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Let $(X, d)$ be a metric space. Let $τ$ be the metric topology on $X$ induced by $d$. For $A ⊆ X$ , let $d(x, A) := \inf_{a∈A} d(x, a) $ for $x ∈ X$

(a) If $f (x) := d(x, A)$ (for a fixed subset A) then show that $f : (X, τ ) → R $ is a continuous mapping.

(b) If every infinite bounded set in $X$ has a limit point (in $X$) and $A$ is closed (in $(X, τ )$ ) then show that for each $x ∈ X$ there exists an element $a_0 ∈ A $ such $d(x, A) = d(x, a_0)$.

I have no clue how should I even approach to this. A detailed solution would be great.

Martin Sleziak
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Rusty
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  • What's the context of this question? Where did you come across it? – Ayman Hourieh Sep 14 '14 at 09:49
  • I am learning Topology. I came across a sheet of problems on topological spaces and continuous functions in which this problem was mentioned. – Rusty Sep 14 '14 at 09:53
  • I think that the question (a) has been asked (and answered) several times on this site; for example here: http://math.stackexchange.com/questions/48850/continuity-of-the-metric-function, http://math.stackexchange.com/questions/418042/show-that-the-function-f-x-to-bbb-r-given-by-fx-dx-a-is-a-continuo – Martin Sleziak Sep 14 '14 at 10:57

1 Answers1

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The part (a) has been asked and answered several times on this site, for example:

For the part (b), you can try proof by contradiction. Assume that $d(x,a)>d(x,A)$ for each $a\in A$. By the definition of infimum we have that for each positive integer $n$ there exists $a_n$ such that $$d(x,a_n)<d(x,A)+\frac1n.$$ Now the set $\{a_n; n\in\mathbb N\}$ is infinite and bounded. (Can you explain why it is infinite and bounded?)

Then it has at least one limit point. Let us denote a limit point of this set by $a_0$. (In fact, this limit point is a limit of a subsequence of $(a_n)_{n=1}^\infty$.) What can you say about $a_0$?

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Martin Sleziak
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