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I was thinking about the following problem: For two disjoint compact subsets $A$ and $B$ of a metric space $(X,d)$ show that $d(A,B)>0.$ I'm having doubt with my attemp. Please have a look and do comment:

Let $d(A,B)=\inf\{d(a,b):a\in A,b\in B\}=0.$ Then $\exists$ sequences $\{a_n\}\subset A,\{b_n\}\subset B$ such that $0\le d(a_n,b_n)<\dfrac{1}{n}~\forall~n.$ Since $A$ is compact, $\exists$ a convergent subsequene $\{a_{r_n}\}$ of $\{a_n\}$ converging to some $a\in A.$ Then $0\le d(a_{r_n},b_{r_n})<\dfrac{1}{r_n}~\forall~n.$ Similarly since $B$ is compact, $\exists$ a convergent subsequene $\{b_{r_{n_m}}\}$ of $\{b_{r_n}\}$ converging to some $b\in B.$ Then $0\le d(a_{r_{n_m}},b_{r_{n_m}})$$<\dfrac{1}{r_{n_m}}~\forall~m.$ We note that $r_{n_m}\ge n_m\ge m>0$$\implies0<\dfrac{1}{r_{n_m}}<\dfrac{1}{m}\to0$$\implies\dfrac{1}{r_{n_m}}\to0.$ Using the sqeezing lemma once again we can see that $\exists$ convergent sequences $\{a_n\}\subset A$ and $\{b_n\}\subset B$ such that $a_n\to a\in A,b_n\to b\in B.$ $\exists$ convergent sequences $\{a_n\}\subset A$ and $\{b_n\}\subset B$ such that $a_n\to a\in A,b_n\to b\in B.$ Then $d(a_n,b_n)\to d(a,b)$ Consequently, $d(a,b)=0\implies a=b,$ a contradiction to $A\cap B=\emptyset.$

Sriti Mallick
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  • This looks valid. You are using the fact that $(x,y)\mapsto d(x,y)$ is a continuous map. – ncmathsadist May 14 '13 at 13:39
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    if $d(A,B)=0$, there's a sequence $(a_n,b_n)$ with $d(a_n,b_n)\to 0$. Since $A\times B$ is compact $d(A\times B)$ is compact & so closed. So there is $(a,b)\in A\times B$ with $d(a,b)=0$. –  May 14 '13 at 13:39
  • Yeah, that last step also uses nondegeneracy of $d$, but looks great. – danzibr May 14 '13 at 14:08
  • @Norbert I don't think that this is a duplicate since the asker is asking for comments on a particular attempt at the question that they have made. – Tom Oldfield May 14 '13 at 14:55

1 Answers1

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Your solution is correct. Here is another one.

Assume $d(A,B)=0$. Consider function $d(x, B)=\inf\{d(x,y):y\in B\}$ It is continuous, and defined on compact metric space $A$. So it attains its minimum, i.e. there exist $a\in A$ such that $d(a,B)=\inf\{d(x,B):x\in A\}=d(A,B)=0$. Since $d(a,B)=0$, then $a\in \operatorname{cl}_X(B)$. Since $B$ is compact it is closed, so $a\in \operatorname{cl}_X(B)=B$. Contradiction, since $A\cap B=\varnothing$.

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Norbert
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  • Thanks. I didn't show the proof that $d:X\times X\to\mathbb R$ is continuous in my attemps. Here it's: Let $d'$ be the product metric on $X\times X$ i.e. $d'((a_1,b_1),(a_2,b_2))=\max{d(a_1,a_2),d(b_1,b_2)}$. Choose $(a,b)\in X\times X$ and $\epsilon>0.$ We are to find a $\delta>0$ such that $(x,y)\in B_{d'}((a,b),\delta)$$\implies|d(x,y)-d(a,b)|<\epsilon.$ Equivalently, $\max{d(x,y),d(a,b)}<\delta$$\implies|d(x,y)-d(a,b)|<\epsilon.$ Isn't choosing $\delta=\epsilon$ serve the purpose? – Sriti Mallick May 14 '13 at 15:37
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    @SritiMallick No this proof inncorrect. if $(x,y)\in B_{d'}((a,b),\delta)$, then $d(x,a)<\delta$, $d(y,b)<delta$. Now from triangle inequality we have $d(x,y)<d(x,a)+d(a,y)<d(a,x)+d(a,b)+d(b,y)<2\delta+d(a,b)$, so $d(x,y)-d(a,b)<2\delta$. Similarly $d(a,b)-d(x,y)<2\delta$, hence $|d(x,y)|-d(a,b)|<2\delta$. Hence for a given $\varepsilon>0$ we need to set $\delta=\varepsilon/2$ – Norbert May 14 '13 at 15:51