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 Post subject: Compact
PostPosted: 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: Compact
PostPosted: Sun May 29, 2016 10:29 pm 

Joined: Mon Jan 18, 2016 4:33 pm
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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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