Do convergence under a metric implies convergence under another metric in the same space?

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Let $(E,d)$ be a metric space and consider the subsets $A,B$ where $A$ is compact and $B$ is closed. Suppose $dist(A,B):=inf_{x\in A, y\in B} d(x,y)=0$. Then I found that there are sequence $\{a_n\}\subseteq A$ and $\{b_n\}\subseteq B$ such that $0\leq \mid a_n - b_n \mid <1/n, \forall n\in \mathbb{N}$. This inequality consider the metric $\mid \cdot \mid$ which is the usual metric in $\mathbb{R}$ ($d:E\times E \rightarrow \mathbb{R}_+ $ in my definitions and, as far as I understand, these $\{a_n\}, \{b_n\}$ are real sequences).

Well, since $A$ is compact, it is sequentially compact, and there is a subsequence such that $a_{n_k} \rightarrow a\in A$. I want to show that $d(a,b_{n_k})\leq d(a,a_{n_k})+d(a_{n_k},b_{n_k}) \rightarrow 0$ as $k\rightarrow 0$. It is clear to me that $d(a,a_{n_k}) \rightarrow 0$. But what about $d(a_{n_k},b_{n_k})$?

My aim here is to show $\exists a\in A: dist(\{a\},B)=0 $ given $dist(A,B)=0$ without using the continuity of $inf$ function.

I dont't know how to proceed since $d$ is not necessarily the usual metric. So I have a sequence of real numbers in $A$ which converges to zero under the usual metric, but I want to show that they converge under the metric $d$ in $(E,d)$.

Answer 1

Convergence (as compactness) depend on the metric, not on the space. So, in somehow your question doesn't makes sense since the fact that $A$ is compact in $(E,|\cdot |)$ doesn't implies that $A$ compact in $(E,d)$.

The only metric spaces that has same convergent sequence are metric induced by normes. Otherwise is not true. For example, take $$d(x,y)=\begin{cases}1&x\neq y\\ 0&x=y\end{cases}.$$ Clearly $(\frac{1}{n})_{n\in\mathbb N^*}$ converges for $|\cdot |$ to $0$ but doesn't converges for $d$.

Answer 2

Let $\{a_n\}, \{b_n\}$ be sequences in $A,B$ respectively, with $d(a_n,b_n)<1/n.$ Since $A$ is compact, $\{a_n\}$ has a convergent sub-sequence $\{a_{f(n)}\}$ converging to $a\in A$, where $f:\Bbb N\to \Bbb N$ is strictly increasing. Then $f(n)\ge n$ so we have $$d(a,b_{f(n)})\le d(a,a_{f(n)})+d(a_{f(n)},b_{f(n)})<$$ $$<d(a,a_{f(n)})+1/f(n)\le$$ $$\le d(a,a_{f(n)})+1/n.$$ And $\lim_{n\to \infty} d(a,a_{f(n)})=0\,$ so $$\lim_{n\to \infty}d(a, b_{f(n)})=0$$ so $a\in \overline B=B$ so $a\in A\cap B.$

Remark: If $A, B$ are not empty, with $A$ compact and $B$ closed but not compact, and $dist(A,B)>0,$ then there need not exist $a\in A$ and $b\in B$ such that $dist (A,B)=d(a,b)$ even if $A$ has only one member.

Example: Let $E=\{p\}\cup \Bbb R$ with $p\not \in \Bbb R.$

For $x,y\in \Bbb R$ let $d(x,y)=\frac {|x-y|}{1+|x-y|}.$

For $x\in \Bbb R$ let $d(x,p)=2-d(x,0).$

Now let $A=\{p\}$ and $B=\Bbb R.$ We have $dist (A,B)=dist (p,B)=1$ but $d(p,x)>1$ for all $x\in B.$

Answer 3

Without using sequences.

If $A\cap B\ne \emptyset$ then we are done. So we show that if $A\cap B=\emptyset$ and $A, B$ are not empty then $d(A,B)>0.$

Suppose $A,B$ are disjoint. Let $$C=\{B_d(a,r/2): a\in A\land r\in \Bbb R^+\land B_d(a,r)\cap B=\emptyset\}.$$ Every $a\in A$ is the center of some member of $C$ because no $a\in A$ belongs to $\overline B=B.$ So $C$ is an open cover of $A.$ Since $A$ is compact and non-empty there exists a finite $$C^*=\{B_d(a_1,r_1/2),...,B_d(a_n,r_n/2)\}\subset C$$ such that $A\subset \cup C^*.$

Let $s=\min (r_1/2,...,r_n/2).$ Then (obviously) $s>0.$

Take any $a\in A$ and take some $j$ such that $a\in B_d(a_j,r_j/2)\in C^*.$ Now no $b\in B$ belongs to $B_d(a_j,r_j),$ so for all $b\in B$ we have $$d(b,a)\ge d(b,a_j)-d(a_j,a)\ge r_j-d(a_j,a)>r_j-r_j/2=r_j/2\ge s.$$ So $dist(A,B)\ge s>0.$