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I need help with the following proof, which my professor added for practice (but not as homework). I am completely lost here.

Let $A$ be a nonempty subset of a metric space $X$. Define $d(\cdot ,A) : X \to [0,\infty)$ by $$d(x,A) = \inf\{d(x,a) : a \in A\}.$$

Prove that $d(\cdot,A)$ is continuous.

Now there are some things that I do know.

  1. The definition of continuous:Suppose X,Y are metric spaces, a ∈ X and f : X → Y. The function f is continuous at a if for every ε>0there is a δ>0such that if dX(a,x) < δ, then dY (f(a),f(x)) < ε. If f is continuous at every point a ∈ X, then f is said to be continuous.

  2. The following theorem:A function f : X → Y is continuous if and only if f−1(U) ⊂ X is open for every open set U ⊂ Y .

Jim Darson
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    Hint: Show that $|d(x,A)-d(y,A)|\le d(x,y)$. – Hagen von Eitzen Nov 08 '13 at 14:59
  • I need more help than that, but thanks for the hint. – Jim Darson Nov 08 '13 at 15:02
  • Thank you for your question! It is good practice on this site to add a bit of information on the context your question came up in, and to share your own work on it. It's also fine if you state that you're completely lost -- the information is helpful for answerers to gauge their answer on. For more information on asking a good question on this site, see here. – Lord_Farin Nov 08 '13 at 15:12

1 Answers1

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In fact it is uniformly continuous. It is easier to prove that.

Let $a\in A$. Then

$$d(x,A)\leq d(x,a)\leq d(x,y)+d(y,a)$$ (Why?) Think about the definition of $d(x,A)$ and the triangle inequality.

Thus we get $d(x,A)\leq d(x,y)+d(y,A)$. (Why?)

Can you continue from here?

Your aim is to show

$$|d(x,A)-d(y,A)|\leq d(x,y)$$

hence one can always pick $\delta = \epsilon$.