Section 6.1 The Theorem
Theorem 6.1. Baire category theorem.
Let \((X,d)\) be a complete metric space. If \((G_n)_{n \in \N}\) is a sequence of dense, open subsets of \(X\text{,}\) then \(\bigcap_{n \in \N} G_n\) is dense in \(X\text{.}\)
Proof.
Let \(p \in X\) and \(\varepsilon \gt 0\) be arbitrary. It suffices to find a point \(q \in \bigcap_{n \in \N} G_n\) with \(d(p, q) \lt \varepsilon\text{.}\)
Set \(r_0 = \varepsilon/2\) and \(x_0 = p\text{.}\) As \(G_1\) is dense in \(X\text{,}\) there exists \(x_1 \in G_1 \cap B_{r_0}(x_0)\text{.}\) Note that \(G_1 \cap B_{r_0}(x_0)\text{,}\) being the intersection of two open sets, is still open by Theorem 1.25. Therefore, there exists \(r_1 \gt 0\) such that \(B_{2r_1}(x_1) \subseteq G_1 \cap B_{r_0}(x_0)\) and \(r_1 \lt r_0/2\text{.}\)
As \(G_2\) is dense in \(X\text{,}\) there exists \(x_2 \in G_2 \cap B_{r_1}(x_1)\text{.}\) The set \(G_2 \cap B_{r_1}(x_1)\) is open and so there exists \(r_2 \gt 0\) such that \(B_{2r_2}(x_2) \subseteq G_2 \cap B_{r_1}(x_1)\) and \(r_2 \lt r_1/2\text{.}\)
Proceeding recursively, we obtain a sequence \((x_n)_{n \in \N}\) in \(X\) and a sequence \((r_n)_{n \in \N}\) of positive numbers such that \(B_{2r_n}(x_n) \subseteq G_n \cap B_{r_{n - 1}}(x_{n - 1})\) and \(r_n \lt r_{n - 1}/2\) for \(n=1, 2, \dotsc\text{.}\) The construction guarantees that
\begin{equation*}
B_{r_n}(x_n) \subseteq B_{2r_n}(x_n) \subseteq B_{r_{n - 1}}(x_{n - 1}) \subseteq \dotsb \subseteq B_{r_0}(x_0) \subseteq B_{2r_0}(x_0) = B_\varepsilon(p).
\end{equation*}
Hence for any \(m, n \in \N\text{,}\) if \(m \gt n\text{,}\) then \(x_m \in B_{r_n}(x_n)\) and thus
\begin{equation*}
d(x_m, x_n) \lt r_n \lt \frac{r_0}{2^n}\text{.}
\end{equation*}
It follows that \((x_n)_{n \in \N}\) is a Cauchy sequence. By the completeness of \(X\text{,}\) there exists a limit \(q = \lim_{n \to \infty} x_n\text{.}\)
The triangle inequality implies that
\begin{equation*}
d(q, x_n) \le d(q, x_m) + d(x_m, x_n) \lt d(q, x_m) + r_n
\end{equation*}
for all \(m, n \in \N\) with \(m \gt n\text{.}\) Letting \(m \to \infty\text{,}\) we conclude that
\begin{equation*}
d(q, x_n) \le r_n\text{.}
\end{equation*}
In particular, this means that \(q \in B_{2r_n}(x_n) \subseteq G_n\) for every \(n \in N\text{.}\) That is,
\begin{equation*}
q \in \bigcap_{n \in \N} G_n.
\end{equation*}
Moreover, we see that \(d(p, q) = d(x_0, q) \le r_0 = \varepsilon/2 \lt \varepsilon\text{,}\) which concludes the proof.
There is an alternative formulation of the theorem involving the following notion.
Definition 6.2.
Let \((X,d)\) be a metric space. A set \(Y \subseteq X\) is nowhere dense if \((\overline{Y})^\circ = \varnothing\text{.}\)
Here \((\overline Y)^\circ\) denotes the interior of the closure of \(Y\text{;}\) see Definition 1.20.
Example 6.3.
The set \(\Z \subset \R\) is nowhere dense in \(\R\text{,}\) since \(\overline{\Z} = \Z\) and \(\overline{\Z}^\circ = \Z^\circ = \varnothing\text{.}\) In contrast, the set \(\Q\) is not nowhere dense in \(\R\text{,}\) as \(\overline{\Q}^\circ = \R^\circ = \R\text{.}\)
Lemma 6.4.
Let \((X,d)\) be a metric space. A subset \(Y \subseteq X\) is nowhere dense if, and only if, the set \(X \setminus \overline{Y}\) is dense in \(X\text{.}\)
Proof.
