Fall 2018 - Problem 5

Solution.

1. Observe that

   $$\bigcup_{N=1}^\infty F_{N,\epsilon} = \bigcup_{N=1}^\infty \bigcap_{n,m \geq N} \set{y \in \br{0,1} \mid \abs{f_n\p{y} - f_m\p{y}} \leq \epsilon} = \br{0, 1},$$

   since $f_n$ converges everywhere to $f$. Notice also that because each $f_n$ is continuous, we see that $F_{N,\epsilon}$ is an intersection of closed sets, hence closed.

   For any $\br{a, b} \subseteq \br{0, 1}$, we see that

   $$\br{a, b} = \bigcup_{N=1}^\infty F_{N,\epsilon} \cap \br{a, b},$$

   but $\br{a, b}$ is complete and each set in the union is closed, so by the Baire category theorem, $F_{N,\epsilon} \cap \br{a, b}$ contains an open interval $I_{a,b}$. Let

   $$D_\epsilon = \bigcup_{a, b \in \br{0,1}} I_{a,b}.$$

   We claim that $D_\epsilon$ is dense: if this were not the case, then there exists $\br{a, b} \subseteq \br{0, 1}$ such that $\br{a, b} \cap D_\epsilon = \emptyset$. But $I_{a,b} \subseteq D_\epsilon$ by construction, and $I_{a,b} \subseteq \br{a, b}$ as well, so this is impossible, and so $D_\epsilon$ is dense. Finally, given any $x \in D_\epsilon$, there exists $I_{a,b} \subseteq F_{N_x,\epsilon}$ for some $N_x \in \N$, and if $n, m > N_x$, we have

   $$I_{a,b} \subseteq F_{n,\epsilon} \implies \sup_{y \in I_{a,b}} \,\abs{f_n\p{y} - f_m\p{y}} \leq \epsilon.$$

   Sending $m \to \infty$, we get the claim.

2. If $f$ were such a function, then by (1), there exists a dense subset $D_{1/4} \subseteq \br{0, 1}$ with the stated properties. Then for any $x \in D_{1/4}$, we get $I \ni x$ and $N_x$ so that if $n > N_x$, then

   $$\sup_{y \in I} \,\abs{f_n\p{y} - \chi_{\Q\cap\br{0,1}}\p{y}} \leq \frac{1}{4}.$$

   For any $q \in \Q \cap I$ and $r \in \p{\R \setminus \Q} \cap I$, we get

   $$\begin{gathered} \abs{f_n\p{q} - 1} = \abs{f_n\p{q} - \chi_{\Q\cap\br{0,1}}\p{q}} \leq \frac{1}{4}\phantom{.} \\ \abs{f_n\p{r} - 0} = \abs{f_n\p{r} - \chi_{\Q\cap\br{0,1}}\p{r}} \leq \frac{1}{4}. \end{gathered}$$

   But by density, we may send $q \to r$ and apply continuity to get

   $$\abs{f_n\p{r} - 1} \leq \frac{1}{4},$$

   which is impossible, since this implies $1 \leq \abs{f_n\p{r} - 1} + \abs{f_n\p{r} - 0} \leq \frac{1}{2}$. Thus, $\chi_{\Q\cap\br{0,1}}$ cannot be the limit of a sequence of continuous functions.