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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.

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Do I appeal to the countability of $\Bbb Q$ and show that $|\Bbb Q| = |\Bbb N|$ implies pointwise convergence? –  Zvpunry Aug 23 '12 at 23:31
    
I don't understand, the set of all rationals in $[0,1]$ is not finite, so how can you list them up to $n$? –  fpqc Aug 24 '12 at 12:08
    
@BenjaLim I think that is the point as to why it does not converge uniformly, but we can always take $n$ sufficiently large as to make the difference less than $\epsilon$. We list up to $n$ simply by AC I believe –  Zvpunry Aug 24 '12 at 12:20
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1 Answer

up vote 2 down vote accepted

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 $f(x)=1$?

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Building off of this, would I suppose to the contrary that there is a sufficiently large $N_0$ so that there is an $x \in [0,1]$ for which $f_{N_0}(x) = 0, f(x) = 1$. Taking $n \ge N_0$, I have that for all $\epsilon > 0$, $|f_n(x) - f(x)| \le \epsilon. This can be done as many times as necessary to achieve arbitrary precision. Is this the lines along which you would approach? –  Zvpunry Aug 23 '12 at 23:58
    
@jmi4: You don't really need prove by contradiction. Answer my questions then you will see. –  timur Aug 24 '12 at 0:01
    
I am not sure what you mean. Can you elaborate further on your hint? I thought I had answered your questions. –  Zvpunry Aug 24 '12 at 2:47
1  
Pick $x\in[0,1]$. If $x$ is irrational, then $f_n(x)=f(x)=0$ regardless of what $n$ is. If $x$ is rational, say $x=r_k$, then $f_n(x)=f(x)=1$ for all $n\geq k$. This is pointwise convergence. –  timur Aug 24 '12 at 14:15
1  
Uniform convergence would mean $\sup_{x\in[0,1]}|f_n(x)-f(x)|$ converges to $0$ as $n\to\infty$. Take any $n$. Then for $x=r_{n+1}$, we have $f_n(x)=0$, and of course $f(x)=1$. This means $\sup_{x\in[0,1]}|f_n(x)-f(x)|\geq1$. As $n$ can be chosen as large as we want, this shows $\sup_{x\in[0,1]}|f_n(x)-f(x)|$ does not go to $0$, so no uniform convergence. –  timur Aug 24 '12 at 14:19
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