mathimatikoi.org
https://www.mathimatikoi.org/forum/

Metric space and distance
https://www.mathimatikoi.org/forum/viewtopic.php?f=15&t=1128
Page 1 of 1

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

  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. ${\rm dist}(x, A) = {\rm dist}(x, \overline{A})$.

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, \(\displaystyle{d(x,A)=d(x,\overline{A})}\).

Page 1 of 1 All times are UTC [ DST ]
Powered by phpBB® Forum Software © phpBB Group
https://www.phpbb.com/