The condition \(\overline{Y}^\circ = \varnothing\) means that for any \(x \in X\) and any \(r \gt 0\text{,}\) there exists \(y \in B_r(x) \setminus \overline{Y}\text{.}\) This is equivalent to the condition \(\overline{X \setminus \overline{Y}} = X\text{,}\) which in turn means that the set \(X \setminus \overline{Y}\) is dense in \(X\text{.}\)
Corollary 6.5.
Suppose that \((Y_n)_{n \in \N}\) is a sequence of nowhere dense subsets of a complete metric space \((X,d)\text{.}\) Then \(X \setminus \bigcup_{n\in\N} Y_n\) is dense in \(X\text{.}\)
Proof.
Define \(G_n = X \setminus \overline{Y_n}\text{.}\) Then each \(G_n\) is open and dense in \(X\) by Lemma 6.4. Hence by Theorem 6.1, the set \(\bigcap_{n \in \N} G_n\) is dense. But
\begin{equation*}
\bigcap_{n \in \N} G_n
= \bigcap_{n \in \N} (X \setminus \overline{Y_n})
= X \setminus \bigcup_{n\in\N} \overline{Y_n}
\subseteq X \setminus \bigcup_{n\in\N} Y_n.
\end{equation*}
Therefore, we conclude that \(X \setminus \bigcup_{n\in\N} Y_n\) contains a dense set, which means that it must itself be dense.
Intuitively, a nowhere dense subset of a metric space may be thought of as a set that does not contain very much of the space. The following corollary asserts that a non-empty, complete metric space cannot be covered by a sequence of such sets.
Corollary 6.6.
A non-empty, complete metric space is not a countable union of nowhere dense subsets.
Proof.
If \((X,d)\) is a complete metric space and \((Y_n)_{n \in \N}\) is a sequence of nowhere dense subsets of \(X\text{,}\) then by Corollary 6.5, the set \(X \setminus \bigcup_{n\in\N} Y_n\) is dense in \(X\text{.}\) If \(X \ne \varnothing\text{,}\) it follows that \(\bigcup_{n\in\N} Y_n \ne X\text{.}\)
The theorem has some deep consequences for familiar objects.
Corollary 6.7.
The space \(\R\) is uncountable.
Proof.
The usual metric on \(\R\) makes it a non-empty, complete metric space. Any singleton (i.e., any set with one element) in \(\R\) is nowhere dense. Hence Corollary 6.6 implies that \(\R\) is not a countable union of singletons.
Exercises Exercises
1. (PS11) Simple examples.
Consider \(\R\) as a metric space with the usual Euclidean metric. In what follows, you do not need to justify your arguments when calculating interiors and closures.
(a)
Let \(x \in \R\text{.}\) Show that \(\{x\}\) is nowhere dense in \(\R\text{.}\)
Solution.
We easily check that \(\{x\}\) is closed, and hence its own closure. Moreover, \(\{x\}^\circ = \varnothing\text{,}\) since \(B_r(x) \not\subseteq
\{x\}\) for all \(r \gt 0\text{.}\)
(b)
Show that \(\set{1/n}{n \in \N}\) is nowhere dense in \(\R\text{.}\)
Solution.
This set is not closed, but we can calculate its closure
\begin{equation*}
\overline{\set{1/n}{n\in\N}} = \{0\} \cup \set{1/n}{n\in\N}\text{,}
\end{equation*}
and the interior of this closure is empty,
\begin{equation*}
(\{0\} \cup \set{1/n}{n\in\N})^\circ = \varnothing\text{.}
\end{equation*}
(c)
Show that \((0,1) \cap \Q\) is not nowhere dense in \(\R\text{.}\)
Solution.
We calculate \(\overline{(0,1)\cap \Q} = [0,1]\text{,}\) and \([0,1]^\circ =
(0,1) \ne \varnothing\text{.}\)
Comment.
If you’re wondering if you’ve calculated a closure or interior correctly, one basic check is that closures must be closed and interiors must be open; see Theorem 1.23 and Theorem 1.22.
2. Nowhere dense sets and set operations.
Let \((X,d)\) be a metric space.
(a)
Suppose that \(U, V \subseteq X\) are open and dense in \(X\text{.}\) Show that \(U \cap V\) is dense in \(X\) as well.
(b)
Suppose that \(A, B \subseteq X\) are nowhere dense in \(X\text{.}\) Show that \(A \cup B\) is nowhere dense in \(X\) as well.
3. Interiors, closures, and complements.
Let \((X,d)\) be a metric space and \(S \subseteq X\text{.}\) Show that \(S^\circ = X \without (\overline{X\without S})\) in two ways:
(a)
Directly from the definitions of interior and closure.
(b)
Using Theorem 1.22 and Theorem 1.23.
Hint.
Show that \(X \without (\overline{X \without S})\) is an open set contained in \(S\text{,}\) while \(X \without S^\circ\) is a closed set containing \(X \without S\text{.}\)
4. (PS11) Baire for linear subspaces.
