# Sequential Equivalence Implies Topological Equivalence

Define two metric spaces $(M,d)$ and $(M,\rho)$ to be equivalent, denoted $d\sim p$, to mean that:

Topological Definition $\forall x\in M: \forall \epsilon>0 \exists \delta_1>0, \delta_2>0: B_{\rho}(x,\delta _1)\subset B_{d}(x,\epsilon)$ and $B_d(x,\delta_2)\subset B_{\rho}(x,\epsilon)$.

This means if a set $U$ is open in $(M,d)$, it's open in $(M,\rho)$ as well, and vice-versa.

Sequential Definition:

$\forall x\in M\;\forall \{x_n\}\subset M, x_n\to x$ in the $d$ metric iff $x_n\to x$ in the $\rho$ metric.

Prove that $d\sim p$ in sequential definition $\implies d\sim p$ in the topological definition.

My professor posed the above question in class as an exercise, and gave us a hint: to try to solve it through a proof by contrapositive. ($\neg q\implies \neg p$). I've negated both statements as follows:

"Attempt":

$\neg q$:

$d\nsim q$ in the topological definition means that $\exists x\in M\exists\epsilon >0\forall\delta_1\forall\delta_2>0 : B_{\rho}(x,\delta_1)\not\subset B_d(x,\epsilon)$ or $B_d(x,\delta_2)\not\subset b_{\rho}(x,\epsilon).$

$\neq p$:

$d\nsim q$ in the sequential definition means that $\exists x\in M\exists \{x_n\}\subset M: x_n\to x$ in $d$ metric but $x_n\not\to x$ in $\rho$ metric or $x_n\to x$ in $\rho$ metric but $x_n\not\to x$ in the $d$ metric.

I'm not sure where to start, so hints would be preferred over complete solutions. Thank you.

## 2 Answers

If $d \sim \rho$ then $(M,d)$ and $(M,\rho)$ have the same open sets, and this implies quite easily that they have the exact same convergent sequences as well. (A sequence converges to $x$ if, for every open set $O$ containing $x$, a tail of the sequence lies in $O$.)

On the other hand, if $(M,d)$ have the same convergent sequences, they have the same closed sets (a set $A$ in a metric space is closed iff all sequences from $A$ that converge have their limit in $A$, so being closed can be purely defined in terms of sequences converging), and thus the same open sets as well (a set is open iff the complement is closed ).

I don't really see any for a contrapositive.

• Maybe the contrapositive is contained in the proof of „sequentially closed $\implies$ closed“. – user87690 Sep 22 '14 at 16:33
• I gave that proof directly (via closed sets)... – Henno Brandsma Sep 22 '14 at 18:01
• Do you mean a set is open iff the complement is closed? – Sujaan Kunalan Sep 22 '14 at 18:32
• @SujaanKunalan edited thx – Henno Brandsma Sep 22 '14 at 18:34
• I mean, you say „a set $A$ in a metric space is closed iff all sequences from $A$ that converge their limit $A$“, i.e. „closed $\iff$ sequentially closed“ holds for metric spaces. But if you want to actually prove the backward implication, you may use a contrapositive. – user87690 Sep 23 '14 at 8:20

$\neg q$:

$d\nsim q$ in the topological definition means that $\exists x\in M\exists\epsilon >0\forall\delta_1\forall\delta_2>0 : B_{\rho}(x,\delta_1)\not\subset B_d(x,\epsilon)$ or >$B_d(x,\delta_2)\not\subset b_{\rho}(x,\epsilon).$

Hint: without loss of generality, assume $\exists x \in M \exists \epsilon>0 : \forall \delta>0 \; B_\rho(x,\delta) \not\subset B_d(x,\epsilon)$

Now find a sequence $x_n$ such that $x_n\to x$ in $\rho$, but not in $d$.

Hope this helps.