TITLE: Understanding proof that if $c_1 d_1(x,y) \leq d_2 (x,y) \leq c_2 d_1 (x,y)$ then $d_1$ and $d_2$ are topologically equivalent metrics
QUESTION [5 upvotes]: Theorem. If there are strictly positive constants $c_1$ and $c_2$ such that
$$c_1 d_1(x,y) \leq d_2 (x,y) \leq c_2 d_1 (x,y)$$
for all $x,y \in X$, then $d_1$ and $d_2$ are topologically equivalent metrics on $X$.
Proof. (for one direction) Let $U$ be open in $(X, d_1)$. Then for $a \in X$ there > exists $\epsilon > 0$ such that
$$\{x \in X; d_1(x,a) < \epsilon\} \subset U.$$
But then
$$\{x \in X; d_2(x,a) < c_1\epsilon\} \subset \{x \in X; d_1(x,a) < \epsilon\} \subset U$$
and $U$ is open in $(X, d_2)$.

The line I don't understand is the one that says $\{x \in X; d_2(x,a) < c_1\epsilon\} \subset \{x \in X; d_1(x,a) < \epsilon\}$. Why is the former a subset of the latter? I don't see the justification of this.

REPLY [2 votes]: We know that $c_{1}d_{1}(x,y) \leq d_{2}(x,y)$ for all $x,y \in X$.
For membership in $\{x \in X: d_{2}(x,a) < c_{1} \epsilon \}$, we have that $d_{2}(x,a) < c_{1} \epsilon$.
Therefore, we conclude that $c_{1}d_{1}(x,a) \leq d_{2}(x,a) < c_{1} \epsilon$, thus $c_{1}d_{1}(x,a) < c_{1} \epsilon$, and thus we can say $d_{1}(x,a) < \epsilon$ since $c_{1}$ is positive.