The book says that statement 2 is a direct consequence of statement 1. I don't see how they prove statement 2 directly from statement 1, can you please help me?

Statement 1:

A complete metric space $(\Omega,\rho)$ is not the union of a countable collection of nowhere dense sets.

Statement 2:

Let $\{G_n\}_{n=1}^\infty$ be a sequence of dense open subsets of a complete metric space. Then $\cap_{n=1}^\infty G_n$ is dense.

So I struggle to see $1\rightarrow 2$. What I do know is that since each $G_n$ is dense, then $G_n^c$ is nowhere dense.(A set is nowhere dense iff the complement og its closure is dense, an $G_n^c$ is already closed, since $G_n$ is open). Hence I do know that $\cap_{n=1}^\infty G_n$ is nonempty, since it's complement is a countable union of nowhere dense sets, and from statement 1, this can't be the whole set.

But I only get that it has atleast 1 element, that is a long way from proving that it is dense.


1 Answer 1


Use the fact that an open subset of a complete metric space is homeomorphic to a complete metric space. Let $U$ be a non-empty open subset of $\Omega$. Then the subspace $U$ has a complete metric $d$ that generates the subspace topology. For $n\in\Bbb N$ let $H_n=G_n\cap U$; each $H_n$ is a dense, open subset of $U$, so by your argument $\bigcap_{n\in\Bbb N}H_n\ne\varnothing$. Thus,

$$\varnothing\ne\bigcap_{n\in\Bbb N}H_n=\bigcap_{n\in\Bbb N}(G_n\cap U)=U\cap\bigcap_{n\in\Bbb N}G_n\;,$$

and $\bigcap_{n\in\Bbb N}G_n$ is dense in $\Omega$.

  • $\begingroup$ Thank you, I know very little topology, so I just have some follow up. I know that a subspace topology is defined has $\{U\cap V\}$, where V is any open set in the bigger topology. And is it this the topology the new metric d creates?, and why do you need a new metric?, we allready have a metric, and are we not supposed to show that our set is dense when considered with that metric? $\endgroup$
    – user119615
    Sep 4, 2015 at 20:12
  • $\begingroup$ @user119615: Yes, $d$ generates the subspace topology that you’ve described. We need a new metric because in general the subspace $U$ won’t be complete in the metric $\rho$, and to apply the Baire category theorem directly, we need a metric in which it’s complete. Fortunately, different metrics can generate the same topology, and the theorem to which I linked ensures that there is a complete metric on $U$ that generates the right topology. Finally, denseness is a purely topological notion, not a metric notion: whether a set is dense in $\Omega$ depends only on the topology of $\Omega$, not ... $\endgroup$ Sep 4, 2015 at 20:18
  • $\begingroup$ ... on the specific metric generating the topology. This is in contrast to completeness, which is a property of the metric. $\endgroup$ Sep 4, 2015 at 20:18
  • $\begingroup$ Thank you, I have to think about this, but interesting that you incorporated topology in the answer, I am trying to learn that aswell. $\endgroup$
    – user119615
    Sep 4, 2015 at 20:20
  • $\begingroup$ @user119615: You’re welcome. The Baire category theorem really is fundamentally a topological theorem; it just happens that one form of it deals with completely metrizable spaces, i.e., topological spaces that admit a complete metric. $\endgroup$ Sep 4, 2015 at 20:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.