I know that this question was asked so many times here, but I have problem in perticular step of first part of the proof!
My attempt:
to prove $(\bar E)'\subset E'$:
let $x\in (\bar E)'$ then $x$ is limit point of $\bar E$. So that every neighborhood of $x$ contains a point of $\bar E$ other than $x$. So that there exists neighborhood say $N_r(x)$ of $x$ such that $y≠x\in N_r(x)$ where $y$ is some point of $\bar E$. So that, $0<d(x,y)<r$.
Now we need to show $x\in E'$. i.e. $x$ is limit point of $E$. i.e. we need to show every neighborhood of $x$ contains some point of $E$ different from $x$. i.e. for some $r>0$ there exists neighborhood $M_r(x)$ such that $z≠x\in M_r(x)$ where $z$ is some point of $E$. Hence we need to show for some arbitrary $r>0$, we have $0<d(x,z)<r$…….*
Clearly from first paragraph of proof, if $y\in E$ we can take $z=y$ and we are done. If $y\in E'$ then it is limit point of $E$. Hence for every $s>0$ there exists neighborhood $K_s(y)$ such that $z≠y\in K_s(y)$ where $z$ is some point of $E$. So that $0<d(z,y)<s$. Let $s=r-d(x,y)>0$. Hence we have by traingle inequality,
$d(x,z)≤d(x,y)+d(y,z)<d(x,y)+s<r$
Hence we had proved * so that $x\in E'$. Hence $(\bar E)'\subset E'$ is proved.
Is above proof is fine? I had seen proof in which the distance $s$ is taken as $s=min(d(x,y), r-d(x,y))$ I don't know why? and isn't is equal to $r-d(x,y)$
Please help….
Best Answer
The set-up of the proof should be:
Let $p \in \overline{E}'$, we want to show that $p \in E'$ so take an arbitary $r$-neighbourhood $N_r(x)$ of $x$ with $r>0$ and we need to find $q \neq p$ such that $d(p,q) < r$ and $q \in E$. (This is then a point of $N_r(p) \cap E \setminus \{p\}$ as required)
To do this we start start with $N_r(p)$ and as $p$ is a limit point of $\overline{E}$ we find some $p' \in \overline{E}$ such that $p' \neq p$ and $d(p,p') < r$ (or $p' \in N_r(p)$). Now, as $N_r(p)$ is itself open (and $p'$ lies in it) and moreover $X\setminus\{p\}$ is open and so $N_r(p)\setminus\{p\}$ is also open, we can find $s>0$ such that $N_s(p') \subseteq N_r(p)\setminus\{p\}$. Now $p' \in \overline{E}$ so there is some $q \in N_s(p') \cap E$ and this $q$ is as required from above: $q \in N_s(p')$ implies $q \neq p$ and $q \in N_r(p)$ and we already knew $q \in E$. QED
The metric is actually quite irrelevant and the only metric-related fact I used was that $\{p\}$ is closed. The $s$ I used to stay inside $N_r(p)$ can be taken as $r- d(p',p)$ (follows from the triangle inequality) and we only need to ensure it's also smaller than $d(p',p)$ to let $p$ be outside it, so take $s < \min(r-d(p',p),d(p',p))$ which is easy to do. This relates to the last part of the question. Long story short, $\overline{E}' \subseteq E'$ holds in any topological space where singletons are closed ($T_1$ spaces), and the reverse is trivial from $E \subseteq \overline{E}$ anyway.