I know that the closure of a set is closed and the proof can be found in Rudin. However this is done by using the fact that the complement of a closed set is open. I am using a different book now and it is asked in the exercises to show that the closure of a set is closed. The only thing mentioned about closed sets so far in this book is that it contains all of its accumulation points.
The accumulation point $x$ of a set $E$ is such that $\forall c>0, (x-c,x+c)\cap {E}$ contains infinitely many points. Now I am trying to establish the above fact using only this definition of a closed set presented so far. However, I am stuck…
If $\bar{E}$ denotes closure, $\bar{E} = E \cup E'$ where $E'$ is the set of all accumulation points of $E$.
Then let $x$ be an accumulation point of $\bar{E}$. I want to show that $x \in \bar{E}$. Since $x$ is an accumulation point, $\forall c>0, (x-c,x+c)\cap {\bar{E}}$ contains infinitely many points.
But $(x-c,x+c)\cap {\bar{E}} = \{(x-c,x+c)\cap E\} \cup \{(x-c,x+c)\cap E'\}$. So either $\{(x-c,x+c)\cap E\}$, or $\{(x-c,x+c)\cap E'\}$, or both contains infinitely many points.
If $\{(x-c,x+c)\cap E\}$ contains infinitely many points, $x\in E' \subset \bar{E}$.
But I am now stuck with the other case left.
Any suggestions/ideas? Help appreciated!
Best Answer
You’re working a bit harder than necessary: it’s enough to show that $(x-c,x+c)\cap E\ne\varnothing$ for each $c>0$. Let $c>0$ be arbitrary. You know that $(x-c,x+c)\cap\operatorname{cl}E\ne\varnothing$, so pick any point $y\in(x-c,x+c)\cap\operatorname{cl}E$. If $y\in E$, you’re done. If not, then $y\in\operatorname{cl}E$. There is an $\epsilon>0$ such that $(y-\epsilon,y+\epsilon)\subseteq(x-c,x+c)$, and $(y-\epsilon,y+\epsilon)\cap E\ne\varnothing$, so ... ?
Added: Just in case it isn’t clear why it’s sufficient to show that $(x-c,c+c)\cap E\ne\varnothing$ for each $c>0$, observe that if this is the case, then necessarily each $(x-c,c+c)\cap E$ is infinite. If you’ve not seen this before, you should try to prove it.