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I want to prove the following result:

Let $(X, d)$ be a metric space. Then $$\mathring E = \{x \in X \mid d(x, X \setminus E) > 0\}$$ where $d(x, A) = \inf\limits_{y \in A} d(x, y)$.

This is a part of my proof. But I'm not sure about a passage.

  • $\mathring E \subseteq \{x \in X \mid d(x, X \setminus E) > 0\}$
    Let $x \in \mathring E$. Then there exists $\varepsilon > 0$ such that $B(x, \varepsilon) \subset E$. We then have that, for every $y \in X \setminus E$, $$d(x, y) \geq \varepsilon > 0.$$ It follows that $d(x, X \setminus E) > 0$.

The other inclusion is much the same. The problem is that the bold passage seems so obvious, but I would like to prove it as well and I don't know how to do it.

How can I prove that if $d(x, a) < \varepsilon$ for every $a \in A$, then $d(x, b) \geq \varepsilon$ for every $b \in X \setminus A$?

Did
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rubik
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    $B(x, \varepsilon) \subset E\iff X\setminus E\subset X\setminus B(x,\varepsilon)\iff\forall y\in X\setminus E,\ldots$ – Did Apr 30 '16 at 09:52
  • If you found the answer, then might I suggest that you either self-answer the question, or ask @Did to turn his comment into an answer. Otherwise this question will go into the unanswered queue. – gebruiker Apr 30 '16 at 10:46
  • @gebruiker Seems reasonable. I'll add an answer, and I'll be open to suggestions to make it better. – rubik Apr 30 '16 at 12:17

1 Answers1

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Here is the complete proof for the first inclusion, with the hint from @Did. The other inclusion is really similar.

Let $x \in \mathring E$. Then there exists $\varepsilon > 0$ such that $$B(x, \varepsilon) \subset E.$$ From that inclusion we deduce $X \setminus E \subset X \setminus B(x, \varepsilon)$. Then it necessarily follows that if $y \in X \setminus E$, it's also true that $y \in X \setminus B(x, \varepsilon)$. Since it's not true that $d(x, y) < \varepsilon$, it must be the case that $d(x, y) \geq \varepsilon > 0$.
We conclude that $d(x, X \setminus E) = \inf_{y \in X \setminus E} d(x, y) \geq \varepsilon > 0$.

rubik
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