Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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?


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.

share|cite|improve this question
Do I appeal to the countability of $\Bbb Q$ and show that $|\Bbb Q| = |\Bbb N|$ implies pointwise convergence? – Moderat 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 $n$? – user38268 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 – Moderat Aug 24 '12 at 12:20
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 $f(x)=1$?

share|cite|improve this answer
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? – Moderat 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. – Moderat Aug 24 '12 at 2:47
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.