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I know that completeness is not a topological property so that if $d_1$ and $d_2$ are two equivalent metric of $X$ then I can state that $X$ is complete with respect $d_1$ if and only if it is complete with respect $d_2$ by the homomorphy of $(X,d_1)$ and $(X,d_2)$ via the identity map. However if $\cal T_1$ and $\cal T_2$ are two equal topology on $X$ then a sequence converges with respect one of theese if and only if it converges with respect the each other (right?) so that I can conclude that a Cauchy sequence $(x_n)_{n\in\omega}$ converges with respect $d_1$ if and only if it converges with respect $d_2$ so that finally $X$ is complete with respect $d_1$ if and only if it is complete with respect $d_2$.

So first of all I ask if the statement is true and so if it is true then I ask if the argumentation I gave are correct; whereas if the result is not true then I ask a counterexample: could someone help me, please?

Antonio Maria Di Mauro
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    While the notion of a convergent sequence does not depend on the choice of (equivalent) metrics, the notion of Cauchy sequence does. Therefore, your overall argument does not fly. – PhoemueX Oct 16 '22 at 17:37
  • @PhoemueX Okay, so is the statement true? if not, do you know a counterexample? – Antonio Maria Di Mauro Oct 16 '22 at 17:39
  • If $X$ is metrizable and compact then every metric for $X$ is complete. That is not hard. The converse: If $X$ is metrizable and non-compact then $X$ has an incomplete metric, is true, but quite hard. BTW, some non-compact metrizable spaces (e.g. $\Bbb R$ ) do also have complete metrics, but some (e.g.$\Bbb Q$) do not. – DanielWainfleet Oct 16 '22 at 22:38

2 Answers2

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$\mathbb{R}$ is complete with respect to the metric $d(x,y)=|x-y|$ but not with respect to the metric $\rho(x,y)=|\arctan x-\arctan y|$. These metrics are equivalent in the sense that they yield the same topology on $\mathbb{R}$. The sequence $x_n=n$ is not Cauchy with respect to $d$ but it is with respect to $\rho$. The completion of $\mathbb{R}$ w.r.t. $\rho$ is the extended real line $\mathbb{R}=\{-\infty,\infty\}\cup\mathbb{R}$.

Mittens
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  • So precisely $x_n=n$ is a not convergent Cauchy sequence with respect $\rho$, right? – Antonio Maria Di Mauro Oct 16 '22 at 18:25
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    Correct! The metrics are equivalent for they generate the same open sets. Thus if a sequence in the common space converges in $d$ then it also converges in $\rho$ and viceversa. – Mittens Oct 16 '22 at 18:33
  • Okay, I upvoted the question. Before to decided what to approve I would like to attend Paul's answer to my comment. Anyway thanks very much for your assistance!!! – Antonio Maria Di Mauro Oct 16 '22 at 19:34
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    @AntonioMariaDiMauro: there is an interesting result that states that if $(X,d)$ is metric space, separable and $d$ is complete (such spaces are called Polish spaces), then any open set $U\subset X$ with the topology inherited from $X$ admits a metric that makes $U$ also a Polish space. – Mittens Oct 16 '22 at 19:46
  • Oh, fantastic! And this can be "married" to my question? – Antonio Maria Di Mauro Oct 16 '22 at 19:50
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    Well this about your favorite Euclidean space, say $\mathbb{R}$ with the standard metric. Then $(0,1)$ admits a metric $d_0$ which is equivalent to the standard metric but under $d_0$, $(0,1)$ is complete: $d_0(x,y)=|x-y|+\Big|\frac{1}{\operatorname{dist}(x,(0,1)^c)}-frac{1}{\operatorname{dist}(y,(0,1)^c)}\Big|$ – Mittens Oct 16 '22 at 19:52
  • Oh, yeah: very fine!!! – Antonio Maria Di Mauro Oct 16 '22 at 19:53
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It is not true.

If we endow all subspaces of $\mathbb R$ with the Euclidean metric $d_e(x,y) = \lvert x - y \rvert$ (which means that they receive their standard topology), then $(\mathbb R,d_e) $ is complete, but $((0,\infty),d_e)$ is not.

The exponential map $\exp : \mathbb R \to (0,\infty)$ is a homeomorphism. Define a metric $d$ on $(0,\infty)$ by $d(x,y)= \lvert \exp(x)- \exp(y) \rvert$. Since $\exp$ is a homeomorphism, the metric $d$ induces the standard topology on $\mathbb R$ which means that $d$ is equivalent to $d_e$. However, the metric space $(\mathbb R, d)$ is not complete. In fact, the sequence $x_n = -n$ is a Cauchy sequence because with $r = \min(n,m)$ we get $$d(x_n,x_m) = \lvert \exp(-n)- \exp(-m) \rvert = \lvert e^{-n}- e^{-m} \rvert = e^{-r}\lvert e^{-n+r}- e^{-m+r} \rvert \\ \le e^{-r}(e^{-n+r}+ e^{-m+r}) \le 2 e^{-r} .$$ But obviously $(x_n)$ does not converge to any $\xi \in \mathbb R$.

