Baire Category Theorem

Prove the Baire Category Theorem: a complete metric space $X$ cannot be written as a countable union $X = \bigcup_{n=1}^{\infty} E_n$ of nowhere dense sets. In other words, a complete metric space is not meagre (first-category) in itself.

X = \bigcup_{n=1}^{\infty} E_n, \qquad \operatorname{int}(\overline{E_n}) = \varnothing \ \ \forall n

\overline{B}(x_{k+1}, r_{k+1}) \subseteq B(x_k, r_k) \setminus \overline{E_{k+1}}, \qquad 0 < r_{k+1} < r_k/2

stage 6:

I_6 = [0.75,\, 0.8125]

, survivor

x \approx 0.78125

— dodged

q_1 = 0

,

q_2 = 1

inside the interval; the rest already lay outside. These nested closed intervals are the proof's nested closed balls

\overline{B}(x_k, r_k)

.

Show proof

Since $E_1$ is nowhere dense, $\operatorname{int}(\overline{E_1}) = \varnothing$ , so $U_1 = X \setminus \overline{E_1}$ is a nonempty open set. Pick $x_1 \in U_1$ ; some $\rho > 0$ has $B(x_1, \rho) \subseteq U_1$ , and with $r_1 = \min(\rho/2, 1/2)$ the closed ball $\overline{B}(x_1, r_1) \subseteq U_1$ with $r_1 < 1$ .
Inductively, suppose $\overline{B}(x_k, r_k)$ is chosen. Then $V = B(x_k, r_k) \setminus \overline{E_{k+1}}$ is a nonempty open set: $B(x_k, r_k)$ is nonempty open, and $B(x_k, r_k) \subseteq \overline{E_{k+1}}$ would give $B(x_k, r_k) \subseteq \operatorname{int}(\overline{E_{k+1}}) = \varnothing$ , a contradiction. Pick $x_{k+1} \in V$ and shrink the radius below $r_k/2$ so the closed ball fits in $V$ : $\overline{B}(x_{k+1}, r_{k+1}) \subseteq B(x_k, r_k) \setminus \overline{E_{k+1}}$ with $0 < r_{k+1} < r_k/2$ .
The sequence $(x_k)$ is Cauchy: for $m > k$ , $x_m \in \overline{B}(x_{k+1}, r_{k+1}) \subseteq B(x_k, r_k)$ , so $d(x_k, x_m) < r_k$ , and $r_k < 2^{-(k-1)} r_1 \to 0$ . Completeness of $X$ gives a limit $x_k \to x \in X$ .
For fixed $k$ , the tail $(x_m)_{m \ge k}$ lies in the closed set $\overline{B}(x_k, r_k)$ , so the limit $x \in \overline{B}(x_k, r_k)$ . Hence $x \in \bigcap_k \overline{B}(x_k, r_k)$ for all $k$ ; with $x \in \overline{B}(x_1, r_1) \subseteq X \setminus \overline{E_1}$ and, for $j \ge 2$ , $x \in \overline{B}(x_j, r_j) \subseteq B(x_{j-1}, r_{j-1}) \setminus \overline{E_j}$ , we get $x \notin E_j$ for every $j$ .
But $x \in X = \bigcup_j E_j$ , so $x \in E_j$ for some $j$ — a contradiction. Hence a complete metric space cannot be a countable union of nowhere dense sets. The intuition: each nowhere dense set leaves a "hole", and completeness traps the limit point obtained by descending through those holes inside the space.

Use the model's nested intervals to prove an irrational exists in $[0, 1]$ : enumerate the rationals of $[0, 1]$ as $q_1, q_2, \dots$ and at each stage take a closed interval $I_k$ dodging $q_k$ whenever it lands inside. Show the lengths $\ell_k = b_k - a_k \to 0$ (using density of the rationals), and that the point $x$ with $\bigcap_k I_k = \{x\}$ differs from every $q_k$ .

q_k \in I_{k-1} \ \Longrightarrow \ \ell_k = \ell_{k-1}/4, \quad q_k \notin I_k

\ell_k \to 0, \qquad \bigcap_{k=1}^{\infty} I_k = \{x\}, \qquad x \ne q_k \ \forall k

stage 14:

I_{14} = [0.80859375,\, 0.8125]

(length

1/256

), survivor

x \approx 0.810546875

— dodged

q_1 = 0

,

q_2 = 1

,

q_7 = 3/4

,

q_{11} = 4/5

inside the interval;

x

differs from all 14 rationals listed.

