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I need to verify something with my proof. Let $A$ be a subset of some metric space $(X,d)$. Take $x\in X$. Consider the following propositions.

(1) $d(x,A)=0 \implies x\in A$

(2) If $\{x_n \}\subset A$ and $x_n \rightarrow x$ then $x\in A$

The claim is that they are equivalent. I started by proving that $(1) \implies (2)$.

Proof: "$\implies$"

Suppose that $(1)$ holds. Now, let $\{x_n\}\subset A$ and $x_n \rightarrow x$. Since $x_n \rightarrow x$, we know that $d(x,x_n) \rightarrow 0$.

(3) thus $\inf\{d(x,x_n):x_n \in A\}=0$.

(4) So we have that $0=\mathrm{dist}(x,A)$. Which means that $x\in A$ by $(1)$

Is it understandable how I obtained $(3)$?

I just realised that (4) might not follow from (3), as A might contain points which are not in $\{x_n\}$ or does that not matter since the infimum in this case can never be less than 0 (the metric always returns non-negative values)?

Can someone please provide hint on how to continue if I'm correct till I get to (4). Note, I don't want a complete solution but rather commentary on my proof and perhaps a hint (on how to continue).

Michael Hardy
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Adeeb
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1 Answers1

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Your proof looks fine to me. You are saying that $x_n \to x$ and that all the $x_n$ are in $A$. So for sure $d(A,x) = \inf\{d(a,x)\mid a\in A\}$ will be zero (I would not that that $d(A,x) = \{d({\color{red}{x_n},x})\mid \color{red}{x_n}\in A\}$. I would replace the $\color{red}{x_n}$ by $a$ because you take the infimum over all of $A$). That is, you can make $d(a,x)$ as small as you want by picking suitable $x_n$ from $A$.

For $(2) \Rightarrow (1)$. So you assume that $\inf\{d(a,x)\mid a\in A\} = 0$. Let $x_n\in A$ be such that $d(x_n,x) < \frac{1}{n}$. Then you get a sequence $\{x_n\}\subseteq A$ such that $x_n \to x$. Then by (2) you will have that $x\in A$.

Thomas
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