Consider a set $A$ on which $2$ metrics are defined: $d_1$ and $d_2$.
Let the two resulting metric spaces $(A, d_1)$ and $(A, d_2)$ be topologically equivalent.
Given the above, is it always the case that the topology induced on $A$ by $d_1$ is necessarily identical to the topology induced on $A$ by $d_2$?
By topological equivalence in the context of metric spaces, I have the definition that $d_1$ and $d_2$ are topologically equivalent iff ($U \subseteq A$ is "$d_1$-open" iff $U \subseteq A$ is "$d_2$-open"). $(1)$
Again, my definition of $d_1$-open is iff it is the neighborhood (under the metric $d_1$) of each of its points.
A neighborhood of a point $x$ is defined as a subset of $A$ which completely contains an open $\epsilon$-ball centered at $x$ for some $\epsilon \in \mathbb R_{>0}$.
The open $\epsilon$-ball of $x$ is the set:
$B_\epsilon (a) := \{x \in A: d_1 (x, a) < \epsilon\}$
… and we are finally down to the bare metal, so to speak.
Now, Bert Mendelson, in his "Introduction to Topology" (3rd ed. 1975, Dover Publications), gives this exercise in his chapter 3, section 2 "Topological Spaces" exercise 3:
"Let $(\mathbb R^n, d)$ and $(\mathbb R^n, d')$ be defined as"
$d(x,y) = \displaystyle \max_{1 \le i \le n} \{|x_i – y_i\}$ (the Chebyshev or chessboard metric)
$d'(x,y) = \left({\displaystyle \sum_{i = 1}^n (x_i – y_i)^2}\right)^{1/2}$ (the usual Euclidean metric)
Prove that the two metric spaces $(\mathbb R^n, d)$ and $(\mathbb R^n, d')$ give rise to the same topological space."
Now this can be straightforwardly addressed (probably, I haven't gone through the motions) of constructing $\tau$ and $\tau'$ corresponding to $d$ and $d'$ respectively, and demonstrating that an arbitrary open set of $\tau$ is in $\tau'$ and an arbitrary open set of $\tau'$ is in $\tau$.
But (as an exercise set in a different work, i.e. W.A. Sutherland's "Introduction to Metric and Topological Spaces") I have already proved that these two spaces are Lipschitz equivalent:
$$d' (x, y) \ge d (x, y) \ge \dfrac 1 2 d' (x, y)$$
and thence, by another result that I have a proof of that Lipschitz equivalent metrics are topologically equivalent in the sense as defined as in $(1)$ above, that they are in fact now proved to be topologically equivalent.
If I could use the result that "two metrics on the same set are topologically equivalent if they give rise to equal topologies (in the sense that the open sets of one are exactly the open sets of the other), then I can answer the question in Mendelson by merely quoting my result above (which I arrived at via that long route via Lipschitz equivalence) that $(\mathbb R^n, d)$ and $(\mathbb R^n, d')$ are proven to be topologically equivalent.
But is it actually the case that "two metrics on the same set are always topologically equivalent if and only if they give rise to equal topologies"?
I have never seen that result stated formally (although I admit I have not read all that widely), and so I am wondering whether, as it has not been so stated, whether it is actually true or not?
Best Answer
I address some of these issues in this older answer.
But in short: yes $(X,d_1)$ and $(X,d_2)$ are exactly topologically equivalent iff $d_1$ and $d_2$ induce the exact same topology on their set $X$. If the metrics are Lipschitz-equivalent (as they will always be on $\Bbb R^n$, if they come from a norm..., as your examples) then they will be topologically equivalent, but the reverse need not need hold. This is related to the so-called uniform structure on a metric space. Two metrics can be non-equivalent for Cauchy sequences and uniform continuity, but still topologically equivalent.
In topology we typically only care about the topology the metric induces. A metric has other uses, but in pure general topology we choose to ignore them.