Let \((X, \n\blank)\) be a normed space and \(L \subseteq X\) a linear subspace.
(a)
Show that the closure \(\overline{L}\) is a linear subspace of \(X\text{.}\)
Hint.
Let \(x,y \in \overline L\) and \(\alpha \in \R\text{.}\) Then there are sequences \(\seq xn\) and \(\seq yn\) in \(L\) converging to \(x\) and \(y\text{.}\) Argue that \((x_n+\alpha y_n)_{n\in\N}\) is a sequence in \(L\) converging to \(x+\alpha y\text{,}\) and hence that \(x+\alpha y \in
\overline L\text{.}\)
Solution.
Let \(x, y \in \overline{L}\) and \(\alpha \in \R\text{.}\) Then there exist two sequences \((x_n)_{n \in \N}\) and \((y_n)_{n \in \N}\) in \(L\) such that \(x = \lim_{n \to \infty} x_n\) and \(y = \lim_{n \to \infty} y_n\text{.}\) Since \(L\) is a linear subspace, we know that \(x_n + \alpha y_n \in L\) for all \(n \in \N\text{.}\) Moreover, thanks to Exercise 1.3.1 we have \(x_n + \alpha
y_n \to x + \alpha y\text{,}\) hence \(x + \alpha y \in \overline L\) as desired.
Rather than appealing to Exercise 1.3.1, we could alternatively have shown \(x_n + \alpha y_n \to x + \alpha y\) directly from the estimate
\begin{equation*}
\n{(x + \alpha y) - (x_n + \alpha y_n)} \le \n{x - x_n} + \abs \alpha \n{y - y_n} \to 0
\end{equation*}
as \(n \to \infty\text{.}\)
(b)
Suppose that \(L^\circ \ne \varnothing\text{.}\) Show that \(0 \in L^\circ\text{.}\) Conclude that \(L = X\text{.}\)
Hint.
Given \(x \in L^\circ\text{,}\) there exists \(r \gt 0\) such that \(B_r(x) \subseteq
L\text{.}\) To see that \(B_r(0) \subseteq L\) as well, argue that any \(y \in
B_r(0)\) can be written as \(y=z-x\) where \(z \in B_r(x)\text{.}\) Finally, to see that \(L=X\text{,}\) pick an arbitrary \(x \in L\) and choose \(\alpha \in \R\) so that \(\alpha x \in B_r(0)\text{.}\)
Solution.
Let \(x \in L^\circ\text{,}\) which we can do because \(L^\circ\) is nonempty. By the definition of \(L^\circ\text{,}\) there exists \(r \gt 0\) such that \(B_r(x) \subseteq
L\text{.}\) We claim that \(B_r(0) \subseteq L\text{.}\) To see this, let \(y \in
B_r(0)\text{.}\) Then the vector \(z = x + y\) belongs to \(B_r(x)\) as \(\n{z-x}=\n{(x+y)-x}=\n y \lt r\text{.}\) Hence \(z \in L\text{,}\) and since \(x \in
L\) as well and \(L\) is a linear subspace, this implies that \(y =
z - x \in L\text{.}\) The claim is proved.
Now let \(x \in X \setminus \{0\}\text{.}\) Then \(w = rx/(2\n x)\) belongs to \(B_r(0)\text{.}\) Again using the fact that \(L\) is a linear subspace, this implies \(x = 2\n xr^{-1} w \in L\text{.}\) Thus \(L = X\) as desired.
(c)
Show that \(L\) is either dense or nowhere dense in \(X\text{.}\)
Hint.
(d)
For every \(n \in \N\text{,}\) suppose that \(L_n \subseteq X\) is a linear subspace such that \(X = \bigcup_{n\in\N} L_n\text{.}\) Show that if \(X\) is complete, then at least one of the spaces \(L_n\) is dense in \(X\text{.}\)
Hint.
Use Corollary 6.6 and the previous part.
Solution.
By the Baire category theorem in the form of Corollary 6.6, at least one of the spaces \(L_n\) is not nowhere dense in \(X\text{.}\) But then Part c implies that \(L_n\) is dense in \(X\text{.}\)
5. A fact about \(\Q\).
Show that \(\Q\) is not a countable intersection of open sets in \(\R\text{.}\)
Hint.
Argue by contradiction. Suppose that \(\Q = \bigcap_{n \in \N} U_n\text{,}\) where \(U_n \subseteq \R\) is open for every \(n \in \N\text{.}\) Suppose that \((q_n)_{n \in \N}\) is an enumeration of \(\Q\text{.}\) Then \(\R =
\bigcup_{n\in\N} (\R \without U_n \cup \{q_n\})\text{.}\) Show that this contradicts the Baire category theorem.)
