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## Metric space and distance

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Riemann
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### Metric space and distance

Let $X$ be a metric space and $A$ be a non empty subspace of $X$. Prove that
1. ${\rm dist}(x, A) = 0$ if-f $x \in \overline{A}$.
2. the function $f(x)={\rm dist}(x, A) \; , \;x \in A$ is continuous.
3. ${\rm dist}(x, A) = {\rm dist}(x, \overline{A})$.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Papapetros Vaggelis
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### 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, $\displaystyle{d(x,A)=d(x,\overline{A})}$.