# Proof of proposition about metric space and closed set.

$$(X,d)$$ : metric space

Proporsion;

If $$d(x,z) \leqq \max\{ d(x, y), d(y,z)\}$$ $$\,\,$$ for $$x,y,z \in X$$ $$\cdots$$ ① holds,

$$U(a, \epsilon ) :=\{ x \in X ; d(x,a) < \epsilon \}$$ (for $$\,\,\ \forall \epsilon >0\,\,$$ and $$\,\,$$ $$a\in X$$) is closed set.

Is this proof correct?

Proof

I'll prove $$U(a,\epsilon) = \overline{U(a,\epsilon)}$$.

For any set $$A, A \subset \overline{A}$$ holds. So I have to prove $$U(a,\epsilon) \supset \overline{U(a,\epsilon)}$$.

For any $$b \in \overline{U(a,\epsilon)} , U(b, \epsilon) \cap U(a, \epsilon)\neq \phi$$.

I can pick up $$c \in U(b, \epsilon) \cap U(a, \epsilon)$$.

Then, $$d(c,b)<\epsilon$$ and $$d(c.a)<\epsilon$$

From ①, $$d(a,b)$$ $$\leqq$$ $$\max\{ d(a, c), d(c,b)\} <\epsilon$$

Therefore, $$b \in U(a, \epsilon)$$

$$U(a,\epsilon) \supset \overline{U(a,\epsilon)}$$

• $U(a,\varepsilon )$ is open. – Surb Oct 28 '20 at 12:57
• Yes, your proof is correct. – Andreas Blass Oct 28 '20 at 15:45