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

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?

Solution 1:

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).

Solution 2:

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.

Solution 3:

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 $\mathbb{R}$.