Point sets

**mathsohard** · January 20th 2011, 05:40 PM

Prove that, if S is open, each of its points is a point of accumulation of S.
How can I prove this??

**Drexel28** · January 20th 2011, 05:54 PM

Originally Posted by mathsohard

Prove that, if S is open, each of its points is a point of accumulation of S.
How can I prove this??

This is not true. Take any discrete space, then any singleton is open yet not a limit point.

**mathsohard** · January 20th 2011, 07:57 PM

This is not true??? hmm... I am using Advanced calculus by taylor and mann and on page 517 #5 they asked me to prove

**Drexel28** · January 20th 2011, 08:05 PM

Originally Posted by mathsohard

This is not true??? hmm... I am using Advanced calculus by taylor and mann and on page 517 #5 they asked me to prove

It's clearly untrue in arbitrary spaces, but since this is an advanced calculus book I'm going to take a wild guess and assume you meant and open subset of $\mathbb{R}^n$ . If so then this is true if $S$ is non-empty. Let $s\in S$ and $U$ any neighborhood of it, then since $S$ is open we know there exists a neighborhood $N$ of $s$ such that $N\subseteq S$ . Note then that $N\cap S$ is a neighborhood of $s$ contained in $s$ . Choose some open ball $s\in B_{\delta}(s)\subseteq S\cap N$ and use the uncountability of open balls in euclidean space to find a point different from $s$ in $B_{\delta}(s)$ .

**Plato** · January 21st 2011, 03:36 AM

Originally Posted by mathsohard

This is not true??? hmm... I am using Advanced calculus by taylor and mann and on page 517 #5 they asked me to prove

@mathsohard, you should give us the total context of a question.
I just looked up that question. Not only is it not a general top-space it is a question about subsets of the real number line.

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