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 Post subject: Compact subset
PostPosted: Mon Jul 11, 2016 7:12 am 
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Prove that \(\displaystyle{S=\left\{x\in\mathbb{R}^{n}: \|x\|=1\right\}}\), where
\(\displaystyle{\|\cdot\|:\mathbb{R}^{n}\longrightarrow \mathbb{R}\,,(x_1,...,x_n)=x\mapsto \|x\|=\sqrt{x_1^2+...+x_n^2}\,\,\,,n\in\mathbb{N}\,\,,n\geq 2}\, ,\)
is a compact subset of \(\displaystyle{\left(\mathbb{R}^{n},\mathbb{T}\right)}\) , \(\displaystyle{\mathbb{T}}\) : is the usual topology of \(\displaystyle{\mathbb{R}^{n}}\) .


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 Post subject: Re: Compact subset
PostPosted: Mon Jul 11, 2016 7:17 am 
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(i) \( \displaystyle S \) is a bounded subset of \( \mathbb{R}^{n} \), since \( \displaystyle S \subset B(0,2) \) , where \( \displaystyle B(0,2) = \left\{ x \in \mathbb{R}^{n} \Big| ||x|| < 2\right\} \)


(ii) \( \displaystyle S \) is a closed subset of \( \mathbb{R}^{n} \) : Let \( \displaystyle (s_{k})_{k\in\mathbb{N}} \) be a sequence of elements of \( \displaystyle S \) which converges to an \( \displaystyle s \in \mathbb{R}^{n} \). Also, suppose that \[ \displaystyle s_{k} = \left( s_{1}^{(k)} , \dots , s_{n}^{(k)} \right) \] and \[ \displaystyle s = \left( s_{1} , \dots , s_{n} \right) \] Since \( \displaystyle s_{k} \longrightarrow s \) , we have that \[ \displaystyle s_{i}^{(k)} \longrightarrow s_{i} , \forall i \in \left\{ 1 , \dots , n \right\} \; \; \; (*) \]
Since for all \( \displaystyle k \in \mathbb{N} \) holds \( \displaystyle s_{k} \in S \), we have that
\[ \displaystyle || s_{k} || = 1 , \; \forall k \in \mathbb{N} \\\implies
|| s_{k} ||^{2} = 1 , \; \forall k \in \mathbb{N} \\ \implies
\left( s_{1}^{(k)} \right)^{2} + \dots + \left( s_{n}^{(k)} \right)^{2} = 1 , \; \forall k \in \mathbb{N} \\\overset{ (*) }{\implies}
\left( s_{1} \right)^{2} + \dots + \left( s_{n} \right)^{2} = 1 \\\implies
|| s ||^{2} = 1 \\ \implies
|| s || = 1 \] This means that \( \displaystyle s \in S \). Thus, we have proved that S is indeed closed, as we claimed in the beginning.

Therefore, \( \displaystyle S \) is a compact subset of \( \mathbb{R}^{n} \).


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 Post subject: Re: Compact subset
PostPosted: Mon Jul 11, 2016 7:18 am 
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A 2nd solution:

To prove that \[S=\bigl\{{\overline{x}\in\mathbb{R}^n\;:\; \|{\overline{x}}\|=1}\bigr\}\] is compact it is enough to prove that \(S\) is closed and bounded subset of \(\mathbb{R}^n\).

\(\color{gray}\bullet\) To prove that \(S\) is closed, it is enough to prove that every accumulation point of \(S\) it is in \(S\). Suppose that there exists a accumulation point \(\overline{a}\notin S\). Then exists a non-constant sequence \(\bigl\{{\overline{x}_{k}}\bigr\}_{k=1}^{\infty}\in S\) which converges to \(\overline{a}\). So for \(\varepsilon=\frac{|{1-\|{\overline{a}}\|}|}{2}>0\) we must have \begin{align*}
(\exists k_0\in{\mathbb{N}})(\forall k\in{\mathbb{N}})&\quad k\geqslant k_0\quad\Rightarrow\\
&\|{\overline{x}_k-\overline{a}}\|\leqslant\varepsilon=\frac{|{1-\|{\overline{a}}\|}|}{2}\quad (1).
\end{align*} But for every \(k\in{\mathbb{N}}\) we have that \[\bigl|{1-\|{\overline{a}}\|}\bigr|=\bigl|{\|{\overline{x}_k}\|-\|{\overline{a}}\|}\bigr|\leqslant\|{\overline{x}_k-\overline{a}}\|\quad(2)\,.\] From \((1)\) and \((2)\) we have
\[\bigl|{1-\|{\overline{a}}\|}\bigr|\leqslant\|{\overline{x}_k-\overline{a}}\|<\frac{|{1-\|{\overline{a}}\|}|}{2}\,,\] a contradiction. So \(S\) is closed.

\(\color{gray}\bullet\) Because for every \(\overline{x},\overline{y}\in S\) holds \[{\rm{d}}(\overline{x},\overline{y})=\|{\overline{x}-\overline{y}}\|\leqslant\|{\overline{x}}\|+\|{\overline{y}}\|\leqslant1+1=2\,,\] \(S\) is bounded.

Finally \(S\) is compact subset of \(\mathbb{R}^n\).

_________________
Grigorios Kostakos


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 Post subject: Re: Compact subset
PostPosted: Mon Jul 11, 2016 7:19 am 
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Thank you for your solutions.

Here is another proof about the closure of \(\displaystyle{S}\) .

We define \(\displaystyle{f:\left(\mathbb{R}^{n},\mathbb{T}\right)\longrightarrow \left(\mathbb{R},\mathbb{K}\right)}\) by \(\displaystyle{f(x)=||x||}\) ,

where \(\displaystyle{\mathbb{K}}\) is the usual topology of \(\displaystyle{\mathbb{R}}\) .

Let \(\displaystyle{x\in\mathbb{R}^{n}}\) . For each \(\displaystyle{y\in\mathbb{R}^{n}}\) holds :

\(\displaystyle{\left|f(y)-f(x)\right|=\left|||y||-||x||\right|\leq ||y-x||\,\,(1)}\) , so :

\(\displaystyle{\left(\forall\,\epsilon>0\right)\,\left(\exists\,\delta=\epsilon>0\right)\,\left(\forall\,y\in\mathbb{R}^{n}\right): ||y-x||<\delta\implies \left|f(y)-f(x)\right|\stackrel{(1)}{\leq} ||y-x||<\delta=\epsilon}\) ,

which means that the function \(\displaystyle{f}\) is continuous at \(\displaystyle{x}\) .

Therefore, the function \(\displaystyle{f}\) is continuous. We observe that :

\(\displaystyle{S=\left\{x\in\mathbb{R}^{n}: ||x||=1\right\}=\left\{x\in\mathbb{R}^{n}: f(x)=1\right\}=f^{-1}\,\left(\left\{1\right\}\right)}\), where

\(\displaystyle{\left\{1\right\}}\) is a closed subset of \(\displaystyle{\left(\mathbb{R},\mathbb{K}\right)}\) .

Due to the fact that the function \(\displaystyle{f}\) is continuous, we have that \(\displaystyle{S=f^{-1}\,\left(\left\{1\right\}\right)}\)

is closed subset of \(\displaystyle{\left(\mathbb{R}^{n},\mathbb{T}\right)}\) .


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