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 $\left \langle{f_n}\right\rangle$ in $C \left({X, Y}\right)$.

Let $D \left({f}\right)$ be the set of points where $f$ is discontinuous.

Then $D \left({f}\right)$ is a meager subset of $X$.

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

Using the oscillation we have, following the convention that we omit the metric when writing the oscillation, that:


 * $\displaystyle D \left({f}\right) = \bigcup_{n \mathop = 1}^\infty \left\{{x \in X: \omega_f \left({x}\right) \ge \frac 1 n}\right\}$

which is a countable union of closed sets.

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

is nowhere dense.

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


 * $\displaystyle A_n = \bigcap_{i, j \mathop \ge n} \left\{{x \in X: d \left({f_i \left({x}\right), f_j \left({x}\right)}\right) \le \epsilon}\right\}$

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

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

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

Consider the sequence $\left\langle{A_n \cap U}\right\rangle$ 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 $\left\langle{A_n \cap U}\right\rangle$, 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:


 * $V \subseteq F_\epsilon^c = \left\{{x \in X: \omega_f \left({x}\right) < 5 \epsilon}\right\}$

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

Since $V \subseteq A_k$:
 * $d \left({f_i \left({x}\right), f_j \left({x}\right)}\right) \le \epsilon$

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

Pointwise convergence of $\left \langle{f_n}\right\rangle$ gives that:
 * $d \left({f \left({x}\right), f_k \left({x}\right)}\right) \le \epsilon$

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:
 * $d \left({f_k \left({x}\right), f_k \left({x_0}\right)}\right) \le \epsilon$

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

By the triangle inequality:


 * $d \left({f \left({x}\right), f_k \left({x_0}\right)}\right) \le 2 \epsilon$

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

Applying the triangle inequality again:


 * $d \left({f \left({x}\right), f \left({y}\right)}\right) \le 4 \epsilon$

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

Thus we have the bound:


 * $\omega_f \left({x_0}\right) \le \omega_f \left({V_{x_0}}\right) \le 4 \epsilon$

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