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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.

user119615
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1 Answers1

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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 $\Omega$.

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Brian M. Scott
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  • 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? – user119615 Sep 04 '15 at 20:12
  • @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 ... – Brian M. Scott Sep 04 '15 at 20:18
  • ... on the specific metric generating the topology. This is in contrast to completeness, which is a property of the metric. – Brian M. Scott Sep 04 '15 at 20:18
  • Thank you, I have to think about this, but interesting that you incorporated topology in the answer, I am trying to learn that aswell. – user119615 Sep 04 '15 at 20:20
  • @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. – Brian M. Scott Sep 04 '15 at 20:21
  • @Matt: You can modify the proof of the first one to show that the countable union of nowhere dense sets in a complete metric space is nowhere dense, in which case the equivalence is trivial. If you simply want to quote the stated version, however, you have to work a bit harder. – Brian M. Scott Sep 04 '15 at 21:10