From the notes: Let $(X, d_1)$ and $(X, d_2)$ denote the two metric spaces, where X is compact and suppose both metrics induce the same topology. Then the identity maps $i : (X, d_1) \to (X, d_2)$ and $j : (X, d_2) \to (X, d_1)$ are continuous and since X compact, they are uniformly continuous. Fix $ε_1 > 0$. Then, by uniform continuity, we can subsequently find an $ε_2 > 0$ and $ε_3 > 0$ such that for all x, y ∈ X: $d_1(x, y) < ε_1 ⇒ d_2(x, y) < ε_2$ and $d_2(x, y) < ε_2 ⇒ d_1(x, y) < ε_3$
My problem is that my understanding of uniform continuity is the other way around, given $\epsilon_2$ we can find such an $\epsilon_1$ and so on. I tried to prove the above my contradiction,which would mean for every $\epsilon_2>0$ there are some x,y in X such that $d_1(x, y) < ε_1 ⇒ d_2(x, y) \geq ε_2$. I tried to show this contradicts uniform continuity of either i or j but I was not successful. Could someone help? Thanks.