# How can I show these two metrics give the same topology?

This question came up while showing the composition of a metric with a certain other function gives another metric. Suppose I have some metric space $(X,d)$ and a continuous, non-decreasing function $f$ on the nonnegative reals. Moreover, suppose that $f(x)=0$ iff $x=0$, and $f$ also satisfies the triangle inequality in that $f (x+y)\leq f(x)+f(y)$.

Using these properties, it is not difficult to show that $f\circ d$ is yet another metric on $X$, so $(X,f\circ d)$ is also a metric space.

I notice the (open) balls given by the metrics are of form $$B_d(x,r)=\{y\in X\mid d(x,y)\lt r\}$$ and $$B_{f\circ d}(x,r)=\{y\in X\mid (f\circ d)(x,y)\lt r\},$$ so it seems that the topologies generated by the base of open balls in each case would probably be the same. I would like to see how one would go about showing the topologies given by these two metrics are indeed the same. Thanks for any insight.

-
It suffices to show that in each open ball in one topology it's possible to find an open ball in the other topology. (The mental picture I have in my head is finding squares in circles, then finding circles in squares, which shows that $\ell^1, \ell^2, \ell^{\infty}$ metrics give the same topology on $\mathbb{R}^2$.) – Qiaochu Yuan Aug 31 '11 at 5:17
In $\mathbb{R}^{n}$ any two norm's are equivalent. – user9413 Aug 31 '11 at 7:00
Thanks @Qiaochu, I had heard those metrics gave the same topology, but didn't quite see why until now. – yunone Aug 31 '11 at 17:40
I was about to post an answer on your last question, so you deleted it in time. Don't feel bad about it, this happens to all of us (and the downvote was completely undeserved, IMO). Here's a more interesting question for you: Is an infinite set in the cofinite topology connected? Challenge: is a countably infinite set with the cofinite topology path-connected? – t.b. Sep 5 '11 at 21:43
Thanks @Theo, that makes me feel better. I just regret that I wasted a few of your minutes writing out an answer. But your first comment was more than helpful enough, for which I am grateful. I will look into these more interesting questions too. – yunone Sep 5 '11 at 21:50

Let $(X,d)$ be a metric space and $f: \mathbb R_+ \rightarrow \mathbb R_+$ a function as you've described. Given $r>0$ and $x \in X$ we want to find an $\varepsilon>0$ such that $B_{f\circ d}(x,\varepsilon)\subset B_d(x,r)$. Set $\varepsilon=f(r)$ and take $w \in B_{f\circ d}(x,\varepsilon)$ then $f(d(x,w))<f(r)$ and since $f$ is non-decreasing it must be the case that $d(x,w)<r$ and therefore $w \in B_d(x,r)$.

To show that we can find $\varepsilon>0$ such that $B_d(x,\varepsilon) \subset B_{f\circ d}(x,r)$ is a bit trickier. Let $y \in (0,r)$ and pick $\varepsilon \in f^{-1}(y)$. Suppose $w \in B_d(x,\varepsilon)$ then $d(x,w)<\varepsilon$ and $f(d(x,w))\leq f(\epsilon)=y<r$ so $w \in B_{f\circ d}(x,r)$ as desired.

I think it's interesting to note that to get the balls to be contained in each other you don't really need the other two properties triangle inequality for $f$. But that it is needed in order to ensure $f\circ d$ is a metric.

-
You used the facts that $f(0)=0$ and $f$ is continuous to ensure that $f^{-1}[\{y\}]\ne\varnothing$. – Brian M. Scott Aug 31 '11 at 7:09
@Brian, thanks again. – JSchlather Aug 31 '11 at 7:25
Just so you know: "triangle inequality for $f$" is usually called "subadditivity of $f$" (after Marcel Riesz who suggested that name to Hille). It is somewhat but not too closely related to convexity. There's quite an impressive body of literature on them. – t.b. Aug 31 '11 at 7:43
Thanks Jacob, this explanation was very clear. Thanks also Brian and Theo for the useful comments. – yunone Aug 31 '11 at 17:39

Here is a Theorem from my book Topology, Sts. Cyril and Methodius University, Skopje, 2002 (in Macedonian) . The proof consists of five rows.

Theorem. X is a set with two topologies T1 and T2 , B1 is a basis for topology T1 and B2 is a basis for topology T2 . If for every B1 in B1 and every point x in B1 there exists B2 in B2 , such that x is in B2 and B2 is a subset of B1, then T1 is a subset of T2 .

Nikita Shekutkovski

-
For those of us who don't have the book, the proof would have been helpful. Regards – Amzoti Mar 21 '13 at 12:16