Question:
Let $\{r_1, r_2, \dots\}$ be the set of rationals in the interval $[0,1]$. For $x \in [0,1]$ and $n \in \Bbb N$, let $f_n(x)$ and $f(x)$ be given by the following:
$$ f_n(x) = \begin{cases}
1 & \text{ if } x= r_1, \dots, r_n \\
0 & \text{ otherwise }
\end{cases} \qquad
f(x) = \begin{cases}
1 & \text{ if } x \text{ rational}\\
0 & \text{ if } x \text{ irrational}
\end{cases} $$
Prove that $f_n \to f$ pointwise, but not uniformly.
My Thoughts:
I'm not sure how to show either convergence result. For pointwise convergence, $| f_n(x) – f(x) |$ becomes $0$ at $x$ irrational or $x \in \{r_1,\dots, r_n\}$, and $1$ at all the rationals not yet enumerated. How can I work this into my proof?
Edit:
Let's suppose to the contrary that there is a sufficiently large $N_0$ so that for some $\epsilon_0$ for all $x \in [0,1]\Rightarrow|f_{N_0}(x) – f(x)| \ge \epsilon_0$. Here is where I am stuck now
Edit 2:
I have found it! I went through my textbook, and I found the following key sentence:
In pointwise convergence, one might have to choose a different $N$ for each different $x$. In uniform convergence, there is an $N$ which works for all $x$ in the set $E$.
So the proofs follow:
Proof of pointwise convergence:
Let $\epsilon > 0$ be given. Then let $x_0$ be the $N$th rational number in $[0,1]$. Taking $n = N$, we have $|f_n(x) – f(x)| \le \epsilon$ for all $x \le x_0$, and we can successively take larger and larger $n$ to always guarantee that $|f_n(x) – f(x)| \le \epsilon$.
Proof of lack of uniform convergence:
Let $\epsilon > 0$ be given. Then if $f_n \to f$ uniformly, there exists an $M$ so that $n \ge M$ implies $|f_n(x) – f(x)| \le \epsilon$ for all $x$. Taking $x$ to be the $(n+1)$th rational number, we have that for sufficiently small $\epsilon$, $|f_n(x) – f(x)| \ge \epsilon$, so $f_n \not \! \to f$ uniformly.
Best Answer
Pick an $x\in[0,1]$. Then can you say that $f_n(x)=f(x)$ when $n$ is sufficiently large?
Take any $n$. Then can you say that there is $x\in[0,1]$ such that $f_n(x)=0$ but $f(x)=1$?