $p$ is a limit point of $E$ if every neighborhood of $p$ contains a point $q$ belonging to $E$ that is different from p , but using this definition we can also prove that if $p$ is a limit point of $E$ then every neighborhood of $p$ contains infintely many points of $E$.

Sketch of the proof is as follows : suppose there was a neighborhood with finitely many points take the minimum of distances between those points and $p$ then take a neighborhood with a diameter less than half of that minimum; this new neighborhood is empty contradicting the fact that $p$ is a limit point ! Excuse my long sentences I'm still working on my Latex .

$\endgroup$
1
  • $\begingroup$ Ok, so requiring the intersection to contain either one point distinct from the point itself, or infinitely many points, is equivalent. Thanks, got it! $\endgroup$ Jul 8, 2016 at 12:11
4
$\begingroup$

Theorem. Every limit point of a every subset of topological space $X$ is an $\omega$-accumulation point of the subset if and only if $X$ is a T1 space, i.e., the singleton $\{x\}$ is closed for each $x\in X$.

(N.B. Here $x\in X$ is a limit point of $A\subset X$ if every neighborhood of $x$ contains a point of $A\setminus\{x\}$. A point $x\in X$ is an $\omega$-accumulation point of a subset $A\subset X$ iff every neighborhood of $x$ contains infinitely many points of $A$. Hausdorff spaces (aka. T2 spaces) are T1 spaces, and metric spaces are Hausdorff.)

Proof: $\Leftarrow$: Let $A\subset X\owns x$. Assume that for some neighborhood $G$ of $x$, the set $G\cap A\setminus\{x\}$ is finite, say $G\cap A=\{a_1,a_2,\ldots,a_n\}$. Then $G':=G\cap\{a_1\}^c\cap\{a_2\}^c\cap\cdots\cap\{a_n\}$ is a neighborhood of $x$ containing no points of $A$, hence then $x$ is not a limit point of $A$.

$\Rightarrow$: If $X$ is not T1, then $x\ne a$ is in the closure of $\{a\}$ for some $a,x\in X$. Then every neighborhood $G$ of $x$ contains $a$ (as otherwise $G^c$ would be a closed set containing $a$ but not $x$). Therefore, $x$ is then a limit point of $\{a\}$ but not an accumulation point, as some (actually, any) neighborhood of $x$ does not contain infinitely many points of $\{a\}$.

$\endgroup$
0
$\begingroup$

If $x_n$ does not have such a nbhd $E_n$ then every nbhd of $x_n$ contains a member of $M$ \ $\{x_n\}.$ So let $U_1$ be any nbhd of $x_n . $ Take $y_1\in (M\cap U_1)$ \ $\{x\}.$ For $m\in N$ let $U_{m+1}=U_m$ \ $\{y_j:j\leq m\} .$ (The inductive hypothesis is that $U_m$ is a nbhd of $x_n$ and that $x_n\not \in \{y_j :j\leq m\}$ .) Take $y_{m+1}\in (M\cap U_{m+1})$ \ $\{x_n\}.$ Then $m\ne m'\implies y_m\ne y_{m'}, $ and $\{y_m: m\in N\}\subset U_1$ \ $\{x_n\}.$

Remark. For a metric space $X,$ and $M\subset X,$ and $p\in X,$ the following are equivalent: (i). $p$ is a limit point of $M .$ (ii).$\;p\in$ Cl$(M$ \ $\{p\} ).$ And the following are equivalent : (i'). $\;p$ is not a limit point of $M.$ (ii'). $\;p \not \in$ Cl$(M$ \ $\{p\}).$ (iii'). $\; p$ has a nbhd $U$ with $U\cap M\subset \{p\}.$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .