Convergence of a sequence of distances from a set

Question

Let $(X,d)$ be a metric space, $A \subset X$ any subset. Suppose that $x_n$ is a sequence of points in $A^c$ that does not converge to any element of $A$.

1) Is it possible that the sequence $d(x_n,A)$ can converge to $0$ ?

2) If the answer to 1) is yes, are there conditions we can put on $X$ and/or $A$ to ensure that this cannot happen?

Many thanks for any help.

Answer 1

If $A$ has two distinct boundary points, say $p$ and $q$, we can find sequences $x_n$, $y_n$ from $X\setminus A$ such that $x_n \to p$, $y_n \to q$, and then the combined sequence $x_n$ with $z_{2n} = p_n, z_{2n+1} = q_n$, so alternating between the two sequences, will have the property that it does not converge in $X$ (or else all subsequences would have the same limit) but $d(z_n, A) \to 0$.

So wanting this property to hold for a set $A$ forces $|\partial A| \le 1$.

Answer 2

A necessary condition is that $A$ is closed. If $A$ does not include a point $x$ in its closure, then the sequence $x_n=x$ converges to $x$ while $d(x,A)=0$.

A sufficient condition is that $A$ is compact. Proof: if $d(x_i,A) \rightarrow 0$, we show that every subsequence of $x_i$ has a convergent subsequence to an element of $A$ which is sufficient. Choose any subsequence of $x_i$ and denote it by $x_i$ again for simplicity. For each $i$, find $y_i \in A$ such that $d(x_i,y_i)<2 d(x_i,A) \rightarrow 0$. Find a convergent subsequence of $y_i$ such that $y_{i_k} \rightarrow y\in A$. Then $x_{i_k} \rightarrow y \in A$.

Another sufficient condition is that $A$ is isolated i.e., there exists $c>0$ such that $d(x,A)>c$ for all $x\notin A$.

Answer 3

EDIT: i misread the question so i didn't considered the extra hypothesis $x_n \nrightarrow a \in A$

Proposition: Given a metric space $(X,d)$ and $\varnothing \subset A \subset X$ then $d(A,X\setminus A) > 0$ if and only if there is no sequence $x_n \in X \setminus A$ such that $d(x_n,A) \to 0$

$(\Rightarrow)$ Given a sequence $x_n$ on $X \setminus A$ we have that $d(A,x_n) \geq d(A, X \setminus A) > 0$ thus the distance does not go to zero

$(\Leftarrow)$ If by contradition $d(A,X \setminus A) = 0$ by definition for all $\varepsilon \in \mathbb{R}^+$ there is a couple of points $x \in X \setminus A$, $y \in A$ such that $$ d(x,y) < \varepsilon $$ Hence for all $n \in \mathbb{N}$ there exists $x_n \in X \setminus A$ and $y_n \in A$ such that $$ d(x_n,A) \leq d(x_n, y_n) < \frac{1}{n} $$ in particular $d(x_n, A) \to 0$, that is a contraddition.

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Now, even if this may seem a tautological equivalent formulation, it tells us a few things. For example $A$ need to be proper clopen (in particular the space need to be disconnected).

One tricky point that fooled me at first is that $A$ being a clopen is also sufficient: That's not true due to the following. Let $$ X = \mathbb{N} \cup \left\{ n + \frac{1}{n} \Bigg| \quad \forall n \in \mathbb{N} \right\} $$ And consider $A = \mathbb{N}$. Now clearly the sequence $\displaystyle x_n = n + \frac{1}{n+1}$ is in $X \setminus A$ but $$ d(x_n,A) = \frac{1}{n+1} \to 0 $$