Theorem: (Baire)

$(X,d)$ is a complete metric space then the intersection of countably many dense, open subsets in the metric topology $\mathcal{T}$ generated by $d$ is dense

In other words,

$$D = \bigcap_{n \in \mathbb{N}} D_n$$ where $D_n$ is a dense, open subset of $X$, is dense

I am asked to prove this raw, no hints whatsoever. I am already hitting my head on several bricks

Idea/Approach: By contradiction, assume that $D$ defined above is not dense, then $(X,d)$ is not complete.

Proof Attempt: (No idea how to start! We will start by stating what we want)

  • We wish to construct a Cauchy sequence $(x_n)$ on $X$, then by completeness of $X$, $x_n \to x$ as $n \to \infty$. Afterwards, we wish to obtain a contradiction such that if $D = \bigcap_{n \in \mathbb{N}} D_n$ is not dense, then $x_n \not\to x$.

(Difficult part is to find out how $D_n$ is related to $x_n$...)

  • Let $\{D_n\}$ be a set of dense, open subset of $X$, where $n \in \mathbb{N}$. By definition, for all $U \in \mathcal{T}$, $D_n \cap U \neq \varnothing$.

  • Let $D = \bigcap_{n \in \mathbb{N}} D_n$ and suppose for contradiction there exists $U \in \mathcal{T}$ such that $D \cap U = \varnothing$ (i.e. $D$ is not dense). Then let $x_n \in D_n \cap U, \forall n \in \mathbb{N}$

  • Then $(x_n)$ is a Cauchy sequence if $\forall \epsilon > 0, \exists N \in \mathbb{N}$, such that $d(x_n, x_m) < \epsilon, \forall n, m > N$

    (At this point it is obvious that without additional assumptions, there is no way $(x_n)$ is Cauchy)


Does anyone see if this approach can still be continued? Any help is appreciated at this point.

  • 1
    $\begingroup$ en.wikipedia.org/wiki/Baire_category_theorem#Proof $\endgroup$ – avs Jul 19 '16 at 0:37
  • $\begingroup$ Note that completion is a reference to se uence limits and you have already produced a se uence of dense subsets. Perhaps you could produce se uences of points , using each dense subset. $\endgroup$ – Jacob Wakem Jul 19 '16 at 0:43
  • $\begingroup$ @BeachedWhale It seems very likely that you are the creator of (metric-topology) tag. I wanted to let you know that I have made a post on meta to discuss this new tag. $\endgroup$ – Martin Sleziak Aug 1 '16 at 10:08
  • $\begingroup$ @MartinSleziak Hi! Yes I like to create new tags at whim, such as Arzela Ascoli. I created metric topology because this is now used a lot in formal teaching as people confuse metric space with topological space and vice versa, mainly because real analysis and topology are taught separately. For example, most people are not introduced to formal definition of topology in real analysis, therefore all intuition about convergence of sequences etc are wrt metric topology, but does not hold in general topology. Perhaps metrizable spaces is more appropriate. But do whatever as you wish to the tag! $\endgroup$ – Carlos - the Mongoose - Danger Aug 1 '16 at 10:47
  • $\begingroup$ Thanks for the response. My impression is that the tag (metric-spaces) is typically used for questions related to metric space, metrizable spaces, topology derived from a metric etc. In any case, if we want to continue this discussion, it would be probably more suitable on meta or in chat. $\endgroup$ – Martin Sleziak Aug 1 '16 at 12:24

Your approach can be carried out; you just have to figure out how to choose the points $x_n$ so that $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence. I’ll get you started.

For $x\in X$ and $r>0$ let $C(x,r)=\{y\in X:d(x,y)\le r\}$, the closed ball of radius $r$ centred at $x$.

Begin by choosing any $x_0\in U\cap D_0$; there is an $r_0>0$ such that $C(x_0,r_0)\subseteq U\cap D_0$. The open ball $B(x_0,r_0)$ intersects $D_1$, so choose $x_1\in B(x_0,r_0)\cap D_1$. There is an $r_1>0$ such that $C(x_1,r_1)\subseteq B(x_0,r_0)\cap D_1$ and $r_1\le\frac12r_0$. (Why?) Similarly, $B(x_1,r_1)\cap D_2\ne\varnothing$, so there is an $x_2\in B(x_1,r_1)\cap D_2$, and there is then an $r_2>0$ such that $C(x_2,r_2)\subseteq B(x_1,r_1)\cap D_2$ and $r_2\le\frac12r_1$. (Again, why?) Now try to complete the blockquoted sentence below that describes how to keep going in this fashion.

Given $x_n\in U$ and $r_n>0$, we know that $B(x_n,r_n)\cap D_{n+1}\ne\varnothing$, so there is an $x_{n+1}\in B(x_n,r_n)\cap D_{n+1}$, and there is then an $r_{n+1}>0$ such that ...

In the end we have a sequence $\langle x_n:n\in\Bbb N\rangle$ such that each $x_n\in U$.

  • Show by induction on $n$ that $x_n\in\bigcap_{k\le n}B(x_k,r_k)$ for each $n\in\Bbb N$.
  • Use this to show that $\langle x_n:n\in\Bbb N\rangle$ is a Cauchy sequence and therefore converges to some $x\in X$.
  • Show that for each $n\in\Bbb N$, $x_n\in\bigcap_{k\le n}D_k$.
  • Use this to show that $x\in\bigcap_{n\in\Bbb N}D_n$.

Note that a proof by contradiction is not needed: the argument shows directly that $\bigcap_{n\in\Bbb N}D_n$ meets every non-empty open $U\subseteq X$.

  • $\begingroup$ Yeah yours is pretty much the same as I eventually did. Although I am good at being terse. $\endgroup$ – Jacob Wakem Jul 19 '16 at 2:42
  • $\begingroup$ I think I understand, more tedious than I thought however, the amount of things you need to check $\endgroup$ – Carlos - the Mongoose - Danger Jul 19 '16 at 3:06
  • $\begingroup$ @BeachedWhale check out my approach in my revised answer. I cut out the fat. $\endgroup$ – Jacob Wakem Jul 19 '16 at 3:14
  • 1
    $\begingroup$ @Jacob: I would say that while your answer provides the intuition, it falls far short of being a proof. That is, what you cut was the actual proof, leaving only the basic idea behind it. $\endgroup$ – Brian M. Scott Jul 19 '16 at 3:16
  • $\begingroup$ @BrianM.Scott I mean it is still readable. Stuff like "much smaller" is easily formalizable (I'm sure it has already been suitably defined by someone and I use that definition freely) and my infinitary sentence can be put into set builder notation in the usual way. $\endgroup$ – Jacob Wakem Jul 19 '16 at 3:33

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.