Prove that, if S is open, each of its points is a point of accumulation of S.

How can I prove this??

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- Jan 20th 2011, 05:40 PMmathsohardPoint sets
Prove that, if S is open, each of its points is a point of accumulation of S.

How can I prove this?? - Jan 20th 2011, 05:54 PMDrexel28
- Jan 20th 2011, 07:57 PMmathsohard
This is not true??? hmm... I am using Advanced calculus by taylor and mann and on page 517 #5 they asked me to prove :(

- Jan 20th 2011, 08:05 PMDrexel28
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 $\displaystyle \mathbb{R}^n$. If so then this is true if $\displaystyle S$ is non-empty. Let $\displaystyle s\in S$ and $\displaystyle U$ any neighborhood of it, then since $\displaystyle S$ is open we know there exists a neighborhood $\displaystyle N$ of $\displaystyle s$ such that $\displaystyle N\subseteq S$. Note then that $\displaystyle N\cap S$ is a neighborhood of $\displaystyle s$ contained in $\displaystyle s$. Choose some open ball $\displaystyle s\in B_{\delta}(s)\subseteq S\cap N$ and use the uncountability of open balls in euclidean space to find a point different from $\displaystyle s$ in $\displaystyle B_{\delta}(s)$.

- Jan 21st 2011, 03:36 AMPlato