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I'm looking for an example of two metrics that induce the same topology, but so that one metric is complete and the other is not (Since it is known that completeness isn't a topological invariant).

Thanks in advance for any hints or ideas.

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The real line and (0,1) are topologically equivalent but the second is not complete. d(x,y) and d(f(x),f(y)) where f is a topological equivalence (homeomorphism) between a complete and an incomplete space will be two metrics that "induce the same topology but one is complete and the other isn't". – T.. Oct 23 '10 at 17:46
It may be worth noting that there can be no compact counterexample, so open intervals and infinite discrete sets, used in all of the examples so far, are natural places to look. – Jonas Meyer Oct 23 '10 at 19:10

7 Answers 7

up vote 35 down vote accepted

The metric space $\{\frac {1}{n} \mid n\in \mathbb{N} \}$ with the usual metric is incomplete (since we don't have zero), and it has the discrete topology.

the same space with the metric $d(x,y)=1 \iff x\neq y$ also has a discrete topology but is complete, since any cauchy sequence will eventually be constant.

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I like this! This is in some sense a minimal counterexample, since every finite metric space has the discrete topology and is complete. – Qiaochu Yuan Oct 23 '10 at 12:29
There is a lot to be said for minimal counterexamples, but it doesn't provide much insight into why the image of a complete space under a homeomorphism can fail to be complete. You could say many fancy words such as "forgetful functor" and "completely metrisable/uniformisable", but in simple terms if you don't mention the words "uniform continuity", you won't have pointed out the central reason why a homeomorphism can fail to preserve completeness. – kahen Oct 23 '10 at 15:27
+1 ; And this is an example for the fact that nonequivalent metric may determine same topology. – HK Lee Nov 10 at 11:07

The interval $M:=]-1,1[$ with the usual line element $ds:=|dx|$ is an incomplete metric space: The sequence $x_n:=1-{1\over n} \ (n\to\infty)$ converges in $\mathbb R$, so it is a Cauchy sequence, but it diverges in $M$. On the other hand, the "hyperbolic metric" defined by $ds:=|dx|/(1-x^2)$ induces the same topology on $M$ but is complete. The latter statement needs of course a proof. Suffice it here to say that now the endpoints $\pm1$ are "infinitely far away".

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Look for a homeomorphism $f: \mathbb R \to \mathbb R_+$ (you should know a good one). Then take the usual metric on $\mathbb R$ and use it to define a metric $d$ on $\mathbb R_+$ by $d(x,y) = |f(x)-f(y)|$ (*). Now what can you say about the relationship between $d$ and the usual metric on $\mathbb R_+$?

EDIT: Made the question here correct and more illustrative.

So the identity $\iota: (\mathbb R_+,d) \to (\mathbb R_+,|\cdot|)$ is a homeomorphism. Now just ask yourself which property, that continuous functions between metric spaces can have, $\iota$ doesn't have? Can you formulate a theorem about what property a homeomorphism must have to preserve completeness?

(*): Pullback and pushforward might be of interest.

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If $(X,d)$ is a connected, complete metric space and $G$ is a nonempty proper open subset of $X$, then $G$ is not complete with the restricted metric (because it is not closed), but $G$ is homeomorphic to a complete metric space, because the map $x\mapsto(x,1/d(x,X\setminus G))$ embeds $G$ onto a closed subspace of $X\times \mathbb{R}$. (This is more or less Lemma 3.1.1 of Arveson's Invitation.)

A topological space that is homeomorphic to a separable complete metric space is called a Polish space. A subspace of a Polish space is Polish if and only if it is a $G_\delta$, while $G_\delta$s in complete metric spaces are not typically complete with the restricted metric. For example, there is a complete metric inducing the usual topology on the set of irrational numbers.

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Consider one complete normed vector space $(V,\|\cdot\|)$, that is, a Banach space, and let $$B = \{ x \in V : \|x\| < 1 \}$$ be the unit ball centered at the origin. Define $f: V \rightarrow B$ by $x \mapsto x /(1+ \|x\|)$. Then f is continuous and is easy to show that the inverse of $f$ is also continuous, so f is a homeomorphism.

Now $B$ isn't complete, since if $x \in V$ and $\| x \| = 1$, then $x_n = (1-1/n)x$ is a cauchy sequence in $B$ that doesn't converge.

As Kahen referred, if we have two metric spaces $X$ and $Y$ with the latter complete and a uniformly continuous homeomorphism $g : X \rightarrow Y$, then $X$ is complete. To prove this you'll need to observe that since $g$ is uniformly continuous then it sends Cauchy sequences to Cauchy sequences.

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As many others have pointed out, the way to do this is to think about pairs of metric spaces that are homeomorphic but not isometric. In many cases, the metric will be complete on one space but not the other. For example:

  • The set $\{1/n \mid n \in \mathbb{N}\}$ is incomplete, but the natural numbers $\mathbb{N}$ are complete.

  • The ray $(0,\infty)$ and the open interval $(0,1)$ are incomplete, but the real line is complete.

  • An open disc in the plane is incomplete, but the entire plane is complete.

  • The punctured plane $\mathbb{R}-\{0\}$ is incomplete, but the infinite cylinder $\mathbb{R} \times S^1$ (where $S^1$ is the circle) is complete.

  • The real line minus the $x$-axis is incomplete, but a disjoint pair of planes is complete.

  • The twice-punctured plane $\mathbb{R}^2 - \{(-1,0),(1,0)\}$ is incomplete, but the graph of $$ z = \frac{1}{(x^2-1)^2 + y^2} $$ in $\mathbb{R}^3$ is complete. (This graph has asymptotic "cusps" at $(-1,0)$ and $(1,0)$.)

In each case, the homeomorphism between the two spaces can be used to define a nonstandard metric on the latter space that makes it incomplete.

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With the usual metric, $(0,1]$ is not complete.

Let us define another metric $\rho$ on this space using the homeomorphism $\phi: [1,\infty) \to (0,1]$. Define $\rho(x,y) = \vert\frac{1}{x}-\frac{1}{y}\vert$. It is easy to check that $\rho$ is indeed a metric. Also, note that the usual metric and $\rho$ are equivalent and hence induce the same topology.

Now consider a Cauchy sequence in $((0,1],\rho)$.

$\rho(x_n,x_m) < \epsilon$ for all $n \geq N$ or,

$\vert\frac{1}{x}-\frac{1}{y}\vert < \epsilon$ for all $n \geq N$.

Hence $(\frac{1}{x_n})$ is a Cauchy sequence in ($[1,\infty),$usual metric), a complete metric space. Let us say, it converges to $y \in [1,\infty)$. It is easy to check that $(x_n)$ converges to $x = \frac{1}{y}$ in $((0,1],\rho)$.

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