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Prove that:

Let $(X, d)$ be a metric space, and let $A$ be a subset of $X$. The function $f_A\colon X\rightarrow \mathbb{R}$, defined by $f_A (x) = d({\{x}\}, A)$, is continuous.

Honestly, I have NO idea where to start. I need to prove that inverse of an open set in $\mathbb{R}$ is open $X$. What it makes it hard to approach is the definition involved: $$d(A, B) = \operatorname{glb}\{d(a, b)\mid a \in A, b \in B\}.$$

Would someone please guide me how to solve this question?

Thank you.

Asaf Karagila
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  • See also http://math.stackexchange.com/questions/48850/continuity-of-the-function-x-mapsto-dx-a-on-a-metric-space and http://math.stackexchange.com/questions/400487/continuity-of-dx-a – Martin Sleziak Jul 15 '15 at 06:44
  • @Martin Sleziak: Neither questions nor (esp.) answers are same as here. I couldn't find my answer from those, but I learnt a lot by answers of here. –  Jul 15 '15 at 07:37
  • And what exactly is the difference? You ask how to show that $x\mapsto d(x,A)$ is continuous. That is the same question as in the post I linked to. (The only minor difference is that in one of them the posts adds a superfluous requirement that $A$ is closed. And one of the questions mentions Lipschitz contidion.) BTW I don't think that slightly different wording of answers should mean that the questions are not duplicates. And Ivo Terek's answer (which is accepted at the moment) is almost the same as Zarrax' answer at one of the posts I linked. – Martin Sleziak Jul 15 '15 at 07:51
  • Pedro Tamaroff's answer in another post about the same question is very similar to the accepted answer, too. And I am pretty sure that you could find a few minor variations of this answer in other linked questions. – Martin Sleziak Jul 15 '15 at 07:56

2 Answers2

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The idea is to use the reverse triangle inequality: $$d(x,a) - d(y,a) \leq d(x,y), $$given $x,y \in X$, $a \in A$. But notice that $d(x,A) \leq d(x,a)$, so we get: $$d(x,A) - d(y,a) \leq d(x,y).$$Reorganize: $$d(x,A)-d(x,y) \leq d(y,a).$$Now take the infimum on $a$: $$d(x,A) - d(x,y) \leq d(y,A).$$Reorganize: $$d(x,A) - d(y,A) \leq d(x,y).$$Since $x$ and $y$ were arbitrary, we can swap the above inequality to obtain: $$d(y,A)-d(x,A) \leq d(y,x) = d(x,y).$$This gives: $$|d(x,A)-d(y,A)| \leq d(x,y).$$Now let $\epsilon > 0$, and check that $\delta = \epsilon > 0$ will work in the definition of continuity. Since $\delta$ does not depend on $x$ and $y$, the function is actually uniformly continuous (which is even better).

Ivo Terek
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Since $f_{A}$ is a map between metric spaces (each with the topology induced by its metric), we can use the $\epsilon$–$\delta$ definition of continuity: $f_{A}$ is continuous at a point $x\in X$ if for all $\epsilon>0$ there exists $\delta>0$ such that $|f_{A}(y)-f_{A}(x)|<\epsilon$ if $|x-y|<\delta$.

So, take any $x\in X$ and any $\epsilon>0$. Consider $y$ near $x$. Intuitively we expect $f_{A}(y)$ to be close to $f_{A}(x)$. We can show this formally using the triangle inequality, which tells us that $d(y,a)\leq d(y,x)+d(x,a)$ for all $a\in A$. Rearranging, $d(y,a)-d(x,a)\leq d(y,x)$. By the same argument, $d(x,a)-d(y,a)\leq d(y,x)$, so that $|d(y,a)-d(x,a)|\leq d(y,x)$. Because this holds for all $a\in A$, we can say that $|f_{A}(x)-f_{A}(y)|\leq d(x,y)$. In particular, $|f_{A}(x)-f_{A}(y)|$ will be less than $\epsilon$ whenever $d(x,y)<\epsilon$. Thus, $f_{A}$ is continuous at $x$. This holds for all $x\in X$, so $f_{A}$ is continuous.

Ben Whitney
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