Two metrics which induce the same topology, question about identity maps and uniform continuity 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.
 A: We first simplify notation by setting $K_1=(X,d_1)$ and $K_2=(X, d_2)$.
We shall show that the identity map on $X$ is uniformly $(d_1,d_2)$-continuous.
Let $\varepsilon>0$ be given. For each $x \in K_2$ we define
\begin{equation}
U_x : = \displaystyle \left\{ y \in X : d_2(\,y,x)<\frac{\varepsilon}{2} \right\}.
\end{equation}
Since the identity map $f(x)=x$ is trivially $(d_1,d_1)$-continuous, we have that for every $x \in K_2$ the set $U_x$ is open in the compact metric space $K_1$ (Definition of Topologically Equivalent Metrics). So the family of open sets $\,\mathcal{U} := \{U_x : x \in K_2\}$ forms an open cover of $K_1$. Thus there is a number $\lambda>0$ such that for every $y \in K_1$ the open ball $B( y, \lambda):= \{z \in X: d_1(z,y)<\lambda\}$ is contained in one of the open sets of $\mathcal{U}$ (Lebesgue's Number Lemma). Therefore
\begin{aligned}
d_2(z,y) & \leq d_2(z,x) + d_2(x,y) \\
& < \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon \; \; \:\; \text{ whenever }\, d_1(z,y)<\lambda \text{, and } z,y \in K_1.
\end{aligned}
Without loss of generality, the identity map is uniformly $(d_2, d_1)$-continuous as well.
