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This is the last question on the exercise sheet and I am having real trouble formalizing my intuitions.

It should be obvious. Since the closure of a set is the set of all points in the universe with distance zero to the set, then there should be no difference between finding the distance between the set and the distance between their closures (since zero is the additive identity). I feel like I could take advantage of the triangle inequality to formalize the picture in my head, but I can't seem to grasp it concretely.

Working Definitions:

$$ d(x, A) = \inf_{a \in A}\{d(x,a)\} $$ $$ \overline{A} = \{x \in X : d(x, A) = 0\} $$ $$ \operatorname{dist}(A, B) = \inf_{b \in B}\{d(b, A)\} $$

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    $\begingroup$ You probably want inf over $x\in B$ in the definition of dist$(A,B)$. $\endgroup$
    – T. Eskin
    Mar 3, 2013 at 12:39
  • $\begingroup$ Right, thanks. My brain is starting to mush. $\endgroup$ Mar 3, 2013 at 12:49

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Hints:

"$\leq$": Since $A\subseteq \overline{A}$ and $B\subseteq \overline{B}$ then $d(\overline{A},\overline{B})\leq d(a,b)$ for all $a\in A$, $b\in B$. Take infimum.

"$\geq$": Prove by triangle-inequality that for any $x,y\in X$ we have \begin{equation*} d(A,B)\leq d(A,x)+d(x,y)+d(y,B), \end{equation*} and conclude that for any $a\in\overline{A}$ and $b\in\overline{B}$ we have $d(A,B)\leq d(a,b)$. Take infimum.

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  • $\begingroup$ I've been looking over this, but I don't see how we can say $d(\overline{A},\overline{B}) \leq d(a, b)$, whereas I can justify $d(\overline{A},\overline{B}) \leq d(A, B)$. This is equivalent to taking the infimum over $a \in A, b \in B$, yes? $\endgroup$ Mar 3, 2013 at 13:03
  • $\begingroup$ @vermiculus. The point is to show that $d(\overline{A},\overline{B})\leq d(a,b)$ is true for all $a\in A,\,b\in B$. Thus, by taking infimum over all $a\in A$ and $b\in B$ in this inequality, we obtain $d(\overline{A},\overline{B})\leq d(A,B)$. But out of curiosity, how do you justify $d(\overline{A},\overline{B})\leq d(A,B)$ without showing that $d(\overline{A},\overline{B})\leq d(a,b)$ for all $a\in A$, $b\in B$? $\endgroup$
    – T. Eskin
    Mar 3, 2013 at 13:07
  • $\begingroup$ $A \subset B \subset X \implies d(x \in B^\complement, A) \geq d(x \in B^\complement, B)$ (I'm not really sure how it follows from there, but that is my honest reasoning. Brain dead. My feeble mind understands the proof now, once I convinced myself that taking such an infimum is exactly the definition. (Really, really brain dead.) $\endgroup$ Mar 3, 2013 at 13:13
  • $\begingroup$ I know my way isn't the greatest (or even logical :P), but I think the professor said something to the effect of the result during lectures, but I can't find it in my notes. Still have to prove that $d(\overline{A},\overline{B}) \leq d(a,b)$, correct? $\endgroup$ Mar 3, 2013 at 13:16
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    $\begingroup$ @vermiculus. Great. Let me know if some points need further clarification. $\endgroup$
    – T. Eskin
    Mar 3, 2013 at 13:28

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