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In a metric space $(X,d)$, define $d(x,A) = \inf \{d(x,y):y\in A \}$. I am trying to prove that the set $A_{\epsilon} = \{ x \in X : d(x,A) < \epsilon \}$ is an open set for every $\epsilon>0$. I know this is a very simple consequence of the function $f(x) = d(x,A)$ being continuous. However, I am trying to prove this WITHOUT using this fact and going straight to the simple definitions.

I have been trying to use the fact that $\vert d(x,A) - d(y,A) \vert \leq d(x,y)$ as I believe somewhere near that inequality lies the answer, however I am stuck. Has anyone tried to produce such proof?

Martin Sleziak
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MathUser
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Given $\epsilon >0$ and $A_\epsilon = \{x\in X, d(x, A) < \epsilon\}$.

Let $x\in A_\epsilon$, then there is $y\in A$ such that $d(x, y) < \epsilon$. Let $\delta >0$ be small such that $d(x, y) + \delta < \epsilon$. Then for all $z\in B_x(\delta) = \{z\in X: d(x, z) < \delta\}$,

$$d(z, y) \leq d(z, x) + d(x, y) < \delta + d(x, y) <\epsilon \Rightarrow d(z, A) < \epsilon\ .$$

Thus $B_x(\delta) \subset A_\epsilon$ and $A_\epsilon$ is open.

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    It may be obvious, but I don't see how there is always such $y\in A$. – MathUser Apr 23 '14 at 00:51
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    Nevermind, here is the argument I did not see at first. Since $x\in A_{\epsilon}$ then $d(x,A) < \epsilon$ so we have that $\epsilon - d(x,A) > 0$. By definition of infimum this implies that, for $\bar{\epsilon} = \epsilon - d(x,A)>0$ there is such $y\in A$ such that $d(x,A) \leq d(x,y) < d(x,A) + \bar{\epsilon}$ now substituting the value of $\bar{\epsilon}$ we get that $d(x,y) < d(x,A) + \bar{\epsilon} = d(x,A) + (\epsilon - d(x,A)) = \epsilon$. – MathUser Apr 23 '14 at 01:14