Baire-Osgood Theorem

Theorem
Let $X$ be a Baire space.

Let $Y$ be a metrizable topological space

Let $f: X \to Y$ be a mapping which is the pointwise limit of a sequence $\sequence {f_n}$ in $\map C {X, Y}$.

Let $\map D f$ be the set of points where $f$ is discontinuous.

Then $\map D f$ is a meager subset of $X$.

Proof
Let $d$ be a metric on $Y$ generating its topology.

Let $\map {\omega_f} x$ denote the oscillation of $f$ at $x$.

We have:


 * $\ds \map D f = \bigcup_{n \mathop = 1}^\infty \set {x \in X: \map {\omega_f} x \ge \frac 1 n}$

which is a countable union of closed sets.

Since we have this expression for $\map D f$, the claim follows if we can prove that for all $\epsilon \in \R_{>0}$ the closed set:
 * $F_\epsilon = \set {x \in X: \map {\omega_f} x \ge 5 \epsilon}$

is nowhere dense.

Let $\epsilon \in \R_{>0}$ be given and consider the sets:


 * $\ds A_n = \bigcap_{i, j \mathop \ge n} \set {x \in X: \map d {\map {f_i} x, \map {f_j} x} \le \epsilon}$

which are closed because $d$ and the $f_i$ are continuous.

Because $\sequence {f_n}$ is pointwise convergent, it is pointwise Cauchy with respect to any metric generating the topology on $Y$, so $\ds \bigcup_{n \mathop = 1}^\infty A_n = X$.

Given a nonempty open $U \subseteq X$ we wish to show that $U \nsubseteq F_\epsilon$.

Consider the sequence $\sequence {A_n \cap U}$ of closed subsets of $U$.

The union of these is all of $U$.

As $U$ is an open subspace of a Baire space, it is a Baire space.

So one of the elements of $\sequence {A_n \cap U}$, say $A_k$, must have an interior point, so there is an open $V \subseteq A_k \cap U$.

Because $U$ is open in $X$, $V$ is open in $X$ as well.

We will show that:


 * $V \subseteq F_\epsilon^c = \set {x \in X: \map {\omega_f} x < 5 \epsilon}$

This will show that:
 * $V \nsubseteq F_\epsilon$ and thus that $U \nsubseteq F_\epsilon$

Since $V \subseteq A_k$:
 * $\map d {\map {f_i} x, \map {f_j} x} \le \epsilon$

for all $x \in V$ and all $i, j \ge k$.

Pointwise convergence of $\sequence {f_n}$ gives that:
 * $\map d {\map f x, \map {f_k} x} \le \epsilon$

for all $x \in V$.

By continuity of $f_k$ we have for every $x_0 \in V$ an open $V_{x_0} \subseteq V$ such that:
 * $\map d {\map {f_k} x, \map {f_k} {x_0} } \le \epsilon$

for all $x \in V_{x_0}$.

By the triangle inequality:


 * $\map d {\map f x, \map {f_k} {x_0} } \le 2 \epsilon$

for all $x \in V_{x_0}$.

Applying the triangle inequality again:


 * $\map d {\map f x, \map f y} \le 4 \epsilon$

for all $x, y \in V_{x_0}$.

Thus we have the bound:


 * $\map {\omega_f} {x_0} \le \map {\omega_f} {V_{x_0} } \le 4 \epsilon$

showing that $x_0 \notin F_\epsilon$ as desired.