# Are strongly equivalent metrics mutually complete?

Maybe I'm missing something, but I can't seem to find any references to my exact question. If two metrics, $d_1(x,y)$ and $d_2(x,y)$ are strongly equivalent, then there exists two positive constants, $\alpha, \beta$, such that

$$\alpha d_1(x,y) \leq d_2(x,y) \leq \beta d_1(x,y).$$

My question is: if one metric is complete, must the other be complete as well? I suspect no, but I haven't seen any explicit counterexamples.

Conversely, if you have one metric that is complete, and another metric that is not, does this imply that the two cannot be strongly equivalent?

Let $X_i = (X,d_i)$. Assume without loss that $X_1$ is complete and let $\{x_n\}$ be a Cauchy sequence in $X_2$, i.e. for every $\epsilon >0$ there is $N$ such that for $n,m \ge N$ we have $d_2(x_n,x_m) \le \epsilon$. If we now apply the inequality that we assumed on the distance functions we get $$d_1(x_n,x_m) \le \frac{1}{\alpha}d_2(x_n,x_m) \le \frac{1}{\alpha}\epsilon.$$ This shows that $\{x_n\}$ is a Cauchy sequence in the complete metric space $X_1$, hence it admits a limit $x \in X$, or in other words for every $\eta > 0$ there is $M$ such that if $n \ge M$ then $$d_1(x,x_n) \le \eta.$$We can finally apply the other branch of the inequality to show that the sequence $\{x_n\}$ converges to $x$ with respect to $d_2$ as well. Indeed for $n\ge M$ we have $$d_2(x,x_n) \le \beta d_1(x,x_n) \le \beta\eta.$$ This proves that every Cauchy sequence in $X_2$ is convergent, which by definition gives the completeness of $X_2$.$\blacksquare$

As I am writing this I have noticed that Daniel Fischer has already posted an answer. I am going to post mine anyway since I think it offers a slightly different point of view.

• Thanks! Both answers are great, but I really love a nice explicit epsilon proof. :) – Patch Aug 25 '15 at 21:21

Such metrics are also called Lipschitz equivalent, the inequality states that the identity map from the space endowed with one metric to the space endowed with the other is a Lipschitz map (in both directions). That ensures that the space is complete in one metric if and only if it is complete in the other metric.

In fact, that holds for a weaker concept of equivalence, uniform equivalence of metrics. Two metrics $d_1,d_2$ on a space $X$ are uniformly equivalent if the identity map $(X,d_i) \to (X,d_j)$ is uniformly continuous for both choices of $\{i,j\} = \{1,2\}$. With formulae, the metrics are uniformly equivalent if for every $\varepsilon > 0$ there are $\delta_1, \delta_2 > 0$ such that

$$\bigl(d_1(x,y) \leqslant \delta_1 \implies d_2(x,y) \leqslant \varepsilon\bigr) \land \bigl(d_2(x,y) \leqslant \delta_2 \implies d_1(x,y) \leqslant \varepsilon\bigr).$$

Since uniformly continuous maps - in particular Lipschitz continuous maps - map Cauchy sequences to Cauchy sequences, uniform equivalence of the metrics implies that a sequence is a $d_1$-Cauchy sequence if and only if it is a $d_2$-Cauchy sequence. Since uniformly equivalent metrics (and hence in particular Lipschitz equivalent metrics) induce the same topology (i.e. they are topologically equivalent), a sequence is $d_1$-convergent if and only if it is $d_2$-convergent.

Hence, if the metrics are uniformly equivalent, the space is complete in one metric if and only if it is complete in the other. And by contrapositive, if you have two metrics on a space such that the space is complete in one but not in the other, the two metrics aren't uniformly equivalent, and a fortiori not strongly equivalent.

• Thank you! This is very helpful, and addresses some other important points. – Patch Aug 25 '15 at 21:23

It's fairly easy to see that equivalent metrics define the same topology on the space (i.e. the open sets are the same); also the Cauchy sequences in $(X,d_1)$ and $(X,d_2)$ are also the same. Suppose $(X,d_1)$ is complete, then any Cauchy sequence $\{x_n\}$ has a limit $x$ in $X$. This means that for an arbitrarily small $\epsilon >0$, there is an open ball $B$ with radius $\epsilon$ and a sufficiently large integer $N$ such that $B$ contains $x$ and $x_n$ for all $n>N$. Since the open sets are the same in $(X,d_2)$, there is an open ball $B'$ with a different (albeit arbitrarily small) radius $\epsilon'$ that contains $x$ and $x_n$ for all $n>N$, so $x$ is also a limit of $\{x_n\}$ in $(X,d_2)$, so $(X,d_2)$ is complete. Similarly the completeness of $(X,d_2)$ implies the completeness of $(X,d_1)$.