Update:

The equivalence of $d$ and $d_e$ follows from the following general observation:

  • Let $(X_1,d_1)$ and $(X_2,d_2)$ be metric spaces with metric topologies $\tau_1, \tau_2$ and let $h : (X_1,\tau_1) \to (X_2,\tau_2)$ be a homeomorphism. Then $d'_1(x,y) = d_2(h(x),h(y))$ defines a metric on $d'_1$ on $X_1$ which is equivalent to $d_1$.

That $d'_2$ is a metric is easily verified (the proof works for any injective, not necessarily continuous, function $h : X_1 \to X_2$). The function $h$ is by definition an isometry $h :(X_1,d'_1) \to (X_2,d_2)$, in particular a homeomorphism $h : (X_1, \tau'_1) \to (X_2,\tau_2)$. This shows that $$id = h^{-1} \circ h : (X_1, \tau_1) \stackrel{h}{\to} (X_2,\tau_2) \stackrel{h^{-1}}{\to} (X_1, \tau'_1)$$ is a homeomorphism which means that $\tau_1 = \tau'_1$.

Let us next observe

  • If $(X_1,d'_1)$ is complete iff $(X_2,d_2)$ is complete.

This is obvious because $h$ is an isometry $h :(X_1,d'_1) \to (X_2,d_2)$.

In my example the metric space $((0,\infty),d_e)$ is not complete since each sequence $(\xi_n)$ in $(0,\infty)$ converging to $0$ when regarded as a sequence in $\mathbb R$ is a non-convergent Cauchy-sequence in $((0,\infty),d_e)$ since $0 \notin (0,\infty)$. Thus also $(x_n) = (\exp^{-1}(\xi_n)) = (\ln \xi_n)$ is a non-convergent Cauchy-sequence in $(\mathbb R,d)$. In my example I took $\xi_n = e^{-n}$ which gives $x_n = -n$.

Paul Frost
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  • Okay, only one request: could you exhibit a Cauchy sequence with respect $d'$ which not converges, please? Forgive the bother. – Antonio Maria Di Mauro Oct 16 '22 at 17:50
  • @AntonioMariaDiMauro See my update. – Paul Frost Oct 16 '22 at 22:51
  • Sorry, but I do not understand completely the last inequality: precisely I did not understand why the inequality $$e^{-r}(e^{-n+r}+e^{-m+r})\le e^r$$ holds: then why this implies that $$ d(x_h,x_k)\le\epsilon $$ for any $h,k\ge r$? Perhaps did you inted to wrote $$e^{-r}(e^{-n+r}+e^{-m+r})\le 2 e^{-r}$$ instead? Indeed for any $\epsilon\in\Bbb R^+$ there exists $n_0\in\Bbb N$ such that $$ e^{-n_0}\le\frac{\epsilon}2 $$ so that the inequality follows observing that $$ e^l\le e^{n_0} $$ for any $l\ge n_0$. – Antonio Maria Di Mauro Oct 17 '22 at 07:43
  • Moreover, since the exponential is continuous then for any $\epsilon\in\Bbb R^+$ there exists $\delta_\epsilon\in\Bbb R^+$ such that $$|e^x-e^y|<\epsilon$$ for any $y\in(x-\delta_\epsilon,x+\delta_\epsilon)$ which proves that $$(x-\delta_\epsilon,x+\delta_\epsilon)\subseteq B(x,\epsilon)$$ so that then inclusion $$\mathcal T_d\subseteq\mathcal T_e$$ holds. However now I am not able to prove the reverse inclusion. So could you explain better, please? Forgive my confusion. – Antonio Maria Di Mauro Oct 17 '22 at 07:43
  • @AntonioMariaDiMauro Concerning $e^r$ you are right. For the other question see my update. – Paul Frost Oct 17 '22 at 08:59
  • @AntonioMariaDiMauro Oliver Diaz' answer uses the homeomorphism $\arctan : \mathbb R \to (-1,1)$. – Paul Frost Oct 17 '22 at 09:08
  • Okay, all clear now: question upvoted and approved. Thanks very much for your assistance!!! See you soon...;-) – Antonio Maria Di Mauro Oct 17 '22 at 11:23
  • Hi professor, perhaps here I found an answer to a question I posed: could I ask your opinion about its correctness? – Antonio Maria Di Mauro Oct 17 '22 at 19:08
  • @AntonioMariaDiMauro It is correct. See my answer. – Paul Frost Oct 18 '22 at 00:23