I am trying to show two equivalent metrics $p$ and $d$ on a set $X$ have the same convergent sequences. $p$ and $d$ are such that $kd(x,y) \leq p(x,y) \leq td(x,y)$ for every $x, y \in X$, $k$ and $t$ are positive constants.
Here's what I am doing –
As $p$ and $d$ are equivalent metrics they generate the same open sets.
Let $A$ be an open set generated by both $p$ and $d$.
As $A$ is open $X\backslash A$ is closed.
As $X\backslash A$ is closed, we can take a convergent sequence $(x_n) \in X\backslash A$ for all $n$, and it will converge to $x \in X\backslash A$.
Im not sure what to do now…can you just say it will converge to the same $x$ regardless of the metric being $p$ or $d$? I don't think I can. Should I be bringing open balls into it? Is there a need to use the positive constants $k$ and $t$?
Best Answer
If you’ve really already proved that metrics $p$ and $d$ related in that way generate the same open sets, you’re practically done, but you’re trying to make it much too complicated. Suppose that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ with respect to $d$; you want to show that it converges to $x$ with respect to $p$ as well. Let $U$ be an open nbhd of $x$. Then since $\langle x_n:n\in\Bbb N\rangle\underset{d}\longrightarrow x$, there is an $m\in\Bbb N$ such that $x_n\in U$ for all $n\ge m$. But that’s also exactly what it means for $\langle x_n:n\in\Bbb N\rangle$ to converge to $x$ with respect to $p$, so $\langle x_n:n\in\Bbb N\rangle\underset{p}\longrightarrow x$.
It’s in the proof that $d$ and $p$ generate the same topology that you would use the constants $k$ and $t$. But it’s not necessary to prove first that $d$ and $p$ generate the same topology: you can prove this result directly.
Suppose that $\langle x_n:n\in\Bbb N\rangle\underset{d}\longrightarrow x$. Then for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $d(x_n,x)<\epsilon$ for each $n\ge m_\epsilon$. This immediately implies that $p(x_n,x)<t\epsilon$ for each $n\ge m_\epsilon$. Thus, for each $n\ge m_{\epsilon/t}$ we have $p(x_n,x)<t\cdot\frac{\epsilon}t=\epsilon$, and it follows that $\langle x_n:n\in\Bbb N\rangle\underset{p}\longrightarrow x$. The opposite implication is proved similarly, using the fact that $d(x,y)\le\frac1kp(x,y)$ for all $x,y\in X$.