Let $M\neq \emptyset$ and $(M, d_1), (M, d_2)$ be two metric spaces. I want to prove the equvalence of 1. and 2., where
- $A \subset M$ is open with respect to $d_1 \iff A \subset M$ is open with respect to $d_2$.
- For $(x_k)_{k\in \mathbb{N}}\subset M$ and $x \in M$ holds: $d_1(x_k,x) \rightarrow 0 \iff d_2(x_k,x) \rightarrow 0$.
For $i = 1,2$, let's denote $A$ being open in $M$ w.r.t. $d_i$, by $A \subset^{open}_i M$.
I have looked at these hints, and I do have these statements at my disposal.
Suppose 1., then for $A^c := M\setminus A$ we have
\begin{align}
&A \subset^{open}_1 M \iff A \subset^{open}_2 M \\
\iff \quad &A^c \subset^{closed}_1 M \iff A^c \subset^{closed}_2 M \\
\iff \quad \big(&\forall (x_k)_k \subset A^c \text{ s.t. } d_1(x_k,x) \rightarrow 0 : x \in A^c \\
& \iff \forall (x_k)_k \subset A^c \text{ s.t. } d_2(x_k,x) \rightarrow 0 : x \in A^c\big).
\end{align}
Since $\emptyset \subset^{open}_i M$ for $i=1,2$, 2. yields for $\emptyset^c = M$. Is this a valid proof? It does not seem too convincing. Since we are resorting to looking at a special case at the end, I am unsure if both implications hold up by this argumentation.
Best Answer
Let me explain the proof you gave in more detail. The overly use of quantifiers makes the thing hard to read.
It is obviously enough to show that both metric spaces have the same closed sets.
Suppose $F$ is $d_1$-closed. We show that $F$ is $d_2$-closed as well.
To see that $F$ is $d_2$-closed, choose $x \in \overline{F}^{d_2}$. Then there is a sequence $(x_n)_n$ in $F$ such that $d_2(x_n,x) \to 0$. But by assumption $d_1(x_n,x)\to 0$ as well, so we obtain $x \in \overline{F}^{d_1}= F$ since $F$ is $d_1$-closed.
We thus have proven that $F= \overline{F}^{d_2}$, so $F$ is $d_2$-closed.
We thus have shown that every $d_1$-closed set is also $d_2$-closed. Similarly, we prove that every $d_2$-closed set is also $d_1$-closed. So both metrics have the same closed sets.
We prove the other implication. Suppose both spaces share the same opens.
Suppose $d_1(x_n,x) \to 0$. We show $d_2(x_n,x) \to 0$.
Let $\epsilon > 0$. Note that $B_{d_2}(x, \epsilon)$ is a $d_1$-open set containing $x$. Since $x_n \to x$, there is $n_0$ such that $x_n \in B_{d_2}(x, \epsilon)$ if $n \geq n_0$. We thus have proven that
$$\forall \epsilon > 0: \exists n_0: \forall n \geq n_0: d_2(x_n, x) < \epsilon$$
and we are done. Similarly, you show that $d_2(x_n,x) \to 0 \implies d_1(x_n,x) \to 0$.
Be careful, this no longer holds for arbitrary topological spaces!