Baire-Osgood Theorem

Theorem
Suppose $X$ is a Baire space, $Y$ a metrizable space and $f: X \to Y$ is a function which is the pointwise limit of a sequence $(f_n)$ in $C(X,Y)$. Then $D(f)$, the set of points where $f$ is discontinuous, 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 oscillaiton, that:


 * $\displaystyle D(f) = \bigcup_{n=1}^\infty \left\{x \in X \mathrel{}\middle\vert\mathrel{} \omega_f(x) \geq \tfrac1n\right\}$,

which is a countable union of closed sets. Since we have this expression for $D(f)$, the claim follows if we can prove that for all $\epsilon>0$ the closed set $F_\epsilon = \left\{x \in X \mathrel{}\middle\vert\mathrel{} \omega_f(x) \geq 5\epsilon\right\}$ is nowhere dense.

Let $\epsilon > 0$ be given and consider the sets


 * $\displaystyle A_n = \bigcap_{i,j\geq n} \left\{x \in X \mathrel{}\middle\vert\mathrel{} d(f_i(x),f_j(x) \leq \epsilon\right\}$

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

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

Given a nonempty open $U \subseteq X$ we wish to show that $U \nsubseteq F_\epsilon$. Consider the sequence $(A_n \cap U)$ of closed subsets of $U$. Since the union of these is all of $U$ and $U$—being an open subspace of a Baire space—is a Baire space, one of them, 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:


 * $\displaystyle V \subseteq F_\epsilon^c = \left\{x \in X \mathrel{}\middle\vert\mathrel{} \omega_f(x) < 5\epsilon\right\}.$

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

Since $V \subseteq A_k$ we have $d(f_i(x),f_j(x)) \leq \epsilon$ for all $x \in V$ and all $i,j \geq k$. Pointwise convergence of $(f_n)$ gives that $d(f(x),f_k(x)) \leq \epsilon$ 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 $d(f_k(x),f_k(x_0)) \leq \epsilon$ for all $x \in V_{x_0}$.

By the triangle inequality we have that


 * $\displaystyle d(f(x),f_k(x_0)) \leq 2\epsilon$

for all $x \in V_{x_0}$, and applying the triangle inequality again we get


 * $\displaystyle d(f(x),f(y)) \leq 4\epsilon$

for all $x,y \in V_{x_0}$. Thus we have the bound


 * $\displaystyle\omega_f(x_0) \leq \omega_f(V_{x_0}) \leq 4\epsilon$

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