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 \(\displaystyle{\leftd(x,A)d(y,A)\right\leq d(x,y)\,,\forall\,x\,y\in X}\)
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, \(\displaystyle{d(x,A)=d(x,\overline{A})}\).
