# Baire's Theorem: Examples for open dense subsets

Theorem (Baire): Let $$(X,d)$$ be a complete metric space and $$(D_n)_{n \in \mathbb{N}}$$ a family of open dense subsets of $$X$$. Then $$\bigcap_{n \in \mathbb{N}} D_n$$ is also dense in $$X$$.

This is the first theorem we proved in our functional analysis lecture and we were told it's essential for the proofs of fundamental theorems of functional analysis.

My question is simpler: I'm struggling to find open dense subsets in a "normal" metric space.

@copper_hat's comment here suggests uninting $$\varepsilon$$-balls with rational centres but are there other good examples?

Fix $$n \in \mathbb{N}$$. Take $$(S_i)_{i \in \mathbb{N}}$$ affine subspaces i.e. $$S_i = V_i + p_i$$ with $$V_i$$ a linear subspace and $$p_i$$ a point. Then each $$S_i$$ is closed and has no interior (this is a fun excersice of linear algebra / functional analysis). Thus each $$S_i^c$$ is open and moreover, it is dense:

$$\overline{S_i^c} = (S_i^°)^c = \emptyset^c = X.$$

Thus $$\bigcap_{i \geq 1}S^c_i$$ is dense. In the former calculation, we showed that the complement of a closed set with no interior is open and dense. Note that the converse also holds, thus giving the following equivalent formulation of Baire's theorem,

Theorem(Baire): let $$X$$ be a complete metric space and $$(F_n)_{n \geq 1} \subset X$$ be a sequence of closed sets in with empty interior. Then $$\bigcup_{n \geq 1} F_n$$ has empty interior.

In particular, this says that we can't write $$X$$ as the countable union of closed sets with no interior. Thus, in this setting, let's rephrase the original comment: each affine subspace is closed and has no interior, thus their union cannot be $$\mathbb{R}^n$$. This proves:

Proposition. An euclidean space cannot be written as a countable union of affine subspaces.

In general this is the type of argument Baire's theorem is used for. Another useful consequence is that if $$X$$ is written as a countable union of closed sets, then some of them has to have non-empty interior. This for example allows us to prove the principle of uniform boundedness, one of the key results you will study.

As a final remark, the example given here lets you prove that

Proposition. Given $$n,m \geq 1$$ there is a set $$A \subset \mathbb{R}^n$$ of size $$m$$ such that any subset of $$n+1$$ points of $$A$$ is affinely independent.

There are quite simpler approaches, but let's go with the one that uses Baire.

Proof. We can fix $$n$$ and use induction on $$m$$. The base case is trivial. As for the inductive step, we have $$A$$ of size $$m$$ with all its subsets of size $$n+1$$ affinely independent. It suffices to see that we can add a point to $$A$$ such that this keeps holding. Suppose not: then, for any point $$p \in \mathbb{R}^n$$ we pick, $$A \cup \{p\}$$ must contain an affinely dependent subset $$S_p \subset A \cup \{p\}$$ of $$n+1$$ elements. Note also that $$p \in S$$, because otherwise $$S \subset A$$ and thus $$S$$ would be affinely independent by inductive hypothesis. If we write $$S_p = T_p \cup \{p\}$$, then $$S_p$$ being affinely dependent says that in particular, $$p$$ is contained in the affine subspace generated by the points of $$T_p$$. To sum up: for any point $$p$$, there is a subset $$T \subset A$$ such that $$p \in V_T$$ where $$V_T$$ is the (affine) subspace generated by $$V$$. In particular,

$$\mathbb{R}^n \subset \bigcup_{T \subset A, |T| = n}V_T$$

which is absurd because the right hand side is a countable (it is finite, even, as $$A$$ is finite) union of closed sets of empty interior, and by Baire's theorem this implies it has empty interior. The contradiction comes from supposing that we can't add a point to $$A$$ without it losing its properties. Hence a point can be added to $$A$$, which is now of size $$m+1$$. This completes the inductive step (and the proof) $$\square$$.

Let $$a\in A$$, then $$X\setminus \{a\}$$ is an open dense subset (as far as singleton are closed).

• Note that this remark comes in handy if you want to prove that a countable dense set (like $\mathbb{Q}$ in $\mathbb{R}$) cannot be written as a countable intersection of open sets – charmd Apr 10 at 18:53
• This is true iff $a$ is not an isolated point of $X$. – Henno Brandsma Apr 11 at 5:45