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 Author: Riemann [ Wed Mar 01, 2017 2:47 pm ] Post subject: Metric space and distance Let $X$ be a metric space and $A$ be a non empty subspace of $X$. Prove that${\rm dist}(x, A) = 0$ if-f $x \in \overline{A}$.the function $f(x)={\rm dist}(x, A) \; , \;x \in A$ is continuous.${\rm dist}(x, A) = {\rm dist}(x, \overline{A})$.

 Author: Papapetros Vaggelis [ Wed Mar 01, 2017 5:28 pm ] Post subject: Re: Metric space and distance i. Suppose that $$\displaystyle{d(x,A)=\inf\,\left\{d(x,y)\geq 0\,,y\in A\right\}=0}$$.For every $$\displaystyle{n\in\mathbb{N}}$$, there exists $$\displaystyle{y_n\in A}$$ such that$$\displaystyle{d(x,y_n)<\dfrac{1}{n}}$$.The sequence $$\displaystyle{\left(y_n\right)_{n\in\mathbb{N}}\subseteq A}$$ satisfies the relation$$\displaystyle{d(x,y_n)<\dfrac{1}{n}\,,\forall\,n\in\mathbb{N}}$$, so $$\displaystyle{y_n\to x}$$and then $$\displaystyle{x\in \overline{A}}$$.On the other hand, suppose that $$\displaystyle{x\in \overline{A}}$$. Let $$\displaystyle{\epsilon>0}$$.Then, $$\displaystyle{B(x,\epsilon)\cap A\neq \varnothing}$$, so there exists $$\displaystyle{y\in X}$$such that $$\displaystyle{y\in A}$$ and $$\displaystyle{d(x,y)<\epsilon=0+\epsilon}$$.Since, $$\displaystyle{d(x,y)\geq 0\,,\forall\,y\in A}$$ and $$\displaystyle{\left(\forall\,\epsilon>0\right)\,\left(\exists\,y\in A\right)\,,d(x,y)<0+\epsilon}$$,we get $$\displaystyle{d(x,A)=\inf\,\left\{d(x,y)\geq 0\,,y\in A\right\}=0}$$.ii. Using the fact that $$\displaystyle{\left|d(x,A)-d(y,A)\right|\leq d(x,y)\,,\forall\,x\,y\in X}$$we have that $$\displaystyle{f}$$ is continuous.iii. If $$\displaystyle{y\in A}$$, then $$\displaystyle{y\in \overline{A}}$$ and then$$\displaystyle{d(x,y)\geq d(x,\overline{A})}$$, so $$\displaystyle{d(x,A)\geq d(x,\overline{A})}$$.Now, let $$\displaystyle{y\in \overline{A}}$$. There exists a sequence $$\displaystyle{\left(y_n\right)_{n\in\mathbb{N}}\subseteq A}$$such that $$\displaystyle{y_n\to y}$$.So, $$\displaystyle{d(x,y_n)\geq d(x,A)\,,\forall\,n\in\mathbb{N}}$$ and with limits,$$\displaystyle{\lim_{n\to \infty}d(x,y_n)\geq d(x,A)\iff d(x,y)\geq d(x,A)}$$.So, $$\displaystyle{d(x,y)\geq d(x,A)\,,\forall\,y\in \overline{A}\implies d(x,\overline{A})\geq d(x,A)}$$.Finally, $$\displaystyle{d(x,A)=d(x,\overline{A})}$$.

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