Why is it that $\mathbb{Q}$ cannot be homeomorphic to any complete metric space?

Certainly $\mathbb{Q}$ is not a complete metric space. But completeness is not a topological invariant, so why is the above statement true?

  • 74,358
  • 9
  • 254
  • 333
  • 569
  • 5
  • 3
  • 3
    Don't forget to accept the answer you find most useful (if there is one); click on the checkmark on the left of the answer. If you think none of the answers are enough, then you might want to ask further questions so they can be clarified. – Arturo Magidin Dec 29 '10 at 22:01

5 Answers5


By the Baire Category Theorem, a space that is homeomorphic to a complete metric space must be a Baire Space: the intersection of a countable family of open dense sets must be dense. But $\mathbb{Q}$ does not have this property, because it is countable and no point is isolated: for each $q\in\mathbb{Q}$, let $\mathscr{O}_q = \mathbb{Q}\setminus\{q\}$. This is open (since $\{q\}$ is closed in $\mathbb{R}$, hence in the induced topology of $\mathbb{Q}$) and dense, since every open ball with center in $\mathbb{q}$ intersects $\mathscr{O}_q$. But $$\bigcap_{q\in\mathbb{Q}} \mathscr{O}_q = \emptyset$$ is not dense. Hence $\mathbb{Q}$ is not a Baire space, hence cannot be homeomorphic to a complete metric space (or even to an open subset of a complete pseudometric space).

Arturo Magidin
  • 356,881
  • 50
  • 750
  • 1,081
  • I had an intuitive argument using the Baire Category Theorem; thank you for writing it out explicitly. – Tyler Dec 28 '10 at 03:34
  • Thank you. Any hint how to show Baire is a topological property? – user Dec 28 '10 at 03:50
  • 2
    @user: What exactly are you having trouble with in proving that being a Baire space is invariant under homeomorphisms? Being an open subset and being a dense subset are both invariant under homeomorphisms, and since homeomorphisms are bijections, the image of the intersection is the intersection of the images. – Arturo Magidin Dec 28 '10 at 03:53
  • 5
    Note that just countable is not enough: you need every singleton to be nowhere dense. If a space has isolated points, it can be complete, like $\mathbf{N}$ or a convergent sequence. – Henno Brandsma Dec 28 '10 at 04:14
  • @Henno: Yes, good point; I did not express that very well. For the particular choice I made I need the complement of every singleton to be open and dense, *and* for the space to be countable. – Arturo Magidin Dec 28 '10 at 04:56

A complete metric space is not a countable union of nowhere dense sets, but $\mathbb{Q}$ is (each singleton is nowhere dense). Since nowhere denseness is a topological property, $\mathbb{Q}$ cannot be homeomorphic to a complete metric space.

  • 13,592
  • 5
  • 26
  • 73

There is another way to test whether a metric space is homeomorphic to a complete metric space (that is, whether the space is completely metrizable: there is a complete metric that induces the same topology). This is known as the Choquet game or the strong Choquet game. It was introduced by Choquet in his Lectures on analysis in 1969; a modern treatment can be found in Classical descriptive set theory by Kechris.

The idea is that each metric space is associated with a two-player game of perfect information, of countable length, such that the metric space is homeomorphic to a complete separable metric space if and only if the second player has a winning strategy in the game. All the details and proofs can be found in Kechris' book.

It is easy to show, once you have all the definitions, that the first player has a winning strategy on this game when it is played on $\mathbb{Q}$, which means the second player does not have a winning strategy, so $\mathbb{Q}$ is not completely metrizable. This gives an alternate way to prove that the rationals are not $G_\delta$ in the reals (cf. the answer by Jonas Meyer). First we prove that the rationals are not completely metrizable, by Choquet's method. Then we prove that a metric space is completely metrizable if and only if it is $G_\delta$ in its completion (actually, this is used in the proof of Choquet's result, so we had to prove it already). Since the completion of $\mathbb{Q}$ is $\mathbb{R}$, this shows that $\mathbb{Q}$ is not $G_\delta$ in $\mathbb{R}$.

