Prove that two metrics generate the same topology.

Question

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?

Answer 1

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.