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Let $A,B \subseteq \mathbb{R}^d$ be non-empty sets. Define their distance to be

$$ d(A,B) = \inf \{ \|x-y\| : x \in A, \; \; y \in B \} $$

For any $A,B$, I want to prove that $d(A,B) = d( \overline{A}, \overline{B} )$.

My Attempt

Put $\alpha = d(A,B)$. Therefore, $\alpha = \|x_0 - y_0 \|$ for some $x_0 \in A$ and some $y_0 \in B$. But, notice $x_0 \in \overline{A}$ and $y_0 \in \overline{B}$ by definition. hence

$$ \| x_0 - y_0\| \geq \inf\{ \|x' - y'\| : x' \in \overline{A}, \; \; y' \in \overline{B} \} = d( \overline{A}, \overline{B} )$$

$$\therefore d(A,B) \geq d( \overline{A}, \overline{B} )$$

I am stuck trying to show the other direction: $d(A,B) \leq d( \overline{A}, \overline{B} )$

Can someone help me? thanks a lot

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    $\begingroup$ You must not assume $\alpha$ is attained. This you can only do if the closure of at least one of the sets is compact. $\endgroup$ – Thomas Feb 15 '14 at 10:51
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    $\begingroup$ 'Looking for an answer drawing from credible and/or official sources.' is like asking 'I must have a proof from a textbook that $\sin x e^x x^7$ is differentiable at the point $\pi$', when you could have done it by hand yourself? $\endgroup$ – Lost1 Feb 17 '14 at 13:09
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    $\begingroup$ why did you ask the question again after you asked it math.stackexchange.com/questions/673213/… and accepted an answer which you clearly did not understand? $\endgroup$ – Lost1 Feb 18 '14 at 20:02
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Let's try a more detailed description, but similar to Hagen's answer (as it is the most straightforward approach). $$ \mathrm{d}(\overline{A},\overline{B})=\inf\{\mathrm{d}(a,b):a\in\overline{A},b\in\overline{B}\}\tag{1} $$ and $$ \mathrm{d}(A,B)=\inf\{\mathrm{d}(a,b):a\in A,b\in B\}\tag{2} $$ Since $A\subset\overline{A}$ and $B\subset\overline{B}$, $\mathrm{d}(\overline{A},\overline{B})$ is the infimum over a larger set than $\mathrm{d}(A,B)$. Therefore, $$ \mathrm{d}(\overline{A},\overline{B})\le\mathrm{d}(A,B)\tag{3} $$ Now, suppose that for $\epsilon\gt0$, $$ \mathrm{d}(A,B)=\mathrm{d}(\overline{A},\overline{B})+\epsilon\tag{4} $$ that is, the inequality in $(3)$ is strict.

Definition $(1)$ says that we can find an $\overline{a}\in\overline{A}$ and a $\overline{b}\in\overline{B}$ so that $$ \mathrm{d}(\overline{a},\overline{b})\lt\mathrm{d}(\overline{A},\overline{B})+\frac\epsilon3\tag{5} $$ By the definition of closure, we can find $a\in A$ so that $$ \mathrm{d}(a,\overline{a})\lt\frac\epsilon3\tag{6} $$ and $b\in B$ so that $$ \mathrm{d}(b,\overline{b})\lt\frac\epsilon3\tag{7} $$ By $(5)$, $(6)$, $(7)$, and the triangle inequality, we get $$ \begin{align} \mathrm{d}(a,b) &\le\mathrm{d}(a,\overline{a})+\mathrm{d}(\overline{a},\overline{b})+\mathrm{d}(b,\overline{b})\\[3pt] &\lt\frac\epsilon3+\mathrm{d}(\overline{A},\overline{B})+\frac\epsilon3+\frac\epsilon3\\ &=\mathrm{d}(\overline{A},\overline{B})+\epsilon\tag{8} \end{align} $$ But by $(2)$ $$ \begin{align} \mathrm{d}(A,B) &\le\mathrm{d}(a,b)\\ &\lt\mathrm{d}(\overline{A},\overline{B})+\epsilon \end{align}\tag{9} $$ which contradicts $(4)$. Therefore, $$ \mathrm{d}(\overline{A},\overline{B})=\mathrm{d}(A,B)\tag{10} $$

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  • $\begingroup$ thank you very very much for your answer. It is really helpful. $\endgroup$ – user124140 Feb 19 '14 at 11:56
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Note that the function $\Phi:\mathbb R^d\times \mathbb R^d\to\mathbb R^+$ defined by $\Phi(x,y)=\Vert x-y\Vert$ is continuous, and that $\overline A\times \overline B=\overline{A\times B}$. So, in order to prove what you want, it is enough to show that if $\Phi:X\to\mathbb R^+$ is a continuous function on a topological space $X$, then $\inf\, \Phi(M)=\inf\, \Phi (\overline M)$ for any (nonempty) set $M\subset X$. Now, you have $\Phi(M)\subset \Phi(\overline M)\subset \overline{\Phi(M)}$, the second inclusion being true because $\Phi$ is continuous. So $\inf\, \overline{\Phi(M)}\leq \inf\, \Phi(\overline M)\leq \inf\, \Phi(M)$. To conclude, it remains to observe that for any (nonempty) set $E\subset\mathbb R^+$ you have $\inf\, E\in \overline E$, so that $\inf \, E=\inf\overline E$. Aplying this with $E=\Phi(M)$, this gives $\inf\, \Phi(\overline M)=\inf \,\Phi(M)$ as required.

