Does homeomorphic metric spaces preserves completeness?I mean two metric space which are homeomorphic and one of them is complete$\Rightarrow$ another one is also complete?


No. $\mathbb{R}$ and $(0,1)$ (with their usual metrics) are homeomorphic, but one is complete and the other isn't.

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    $\begingroup$ I will add the remark that the homeomorphism shows that the spaces are completely metrizable (that is, there is a metric which generates the same topology and it is complete). Such metric can be defined via the homeomorphism function by a transfer of structure. $\endgroup$ – Asaf Karagila Apr 20 '12 at 15:22

The answer is no... and a bit yes. No, because, the homeomorphism by itself is a topological notion and does not say anything about the choice of the metric on the other space, and the completeness of the other space may depend on the choice of its metric. Yes, because if a space is homeomorphic to a complete metric space then it is metrizable, and there exists at least one complete metric on it.

In fact, you do not need 2 homeomorphic metric spaces: one topological space may have two different metrics inducing its topology such that the space is complete under one of them and not complete under the other.

Moreover, there is a specific term for this: a metrizable topological space is called complete-metrizable, if there is at least one complete metric inducing its topology.

Given a metrizable topological space $\mathbb X=(X,\mathcal T)$, and $\mathbb{Y}=(Y,\mathcal{T}')$ homeomorphic to it, we have 3 cases:

  1. $\mathbb X$ is not complete under any metric (not complete-metrizable). Then $\mathbb{Y}$ is not complete-metrizable either. In this case the best one can hope is to complete the space under some metric. The completion, of course, depends on the metric. For example, $\mathbb{Q}$ is not complete-metrizable. The completion of $\mathbb{Q}$ under the standard metric gives you real numbers. Besides the standard metric on $\mathbb{Q}$ there is another class of metrics, the completion under which leads to so-called $p$-adic numbers.

  2. $\mathbb X$ is complete under some metrics, but not complete under others. Then $\mathbb{Y}$ is complete-metrizable but not complete under some metrics inducing its topology. For example, the discrete topology on a countable set.

  3. $\mathbb X$ is complete under every metric. Then, so is $\mathbb{Y}$. For example, a finite set in discrete topology.


The answer is not always. If a space is homeomorphic to a complete metric space we say that it is completely metrizable, namely there exists a metric which is complete and generates our given topology. We know that a space is completely metrizable if and only if it is a $G_\delta$ set (intersection of countably many open sets) in every complete metric space embedding it, in particular in its metric completion.

However if the metric space is compact then it is always complete. To see this note that a Cauchy sequence in a compact metric space is an infinite set and therefore has a limit point, however being a Cauchy sequence it can have at most one limit point - the limit of the sequence. Therefore every Cauchy sequence is convergent and thus the metric is complete.


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