1
$\begingroup$

Consider the metric $d: X \times X \to \mathbb{R}$, and the metric $\min\{d,1/2\} = d \land 1/2$. Prove that these metrics induce the same topology.

Attempt:

Let $a \in X, \epsilon > 0$

Then $B_d(a, \min\{{\epsilon},1/2\}) \subseteq B_{d \land1/2}(a,\epsilon)$ and $B_{d \land1/2}(a, \epsilon) \subseteq B_d(a, \epsilon)$.

This ends the proof. Is this correct?

$\endgroup$
1
$\begingroup$

Assuming that you know that, given two metrics $d$ and $d'$ defined on the same space $X$, $d$ and $d'$ are equivalent if and only if for each $x\in X$, and for each $\varepsilon>0$ there are $\delta,\delta'>0$ such that $B_d(x,\delta)\subset B_{d'}(x,\varepsilon)$ and that $B_{d'}(x,\delta')\subset B_d(x,\varepsilon)$, then, yes, it is correct.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, that's exactly what I use. Thanks. $\endgroup$ – user370967 May 10 '18 at 18:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy