TITLE: Equivalent metrics determine the same topology
QUESTION [21 upvotes]: Suppose that there are given two distance functions $d(x,y)$ and $d_1 (x,y)$ on the same space $S$. They are said to be equivalent if they determine the same open sets. 
Show that $d$ and $d_1$ are equivalent if to every $\varepsilon$ there exists a $\delta>0$ such that $d(x,y)<\delta$ implies $d_1 (x,y)<\varepsilon$ and vice versa.
I have no idea how to handle this kind of problem.

REPLY [12 votes]: If you're familiar with the concepts of topologies and topological spaces, I think you will find these other ideas useful. Recall that

DEFINITION. Given a topological space $(X,\mathfrak I)$, a collection $\mathscr B$ of subsets of $\mathfrak I$ is said to be a basis for the toplogy on $X$, or a basis for the open sets, or a basis for $\mathfrak I$, iff every element of $\mathfrak I$ is a union of elements of $\mathscr B$.

This will be useful later

LEMMA Let $\mathscr B$ be a collection of elements of $\mathfrak I$. Then $\mathscr B$ is a basis for $\mathfrak I$ if and only if for each $x\in X$ and $x\in O$ an open set, there is an element $B$ of $\mathscr B$ for which $x\subset B\subset O$.

PROOF First suppose $\mathscr B$ is a basis. Then each open set is a union of basic elements, say $O=\bigcup_{\alpha \in A} B_\alpha$, so given an open set $O$, and $x\in O$, it follows $x\in B_\alpha\subset O$ for some index. Conversely, suppose that for each $x\in X$ and $x\in O$ an open set, there is an element $B$ of $\mathscr B$ for which $x\subset B\subset O$. Then for each $x\in O$ we may set $$O=\bigcup_{x\in O}B_x$$
so each set is a union of elements of $\mathscr B$, and $\mathscr B$ is a basis.
Now, we have this useful criterion for comparing topologies in terms of their bases:

THEOREM Let $\mathscr B$ and $\mathscr B'$ be bases for the topologies $\mathfrak I$ and $\mathfrak I$ respectively, on $X$. Then the following statements are equivalent:
$(1)$ $\mathfrak I'$  is finer than $\mathfrak I$, that is $\mathfrak I\subset \mathfrak I'$
$(2)$ For each $x\in X$, and each $B\subset \mathscr B$ for which $x\in B$, there is a $B'\subset \mathscr B'$ for which $x\in B'\subset B$.

PROOF
$(2)\Rightarrow (1)$ We want to show that whenever $O\in \mathfrak I$ then $O\in \mathfrak I'$. Since $\mathscr B$ is a basis for $\mathfrak I$, given any point $x\in O$ there is a $B$ for which $x\subset B\in O$. Condition $(2)$ says that there is a $B'\in \mathscr B'$ for which $x\in B'\subset B \subset O$, that is $x\in B'\subset  O$. Thus $O\subset \mathfrak I'$ by the previous LEMMA and the defintion of basis.
$(1)\Rightarrow (2)$ Let $x\in X$, $B\subset \mathscr B$, and $x\in B$. Now, $B\subset \mathfrak I$ ($B$ is itself an open set), and $\mathfrak I\subset \mathfrak I'$ by hypothesis, so $B\subset \mathfrak I$. Since $\mathfrak I'$ is generated by $\mathscr B'$, there is an element of $\mathscr B'$, call it $B'$; for which $x\in B'\subset B$.
So, the definition is

DEFINTION Two bases on a set $X$ are equivalent if they generate the same topology$\mathfrak I$, or collection of open sets.  (Note a metric defines a base, namely, the collection of all open balls)

An immediate corollary of the last theorem is:

COROLLARY Two bases $\mathscr B'$ and $\mathscr B$ are equivalent if and only if for each $B\in \mathscr B$ and $x\in B$ there is a $B'\in \mathscr B'$ for which $x\in B'\subset B$, and viceversa.

...which is what you're asked to prove.