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 Post subject: CompactPosted: Sun May 29, 2016 10:28 pm
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Joined: Mon Nov 09, 2015 1:52 pm
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Let $\displaystyle{\left(X,\mathbb{T}\right)}$ be a topological space. If $\displaystyle{\left(x_{n}\right)_{n\in\mathbb{N}}}$ is

a sequence of elements of $\displaystyle{X}$ such that $\displaystyle{x_{n}\longrightarrow x\,,n\longrightarrow +\infty}$, then prove

that the set $\displaystyle{A=\left\{x_{n}:n\in\mathbb{N}\right\}\cup\left\{x\right\}\subseteq X}$ is a compact subset of $\displaystyle{\left(X,\mathbb{T}\right)}$ .

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 Post subject: Re: CompactPosted: Sun May 29, 2016 10:29 pm

Joined: Mon Jan 18, 2016 4:33 pm
Posts: 4
Hello!

Let $A=\{x_n,x\}\subset X$ and $\{V_i\}_{i\in I}$ an open cover of $A$,i.e. $V_i\in\mathbb T\quad\forall i\in I$ and $A\subset\bigcup_{i\in I}V_i$.

It follows that $\forall n\in\mathbb N\quad\exists i_n\in I$ such that $x_n\in V_{i_n}$ and also $\exists i_x\in I$ such that $x\in V_{i_x}$.

The previous means that $A\subset(\bigcup_{n\in\mathbb N} V_{i_n})\cup V_{i_x}$.

Since $V_{i_x}$ is open it contains a neighborhood $U_x$ of x,and from the convergence of ${x_n}$ we have that $x_n\in U_x\quad\forall n\geq N$ for some $N\in\mathbb N$.

It follows that:$A\subset(\bigcup_{n=0}^{n=N} V_{i_n})\cup U_x\subset(\bigcup_{n=0}^{n=N} V_{i_n})\cup V_{i_x}$ which means that $\{V_i\}_{i\in I}$ has a finite subcover.

By definition we deduce that $A$ is a compact subset of $(X,\mathbb T)$

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