About distance between a set and element

Question

Ploblem:

For every nonempty set $A$ of a metric space $M$, define $A^*=\{x∈M: d(x,A)=0\}$. Prove that $(A^*)^* = A^*$.

Now by definition, the distance between $x$ and $A$ is $d(x,A)=\inf\{d(a,x):x∈A\}$. My question is how are the elements in $\inf\{d(a,x):x∈A\}=0$ ? how can I take an element of this set?

Note that $A \subset A^$ to start. This gives $A^ \subset (A^)^$ 'for free'. — copper.hat, Dec 20 '16 at 22:33

Guest · Accepted Answer · 2016-12-20T23:19:13.340

1

Lemma: $A^*$ is closed in $M$.

Proof: We show that $(A^*)^c$ is open in $M$. We have $$ (A^*)^c=\{x\in M:d(x,A)>0\} $$ Let $x\in(A^*)^c$. Then $\gamma:=d(x,A)>0$ and $B_M(x,\gamma/2)\subset(A^*)^c$. Indeed, for all $y\in B_M(x,\gamma/2)$ and for all $a\in A$, we have \begin{align} d(y,a)&\geq d(x,a)-d(x,y)\\ &>\gamma-\frac{\gamma}{2}\\ &=\frac{\gamma}{2} \end{align} hence $d(y,A)>\gamma/2>0$.

Proof of $\subset$: Let $x\in(A^*)^*$. Then $d(x,A^*)=0$ so there exists a sequence $(a_n)\subset A^*$ converging to $x$. Since $A^*$ is closed, $\lim a_n=x\in A^*$.

edited Dec 20 '16 at 23:19

answered Dec 20 '16 at 23:05

Guest

4,158

Thank you 'Guest', really appreciate it :) – Jacob S. Dec 20 '16 at 23:16
@Jacob Schwartz You are welcome. – Guest Dec 20 '16 at 23:20
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@Jacob Schwartz Alternatively, if you already know that the function $x\mapsto d(x,A)$ is continuous, then the Lemma is a one-liner since $A^*$ is the preimage of the closed subset ${0}$ of $\mathbb{R}$. – Guest Dec 20 '16 at 23:28

About distance between a set and element

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