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  • $\begingroup$ seriously? is this necessary? 10 bucks says the OP has no clue what you are on about. $\endgroup$ – Lost1 Feb 17 '14 at 22:48
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Apparently we need to show the following two inequalities:

  1. $d(A,B)\le d(\bar A,\bar B)$ and

  2. $d(A,B)\ge d(\bar A,\bar B)$.

The second inequality is straightforward: Clearly $$ \big\{d(x,y): (x,y)\in A\times B\big\}\subset \big\{d(x,y): (x,y)\in \bar A\times \bar B\big\}, $$ and hence $$ d(A,B)=\inf\big\{d(x,y): (x,y)\in A\times B\big\}\ge \inf\big\{d(x,y): (x,y)\in \bar A\times \bar B\big\}=d(\bar A,\bar B). $$

For the first inequality, assume that $\varrho=d(\bar A,\bar B)$. Then there exist sequences $\{x_n\}\subset \bar A$ and $\{y_n\}\subset \bar B$, such that $d(x_n,y_n)=\varrho.$ But as $x_n\in\bar A$ and $y_n\in\bar B$, there exist $x_n'\in A$ and $y_n'\in B$, such that $$ d(x_n,x_n')<\frac{1}{2n} \quad\text{and}\quad d(y_n,y_n')<\frac{1}{2n}, $$ since arbitrarily near to a point of $\bar E$ it is possible to find a point of $E$. Now $$ d(A,B)\le d(x_n',y_n')\le d(x_n',x_n)+d(x_n,y_n)+d(y_n,y_n')<d(x_n,y_n)+\frac{1}{n}, $$ and letting $n\to\infty$, the right-hand side of the above tends to $\varrho$ and hence $$ d(A,B)\le \varrho=d(\bar A,\bar B). $$ This completes the proof of inequality 1.

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  • $\begingroup$ thank you very much!! I have to wait some time to award you the $100$ points. One question: what definition of closure are you using? $\endgroup$ – user124140 Feb 17 '14 at 19:45
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    $\begingroup$ I am using the following one: $x\in\bar A$ if and only if, for every $\varepsilon>0$, there exists a $y\in A$, such that $d(x,y)<\varepsilon$. -- This definition implies that if $\{x_n\}\subset \bar A$, then there exists a sequence $\{x_n'\}\subset A$, such that $d(x_n,x_n')<1/n$. $\endgroup$ – Yiorgos S. Smyrlis Feb 17 '14 at 19:48
  • $\begingroup$ Amazingly clear, thanks! $\endgroup$ – Broken_Window Sep 22 '14 at 2:39
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The direction $d(A,B)\ge d(\bar A,\bar B)$ has nothing to do specifically with cloures. It follows simply from $A\subseteq \bar A$, $B\subseteq \bar B$ and that we take the infimium over a larger set of values.

Assume $\epsilon:=d(A,B)-d(\bar A,\bar B)>0$. Find points $a'\in \bar A$, $b'\in\bar B$ width $d(a',b')<d(\bar A,\bar B)+\frac\epsilon 3$ and $a\in A$, $b\in B$ with $d(a,a')<\frac\epsilon3$, $d(b,b')<\frac\epsilon3$.

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  • $\begingroup$ I dont understand the second paragraph. Would you mind explaining a little bit less dry ? thanks a lot. $\endgroup$ – user124140 Feb 15 '14 at 11:55
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    $\begingroup$ What's dry about it? $\endgroup$ – Hagen von Eitzen Feb 15 '14 at 13:17
  • $\begingroup$ the way you wrote it. I cannot understand it. :( $\endgroup$ – user124140 Feb 15 '14 at 17:43
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Note, why do you always have $d(A,B)\geq d(\bar{A},\bar{B})$?

if $d(\bar{A},\bar{B}) = d$ then $\exists x_1,x_2,...\in \bar{A}$ and $y_1,y_2,...\in \bar{B}$, for $\epsilon>0$, $\exists N$ such that $d(x_n,y_n)\leq d+\epsilon$ for $n>N$

now find a sequence $x_1',x_2',...\in A$ and $y_1',y_2',...\in B$ which also does the job, i.e. for $\epsilon>0$, $\exists N$ such that $d(x_n',y_n')\leq d+\epsilon$ for $n>N$ (how? use definition of closure)

For a proper proof: Distance between two sets in a metric space is equal to the distance between their closures

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  • $\begingroup$ can you explain a little bit more? thanks a lot $\endgroup$ – user124140 Feb 17 '14 at 1:38
  • $\begingroup$ @Learner which bit needs clarification? I have actually given you a proof, except 2 gaps you need to fill yourself? Normally I'd vote to close your question because you made no attempt yourself. $\endgroup$ – Lost1 Feb 17 '14 at 13:07

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