Show proof

By construction each $I_k$ is a closed interval with $I_{k+1} \subseteq I_k$ ; when $q_k$ lands in $I_{k-1}$ we take $I_k$ of one-quarter the length with $q_k \notin I_k$ (if $q_k$ already lies outside, $I_k = I_{k-1}$ ).
The lengths tend to $0$ . Suppose no more "hits" occur after stage $K$ ; then $I_n = I_K$ for all $n \ge K$ and the length stays positive. But $I_K$ has positive length, so it contains infinitely many rationals, each appearing exactly once in the enumeration — hence some $q_m \in I_K$ has index $m > K$ . Then $I_{m-1} = I_K \ni q_m$ , so stage $m$ is a hit — a contradiction. Hits therefore occur infinitely often, and each one quarters the length, so $\ell_k \to 0$ .
By completeness of $\mathbb{R}$ (the nested interval theorem), nested closed intervals whose lengths tend to $0$ intersect in exactly one point: $\bigcap_k I_k = \{x\}$ . And since $x \in I_k$ while $q_k \notin I_k$ , we have $x \ne q_k$ for every $k$ .
Since $\{q_k\}$ enumerates every rational of $[0, 1]$ , $x$ is an irrational in $[0, 1]$ . Hence $[0, 1] = \bigcup_k \{q_k\}$ — a countable union of points — is impossible, and $[0, 1]$ is uncountable. This is the point-set version of Baire's theorem: a singleton is nowhere dense in $\mathbb{R}$ , so $[0, 1]$ , being complete, cannot be a countable union of points without contradicting Baire.

[Idea-level sketch — not a complete proof] $C[0, 1]$ with the sup norm $\|f\|_\infty$ is a complete metric space. For $m \ge 1$ let $F_m = \{f : \exists x\ \forall y\ |f(x) - f(y)| \le m|x-y|\}$ . Using that each $F_m$ is closed nowhere dense (sketch), explain the idea that "nowhere-differentiable functions are typical" — i.e. comeager. The closed/nowhere-dense verifications below are sketched, not fully proved here.

F_m = \{f \in C[0,1] : \exists x\ \forall y\ |f(x) - f(y)| \le m|x-y|\}

\text{differentiable somewhere} \subseteq \bigcup_{m=1}^{\infty} F_m \quad \text{(meagre)}

stage 10:

I_{10} = [0.796875,\, 0.8125]

, survivor

x \approx 0.8046875

— three shrinkings dodged

q_1

,

q_2

,

q_7

inside. The same nested-ball idea, applied in

C[0,1]

, makes nowhere-differentiable functions typical.

Show proof

[Sketch] $F_m$ is closed. If $f_j \in F_m$ with $f_j \to f$ uniformly, witnessed by $x_j$ — i.e. $|f_j(x_j) - f_j(y)| \le m|x_j - y|$ — pass to a convergent subsequence $x_{j'} \to x$ (Bolzano–Weierstrass); the estimate survives the limit as $|f(x) - f(y)| \le m|x - y|$ , so $f \in F_m$ .
[Sketch] $F_m$ is nowhere dense. Given $f \in F_m$ and $\varepsilon > 0$ , first approximate $f$ within $\varepsilon/2$ by a piecewise-linear $p$ (uniform continuity), then superimpose a fine zigzag of amplitude $<\varepsilon/2$ whose slopes exceed those of $p$ by more than $m$ . At every candidate point $x$ , a nearby tooth of the zigzag breaks the $m$ -Lipschitz bound, so the perturbed $g \notin F_m$ while $\|g - f\|_\infty < \varepsilon$ . Being closed, $F_m$ has empty interior.
A function differentiable at even one point lies in $\bigcup_m F_m$ . If $f'(x_0)$ exists, then $|f(y) - f(x_0)| \le m_1|y - x_0|$ for $|y - x_0| < \delta$ (take $m_1 \ge |f'(x_0)| + 1$ ), while for far points $|y - x_0| \ge \delta$ boundedness gives $|f(y) - f(x_0)| \le 2\|f\|_\infty \le \frac{2\|f\|_\infty}{\delta}|y - x_0|$ . With an integer $m \ge \max(m_1, 2\|f\|_\infty/\delta)$ we get $|f(y) - f(x_0)| \le m|y - x_0|$ for all $y$ — i.e. $f \in F_m$ .
Hence "functions differentiable somewhere" $\subseteq \bigcup_{m=1}^{\infty} F_m$ is a countable union of nowhere dense sets — meagre. Its complement, the nowhere-differentiable functions, is comeager, and by Baire (problem 1) it is not just nonempty but dense. To stress again: the closed/nowhere-dense verifications above are idea-level sketches, not complete proofs.

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Examples & Exercises