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I ran into something which seems like it should be true (and must be true because it is used in the proof of a theorem).

I've extracted the detail that I cannot quite verify.

Let $(x_{n})$ be a sequence in a first countable topological space $X$. Let $y$ be an accumulation point of the set $I:=\{x_{n}:n\geq 1\}$ (that is, every neighborhood of $y$ contains $x_{n}$ for some $n\geq 1$ with the property that $y\neq x_{n}$).

Then $y$ is a limit point of the sequence $(x_{n})$. That is, every neighborhood $U$ of $y$ has the property that $U\cap I$ is infinite.

Note: I'm not sure if First countability is required here, but it was a hypothesis in the theorem (and was indeed used elsewhere in the proof so it probably isn't required) so I included it just in case.

Why I am stuck: If $X$ is a metric space, the argument is simple, assume for a contradiction that the set $U\cap I$ is finite, then take a ball $B:=B(y,\epsilon)$ with $\epsilon$ equal to half of $min\{d(y,x_{n}) : x\in U\cap I, x\neq y\}$. Then $U\cap B$ is a neighborhood of $y$ which contains no points of $I$ except possibly $y$ itself, which is a contradiction.

But since I'm not dealing with a metric space, this argument does not apply.

But I think that I can come up with a more general argument if $\{x\}$ is a closed set for every $x\in X$ (and more importantly for me, its complement is open).

Is this normally a hidden assumption in topological setings? Or is there another argument?

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1  
When you say if point sets are closed, do you mean if singletons are closed, i.e., sets of the form $\{x\}$? That amounts to assuming that the space is $T_1$. –  Brian M. Scott May 15 '12 at 18:30
    
Yes that's what I meant. I'll edit it now. I did a google of point-sets and discovered that such terminology actually refers to something else. –  Kyle Schlitt May 15 '12 at 18:33

3 Answers 3

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Say $X = \{a,b,z\}$, where open sets are $\emptyset$, $X$, $\{a,b\}$, $\{z\}$. So this is certainly not a $T_1$ space. But it is first-countable, since there are only 4 open sets in all. Note every neighborhood of $a$ contains $b$. Here is our sequence: $x_1 = b$, and $x_k = z$ for all $k \ge 1$. Then, according to your definition, $a$ is an accumulation point of the set $I$. But $a$ is not a limit point of the sequence.

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The general argument for a $T_1$-space $X$ goes like this. Suppose suppose that $x$ is an accumulation point of some set $S\subseteq X$. Let $U$ be any open nbhd of $x$, and suppose that $U\cap S$ is finite. Then for each $y\in U\cap S$ there is an open nbhd $B(y)$ of $x$ that does not contain $y$. Now let $$V=U\cap\bigcap_{y\in U\cap S}B(y)\;;$$ $V$ is the intersection of finitely many open nbhds of $x$, so $V$ is an open nbhd of $x$, and clearly $V\cap S\subseteq\{x\}$. This contradiction shows that $U\cap S$ must be infinite for each open nbhd $U$ of $x$.

Note that first countability is not in fact necessary, and neither is countability of $S$. However, as you thought, you do need to be able to separate $x$ from individual points of $S$ by open sets. Without that, there might be a point $y\in S\setminus\{x\}$ such that every open nbhd of $x$ contains $y$; if in addition $x$ has a nbhd $U$ that excludes all other points of $S$, which is certainly possible, $x$ will be an accumulation point of $S$, but $U\cap S$ will be finite. A simple example is to let $X=\{0,1\}$ with the topology whose open sets are $\varnothing,\{0\}$, and $X$: in this space $1$ is an accumulation point of $\{0\}$!

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Thank you for the very simple counter-example! Those are always the best ones. :) –  Kyle Schlitt May 15 '12 at 20:32

I found the (other) answer in another book: Introduction to Topological Manifolds, by John M. Lee.

Using a nested sequence open sets (which is guaranteed by first countability, so it is actually needed!) about the point $y$, we immediately get a sequence which converges to $y$.

Of course the actual question I asked was about the closed-ness of singleton sets, which was answered already by Brian Scott and GEdgar (thanks!).

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