Carl Mummert
  • 77,741
  • 10
  • 158
  • 292

A subspace of a separable complete metric space is homeomorphic to a complete metric space if and only if it is a $G_\delta$. (I have a reference that says the direction relevant to your question is proved on page 197 of Bourbaki's General topology, volume 2, but I don't currently have access to that page. The other direction is proved on page 196.) The set of rational numbers is not a $G_\delta$ in the real numbers. I must admit that the only way I know how to prove that $\mathbb{Q}$ is not a $G_\delta$ is to appeal to Baire's theorem. (Every open set containing the rationals is dense, and the set of irrationals is a countable intersection of dense open sets, so if the set of rationals were a $G_\delta$, then the empty set (obtained by intersecting the sets of rational and irrational numbers) would be a countable intersection of dense open sets, contradicting Baire's theorem.)

Jonas Meyer
  • 50,123
  • 8
  • 190
  • 292
  • Separability isn't required, although the fact that $G_\delta$ means a *countable* intersection of open sets is required. – Carl Mummert Dec 28 '10 at 16:11
  • @Carl: Thanks. I actually don't know how to prove that if a subspace of a completely metrizable space is completely metrizable then it is a $G_\delta$, and I included the separability hypothesis because I had only seen the statement for the separable case (Polish spaces), and that applies here. I better try to prove it or look it up. – Jonas Meyer Dec 29 '10 at 04:57
  • 1
    The Polish case is definitely the more common one. There is a proof of the general result in Kechris' *Classical descriptive set theory* (Theorem 3.11). It goes via an extension lemma attributed to Kuratowski: a continuous function from a subset $A$ of a complete metric space into a complete metric space can be extended to a continuous function on a $G_\delta$ set $G \supseteq A$. – Carl Mummert Dec 29 '10 at 12:12
  • 1
    I'm *very* late to the party here, but as I stumbled over this... I gave an argument for Kuratowski's characterization for completely metrizable subspaces [in this answer here](http://math.stackexchange.com/questions/19344/polish-spaces-and-the-hilbert-cube/19345#19345) via Lavrentiev's theorem (the extension lemma mentioned by Carl). It is quite similar to what Kechris does but expressed in somewhat different language. I'm not aware of a substantially different argument. – t.b. Aug 15 '11 at 18:32

There is a MISTAKE in the "proof" below the the line. That was the original post. See the discussion that follows. I'm leaving it here as it may be instructive for someone, like it was for me. I'm happy to delete this post if that is considered the best course of action in this community. Please, let me know. Thanks to Asaf Karagila and Nate Eldredge for pointing out the mistake.

My mistake was when I wrote "and more importantly, that $\lim z_{m}=0$", because, for example the numbers $d\left(h(y_{m+p}),h(y_{m}\right)$ may be constant (for all positive integers $p$). Counterexamples have already been sketched in the comments below (thanks, Asaf, Nate). Let me provide a counterexample in more detail: in my setup, if $Y=(−\pi/2,\pi/2)$, $X=ℝ$ and $h(y) = \tan y$, then setting $x_{n} = n$ and $y_{n} = h^{−1}(x_{n})$, we obtain $y_{n}$ that increases monotonically towards $\pi/2$, and is thus Cauchy. But $x_{n}$ is not Cauchy, by construction.

If $h$ were an isometry (as pointed out by Asaf) the "proof" below would be correct. But isometry is more than needed; $h$ being Lipschitz-continuous would be sufficient. In the counter-example above, it is easy to see that $h$ is not Lipschitz (with the problem happening near $\pi/2$ and $-\pi/2$)

This proof is more elementary than the ones presented so far (unless I overlooked something).

Let $X$ and $Y$ be metric spaces with distance functions $d$ and $e$, respectively, and let $h: Y \to X$ be a homeomorphism. Suppose $X$ is complete. We will now show that $Y$ is complete, and hence $Y \neq \mathbb{Q}$.

