Given $E=\{ p_n : n \in \mathbb{N}$ and $lim_{n\rightarrow \infty} p_n =p\}$. Prove $Cl(E)=E \cup\{p\}$ and $Cl(E)$ is compact

Question

Where $p_n \rightarrow p$. I'm trying to prove that for $E=\{ p_n : n \in \mathbb{N}$ and $lim_{n\rightarrow \infty} p_n =p \}$, then $Cl(E)=E \cup \{p \}$ and $Cl(E)$ is compact.

Also, I'm currently using the definition of limit points as p is a limit point if $\forall r>0, (E \cap N_r(p)) \backslash \{p\} \neq \emptyset$.

Here's a rough outline for what I have:

Proving $Cl(E)=E \cup \{p \}$:

By a theorem, we konw that $\{ p_n \}$ converges to $p \in E$ iff every neighborhood of p contains $p_n$ for all but finitely many n. So I'm thinking this theorem shows that $\forall r>0, (E \cap N_r(p)) \backslash \{p\} \neq \emptyset$.However, I'm not sure I'm supposed to prove that p is the only limit point.

Proving compact

I'm also still unsure how to prove $Cl(E)$ is compact just by the definition. Since $d(p_n, p) < \epsilon$, does this imply that all open covers $\mathcal{G}$ is somehow bounded by this and we therefore have finite subcovers?

Answer 1

If $\lim_{n\to \infty}p_n=p$ then every nbhd of $p$ contains $p_n$ for all but finitely many $n$ (regardless of whether or not $p\in E=\{p_n: n\in \mathbb N\}.$)

If $q \not \in E\cup \{p\}$ then there are open sets $U, V$ with $p\in U$ and $q\in V$ and $U\cap V=\emptyset.$ Now $$\{n\in \mathbb N: p_n\in V\}\subset \{n\in \mathbb N:p_n\not \in U\},$$ and the RHS, above, is a finite set. So there exists $n_0\in \mathbb N$ such that $$V\cap E\subset \{p_n:n\leq n_0\}.$$ And since $q\not \in E$ we have $q\not \in \{p_n:n\leq n_0\}.$ So $r,$ the least distance from $q$ to any member of $\{p_n:n\leq n_0\},$ is positive.

So $N_r(q)\cap E$ is empty. Hence $q\not \in \bar E.$ And $q$ is not a limit point of $E$.

Answer 2

I'm supposing $p$ is a fixed point... Recall that $Cl(E)=E\cup E'$, where $E'$ is the set of all limit points of $E$, that is $$E'=\{ x:\exists \, \{ x_n\}\subset E, \, \lim_{n\to \infty}x_n=x \}$$ (Where $\{ x_n\}\subset E$ is a sequence of distinct points.) All you have to prove is that $p$ is the only limit point of $E$.

Now, if you want to prove that $Cl(E)$ is compact directly from the definition of a compact set... Let $\{ B_n\}_{n\in \Bbb{N}}$ be an open cover of $Cl(E)$. Since $\lim_{n\to \infty}p_n=p$, for $\epsilon =1$ there is an $N\in \Bbb{N}$ such that... I'll leave the rest of the proof to you... And don't forget to use the fact that $\{ B_n\}_{n\in \Bbb{N}}$ is an open cover of $Cl(E)$.

Answer 3

If $\lim_{n\to \infty}p_n=p$ then every nbhd of $p$ contains $p_n$ for all but finitely many $n$ (regardless of whether or not $p\in E=\{p_n: n\in \mathbb N\}.$)

If $q \not \in E\cup \{p\}$ then there are open sets $U, V$ with $p\in U$ and $q\in V$ and $U\cap V=\emptyset.$ Now $$\{n\in \mathbb N: p_n\in V\}\subset \{n\in \mathbb N:p_n\not \in U\},$$ which is a finite set. So there exists $n_0\in \mathbb N$ such that $$V\cap E\subset \{p_n:n\leq n_0\}.$$ And since $q\not \in E$ we have $q\not \in \{p_n:n\leq n_0\}.$ So $r,$ the least distance from $q$ to any member of $\{p_n:n\leq n_0\},$ is positive.

So $N_r(q)\cap E$ is empty. Hence $q\not \in \bar E.$ And $q$ is not a limit point of $E$.