# Help writing a better proof.

I have a question I am working on that says:

Prove that two metrics $d_1$ and $d_2$ are equivalent if they generate the same open sets (recall that two metrics are equivalent if $d_1(x_n,x)\rightarrow0$ and $d_2(x_n,x)\rightarrow 0$ as $n\rightarrow\infty$).

I just had such a hard time with this because I feel like I am proving a definition and I don't know where to start. Here is what I ended up getting:

Suppose $x_n$ is a sequence that converges to $x\in U$ and that $U$ is an open subset of $M$. Then $$\exists N_1\in \mathbb{N} \text{ s.t. } B_\epsilon^{d_1}(x_n) \text{ is a subset of } U$$ $$\exists N_2\in \mathbb{N} \text{ s.t. } B_\delta^{d_2}(x_n) \text{ is a subset of } U$$ Let $N=max(N_1,N_2)$. Then for every $n \geq N$ $x_n \in B_\epsilon^{d_1}(x)$ and $x_n \in B_\delta^{d_2}(x)$. And so if sequences converge for one metric they converge for the other. Therefore the two metrics generate the same open sets.

Does this make sense? I feel like I'm missing something. Thank you. I frequently miss problems because I skip over the important part of the proof some how.

• how do you define the convergence $x_n \to x$ from the open sets ? use the so-called neighborhoods of $x$ : $\{ U \in Neighborhood(x) \}$ and search for sequences of sets $U_n \supset U_{n+1}$ and $\bigcap_n U_n = \{x\}$. Then if for every $n$ : $x_n \in U_n$, the sequence $(x_n)$ is said to converge to $x$ – reuns Oct 5 '16 at 1:16
• There is a theorem in the book that says "A set $U$ in $(M,d)$ is open iff whenever a sequence $(x_n)$ in $M$ converges to a point $x \in U$, we have $x_n \in U$ for all but finitely many $n$." I am trying to use this theorem. – user197950 Oct 5 '16 at 1:18
• Yes if the $U_n$ are neighborhoods ordered by inclusion and whose limit (the intersection of all the sets) is $\{x\}$ then $x_k \to x$ iff for every $n$, there is a $K$ such that $k \ge K \implies x_k \in U_n$ – reuns Oct 5 '16 at 1:21
• ok this makes sense. But then how do I relate this to my two metrics? I feel like this could be used for one metric or the other, but shouldn't there be a logical step that brings them together as equivalent? – user197950 Oct 5 '16 at 1:23
• If two topologies have the same open sets, then for such a sequence $U_n$ of open sets (in topology $1$) ordered by inclusion and whose intersection is $\{x\}$, it is also a sequence of open sets ordered by inclusion and whose intersection is $\{x\}$ in topology $2$, and hence $x_k \to x$ in topology $1 \implies x_k \to x$ in topology $2$ – reuns Oct 5 '16 at 1:25