I want to show topological equivalence of metrics using the definition I've been given. Which is:

If we take two metric spaces on the same set $X$, with metrics $g_1, g_2$ then they are equivalent iff

$\forall x \in X, \forall \epsilon>0, \exists \delta>0$ such that the open balls nest, i.e. $B_{g_1}(x, \delta) \subset B_{g_2}(x, \epsilon)$ and $B_{g_2}(x, \delta) \subset B_{g_1}(x, \epsilon)$.

After attempting to use this definition on the metric spaces, $(X=\mathbb{R}^n, g_1=\|x-y\|_1)$, $(X=\mathbb{R}^n, g_2= \|x-y\|_{\infty})$ I feel like there is something wrong with the definition.

My attempt:

If we take an arbitrary $y \in B_{g_1}(x, \delta)$, then we have that: $$ g_1(x,y) < \delta \implies \sum_{i=1}^n|x_i - y_i| < \delta $$ then if we take any $\epsilon < \delta$, we must also have that $y \in B_{g_2}(x, \epsilon)$ since the maximum is always less than the sum of all the terms as they are all positive.

Conversely, if we take $y \in B_{g_2}(x, \delta)$, then: $$ g_2(x,y) < \delta \implies \max|x_i - y_i| < \delta $$

but it is impossible for $y$ to be in $B_{g_1}(x, \epsilon)$ when $\epsilon < \delta$, since in the best case, we can tale all the other terms to be zero apart from the maximum term?

Am I misunderstanding something or is the definition incorrect?


The definition is almost correct, but you need two $\epsilon$-$\delta$ clauses:

$(X,d_1)$ is equivalent to $(X, d_2)$ iff

$(1) \forall x\in X: \forall \epsilon>0 :\exists \delta >0: B_{d_1}(x,\delta) \subseteq B_{d_2}(x,\epsilon)$ and

$(2) \forall x\in X: \forall \epsilon>0 : \exists \delta >0: B_{d_2}(x,\delta) \subseteq B_{d_1}(x,\epsilon)$

(1) implies that any $d_2$-open subset is also $d_1$-open

and (2) implies that any $d_1$-open subset is also $d_2$-open.

So together they imply the equality of topologies and hence equivalence. They are necessary, as $x$ must be a $d_1$-interior point of the $d_2$-open set $B_{d_2}(x,\varepsilon)$ which is what (1) says and likewise $x$ must be a $d_2$-interior point of the $d_1$-open set $B_{d_1}(x,\varepsilon)$, which is what (2) says.

So two such proofs are needed. See also this answer of mine which also treats the Euclidean case.

  • $\begingroup$ so are you saying that the $\epsilon, \delta$ can be different from case 1 to case 2? $\endgroup$ – dimebucker Apr 21 '18 at 6:03
  • $\begingroup$ Yes, every side can have a different recipe of getting $\delta$ from $\epsilon$. This is the case for the standard metrics on $\mathbb{R}^n$, e.g. @dimebucker $\endgroup$ – Henno Brandsma Apr 21 '18 at 7:14

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