Let $f_n\in C[0, 1]$. Assume that for every $x\in [0, 1]$ there exists a finite limit $f(x) = \lim f_n(x)$. Apply the Baire category theorem to the sets
$$ A_n = \{x : |f_m(x) – f_n(x)| \leq \frac{1}{3} , \forall m > n\}$$
to prove that f cannot be the salt and pepper function
$$\left\{\begin{matrix}
1 & \text{if} \ x \in \mathbb{Q} \\
0 & \text{otherwise}
\end{matrix}\right.$$
I have to show that the metric space $C[0, 1]$ cannot be written as the countable union of closed , no where dense sets. So the set $A_n$ as given must be proven that is closed , nowhere dense. Right? (I don't know how to approach any of these requirements)
how to show the second part?
Also the function g is not continuous, so it does not belong to $C[0,1]$. What is the role of it in the question?
Best Answer
We have $A_n = \bigcap_{m>n}(f_n-f_m)^{-1}([-1/3,1/3])$, and hence $A_n$ is closed for each $n$. Moreover, for each $x\in [0,1]$, as $(f_n(x))_{n\in\mathbb N}$ is a Cauchy sequence for each $x$, there exists $n$ such that $x\in A_n$. So, $\bigcup_nA_n = [0,1]$. By Baire's theorem, there exists $n_0$ such that $A_{n_0}$ has a non-empty interior; so, there exists an open interval $I\subset A_{n_0}$. Thus, for $x\in I$ we have $|f_m(x)-f_{n_0}(x)|\le 1/3$ for all $m>n_0$. In particular (letting $m\to\infty$), $$ |f(x)-f_{n_0}(x)|\le 1/3\quad\text{for all $m>n_0$ and all $x\in I$.} $$ Now, choose $x_0\in I$ irrational. Assume that $f(x_0) = 0$. That means that $f_{n_0}(x_0)\in [-1/3,1/3]$. Because $f_{n_0}$ is continuous, for $x$ in a neighborhood $J$ of $x_0$ we have $f_{n_0}(x)\in (-2/3,2/3)$. Choose some rational $x_1\in J$. Then $$ |f(x_1)|\le|f(x_1)-f_{n_0}(x_1)|+|f_{n_0}(x_1)| < 1/3+2/3 = 1, $$ so $f(x_1)\neq 1$. So, $f$ cannot be the salt-and-pepper function.