Properties of a dense set in a metric space

Question

Let $(M,d)$ be a metric space and $X \subset M$, with $X$ being a dense set in $M$.

a) Show that $d(p,X)=0$, $\forall p \in M$

b) Show that if $M$ is a limited set, then $\operatorname{diam}(M)=\operatorname{diam}(X)$

c) Let $(x_n) \subset M$ be a sequence. Proof that if $\exists p \in M$ that $\lim\limits_{n \to \infty} d(x_n,x)=d(p,x)$, $\forall x \in X$, then $x_n \to p$.

My thoughts for a):

Since $\forall p \in M$ and $\forall \varepsilon > 0$, we have $B(p,\varepsilon)\cap X \not= \emptyset$, let $\varepsilon = \frac{1}{n}, n \in \mathbb{N}$.

It follows that for any $p \in M$, we have an $x_n \in X$ that $d(p,x_n)< \frac{1}{n}$.

So $d(p,X) = \inf\bigl\{\frac{1}{n}, n \in \mathbb{N}\bigr\} = 0, \forall p \in M$. (right?)

For b) and c) I can't think anything to proceed. Any leads?

Answer 1

$diam(M)=max_{x,y\in M}d(x,y)$. For every $n>0$, there exists $x_n,y_n\in M$ such that $d(x_n,y_n)>diam(M)-{1\over{2n}}$. Since $X$ is dense, there exists $a_n,b_n\in X$ such that $d(a_n,x_n)<{1\over{4n}}$ and $d(b_n,y_n)<{1\over{4n}}$.

We have: $diam(M)-{1\over{2n}}<d(x_n,y_n)<d(x_n,a_n)+d(a_n,b_n)+d(b_n,y_n)<d(a_n,b_n)+{1\over{2n}}$, we deduce that:

$d(a_n,b_n)>diam(M)-{1\over n}$ and $diam(X)\geq diam(M)$.

For $c)$.

Let $x_n$ be a sequence, suppose that there exists $p$ such that $d(x_n,x)=d(x,p)$ for every $x\in X$. Since $X$ is dense, for every $m>0$, there exists $p_m\in X$ such that $d(p,p_m)<{1\over m}$. We have $d(x_n,p)\leq d(x_n,p_m)+d(p_m,p)=2d(p_m,p)<{2\over n}$ implies $x_n=p$. The sequence is a constant.