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 Post subject: Compact subsetPosted: Mon Jul 11, 2016 7:12 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
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 subsetPosted: Mon Jul 11, 2016 7:17 am
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Joined: Tue Nov 10, 2015 8:25 pm
Posts: 314
(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 subsetPosted: Mon Jul 11, 2016 7:18 am
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 457
Location: Ioannina, Greece
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*}
\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 subsetPosted: Mon Jul 11, 2016 7:19 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426

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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