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$(X,d)$ : 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, $b \in U(a, \epsilon)$

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

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

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