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So assume $(X,d)$ is a metric space with $A \subseteq X$ as a subspace. Show that the function $f: X \rightarrow \mathbb{R}$ defined by $f(x)=\inf\{d(x,a)|a\in A\}$is continuous.

My instinct is to use the definition of continuity for metric spaces. Namesly, given any $\epsilon >0$, there is a $\delta$, such that:

$d(x,y) < \delta \implies d(f(x),f(y))< \epsilon$, can I therefore assume that $\mathbb{R}$ has the usual topology or does that change the nature of what's being asked? Thanks

simplesedition
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  • yes I would use an $\epsilon$,$\delta$ kind of proof here. About the last question, yes you should assume that $\mathbb{R}$ has the usual topology... – Yanko Aug 12 '17 at 21:15
  • Yes we assume the usual topology on $\mathbb{R}$. – John Griffin Aug 12 '17 at 21:19
  • i am sure you are expected to assume the usual topology on $\Bbb{R}$. If you allow more open sets in $\Bbb{R}$ than under the usual topology you will have difficulty getting the proof to go through. (There will be lots of special cases when the result will still hold for other topologies, but I am sure you are not being asked to explore these cases.) – Rob Arthan Aug 12 '17 at 21:22
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    PS: the problem should specify that $A$ is non-empty. – Rob Arthan Aug 12 '17 at 21:25
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    One can speak about the continuity or discontinuity of a function only when the domain and the codomain of the function have a topology. When one of the set is $\mathbb{R}$ and nothing is said about the topology, it is always the usual topology. – Gribouillis Aug 12 '17 at 21:26

2 Answers2

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We denote $d(x,A):=\inf\{d(x,a) \mid a\in A\}$. It will be useful to first show that any $x,y\in X$ satisfy $$ d(x,A) \leq d(x,y) + d(y,A). $$ To see this, note that by the triangle inequality we have $$ d(x,A) \leq d(x,a) \leq d(x,y) + d(y,a), \quad a\in A. $$ Now taking the infimum over $a\in A$ yields the desired inequality.

Now let $\varepsilon>0$ and choose $\delta:=\varepsilon$. Then, if $d(x,y)<\delta$, we have by our inequality that $$ |f(x)-f(y)|=|d(x,A)-d(y,A)|\leq d(x,y)<\varepsilon. $$ This completes the proof.

John Griffin
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Fir $x,y\in X $ and $a\in A $,

$$f (x)\le d (x,a)\le d (x,y)+d (y,a) $$

$$\implies f (x) \le d(x,y) + f (y) $$ $$\implies f (x)-f (y)\le d (x,y)$$ $$\implies |f (x)-f (y)|\le d (x,y) $$

$\implies f $ Lipschitz

$\implies f $ uniformly continuous at $X $

$\implies f $ continuous at $X $.

SvanN
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hamam_Abdallah
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