The idea is simple. Pick an arbitrary Cauchy sequence $y_{n}$ in $Y$. Then the sequence $x_{n} = h(y_{n})$ ought to be Cauchy. Now, this image lives in $X$, and thus it converges to some $x \in X$. It must be the case then that $h^{-1}(x)$ is the limit of $(y_{n})$. I strongly recommend you try to prove this without reading the proof below. The key fact here is that the image of a Cauchy sequence by a continuous function is a bounded set. Nothing fancy.

${\bf Claim:}$ Every Cauchy sequence in $Y$ converges to a point in $Y$.

${\bf\it Proof.}$ Let $(y_{n})$ be an arbitrary Cauchy sequence in $Y$. For every positive integer $n$, define $x_{n} = h(y_{n})$. For all $m \in \mathbb{N}$ define

$$ z_{m} = \sup \{ d(h(y_{m+p}), h(y_{m})) : p \in \mathbb{N} \}. $$

Because $d$ is a distance function, $z_{m}$ is always nonnegative. As $d$ and $h$ are both continuous and $(y_{n})$ is Cauchy, it follows that for all $m$, $z_{m} < + \infty$, and, more importantly, that $\lim z_m = 0$. (If it's not obvious, try to write down a detailed proof of this). It follows that the sequence $(x_{n})$ is Cauchy.

Because $X$ is complete and the Cauchy sequence $(x_{n})$ is in $X$, it must be then that there exists $x \in X$ such that $x_{n} \to x$. Define $y = h^{-1}(x)$. Then, for all $n \in \mathbb{N}$,

$$ d(y_{n}, y) = d(h^{-1}(x_n), h^{-1}(x)) $$

and again, as $n \to \infty$, the right-hand side above goes to zero because $x_n \to x$ and $h^{-1}$ is continuous. It follows that $y_{n} \to y$. The proof is completed by observing that we chose the Cauchy sequence $(y_{n})$ arbitrarily.


  • 4
    Homeomorphism is not necessarily an isometry. The irrationals are homeomorphic to a complete metric space; but certainly there are Cauchy sequences of irrationals which converge to a rational number. – Asaf Karagila Oct 23 '12 at 02:13
  • > Homeomorphism is not necessarily an isometry. – Guilherme Freitas Oct 23 '12 at 02:36
  • 1
    Yes. That is what **I** said. You, on the other hand, wrote in your answer that a Cauchy sequence is ought to be mapped to a Cauchy sequence, which is false. Furthermore, the definition of a complete metric space is a metric space where every Cauchy sequence converges. So I'm not sure what you are trying to prove in that claim. – Asaf Karagila Oct 23 '12 at 02:45
  • Sorry, I didn't know about the 5min to comment rule. I'll reply shortly. – Guilherme Freitas Oct 23 '12 at 02:52
  • 2
    A more obvious counterexample is that $(0,1)$, which is not complete in its usual metric, is homemorphic to $\mathbb{R}$ with its usual metric, which is. – Nate Eldredge Oct 23 '12 at 03:04
  • 1
    Agreed. Bit by the "yeah... this should be obvious". My mistake was when I wrote "and more importantly, that $\lim z_{m} = 0$", because, for example the numbers $d(h(y_{m+p}), h(y_{m})$ may be constant (for all positive integers $p$). Counterexamples have already been mentioned (thanks, Asaf, Nate). More explicitly: in my setup, if $Y = (-1, 1)$, $X = \mathbb{R}$ and $h^{-1}(x) = \frac{x}{1+|x|}$, then setting $x_{n} = n$ and $y_{n} = h^{-1}(x_{n})$, we obtain $y_n$ that increases monotonically towards 1, and is thus Cauchy. But $x_{n}$ is not Cauchy, by construction. Thanks, guys. – Guilherme Freitas Oct 23 '12 at 03:26
  • Plus one. I would like to keep wrong proofs on the site for instructional purposes. – Rudy the Reindeer Oct 23 '12 at 06:48