# Showing that two topologies give rise to the same topology

Suppose that for a metric space $(X,d)$ we have another metric $p$ forming the metric space $(X,p)$. Both of these will give rise to two topologies $\tau_{1}$ and $\tau_{2}$. I want to show that $\tau_{1}$ and $\tau_{2}$ give rise to the same topology.

How could one do it generally in this case? I was thinking of taking an open set $U_{d}$ in the metric space $(X,d)$ and showing that this is open in $(X,p)$ and conversely, taking an open set $U_{p}$ in $(X,p)$ and showing its open in $(X,d)$. Would this be the idea to prove that the topologies are equivalent? (I'm sure this can be used with more generality for the case that we have two topological spaces rather than two metric spaces). If not, what would be the correct way?

Thanks for the help.

• That's exactly how I would do it. That would be the first thing I would try even if the two spaces were not metric spaces. – John Douma Sep 14 '15 at 14:08
• Yes, that is the general way of proving it. It's generally enough with $U_d$ and $U_p$ being open balls in their respective metrics. – Arthur Sep 14 '15 at 14:09
• Small linguistic detail: You want to show that the two metrics $d$ and $p$ give rise to the same topology, and you want to show that $\tau_1$ and $\tau_2$ are the same topology. It is wrong to say that you want to show that the two topologies give rise to the same topology. – Arthur Sep 14 '15 at 14:11
• If I am thinking about this properly, you need to make a statement to the effect that neither of the metrics are the discrete metric (or both are the discrete metric). – Clayton Sep 14 '15 at 14:12

In the case of metric spaces it is enough to show that every open ball $B_d(x,r)$ contains an open ball $B_\rho(x,r')$ with the same center but possibly smaller radius (and vice versa).
In general topological spaces it is enough to show that every open subset of one topology containing a point $x$ contains an open subset in the other topology containing $x$ (and vice versa).