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Suppose that $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous on $\mathbb{R}$ and that $f(r)=0$ for all $r \in \mathbb{Q}$. Prove that $f(x)=0$ for all $x \in \mathbb{R}$.

My attempt: Define a sequence $(x_n)$ where $x_n \in \mathbb{Q}$ for all $n \in \mathbb{N}$ and assume that $(x_n) \rightarrow a \not\in \mathbb{Q}$. Since $f$ is continuous, we have $\lim_n{f(x_n)}=f(a)=0$. Since $a$ is arbitrary irrational number, we have $f(a)=0$ for all $a \not\in \mathbb{Q}$. Hence, we proved the statement.

Is my proof valid? or is there any flaw ?

Idonknow
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  • Looks good to me. – Alex Becker Mar 28 '13 at 15:27
  • The idea is fine, the writeup is clear. Minor point. You should have started with "Let $a$ be irrational. There is a sequence $(x_n)$ of rationals such that $\dots$." Whether the proof is acceptable depends on your course context. Has the fat that there is such a sequence been proved? Has the fact that if $f$ is continuous then $\lim f(x_n)=f(\lim x_n)$ been proved? – André Nicolas Mar 28 '13 at 15:36
  • @André Nicolas Are you serious? Does he needs to prove also that irrational numbers exist? Or that the axioms for the real numbers define reals uniquely up to an isomorphism? –  Mar 28 '13 at 15:59
  • @Thus: It's a reasonable point that a course-acceptable proof will only use results that are available to the students. – Cameron Buie Mar 28 '13 at 16:08
  • @Cameron Buie Isn`t it obvious that the original poster in the proof knows the results about which is Andre speaking if he uses them in his proof? –  Mar 28 '13 at 16:11
  • The OP does not need to prove irrationals exist: the result would be vacuously true if they didn't. (That was a joke.) But to be serious, in a first analysis course, one has to be careful to avoid taking for granted "facts" that were treated less formally in an earlier course. So what a good proof is depends very much on which results have already been rigorously established. – André Nicolas Mar 28 '13 at 16:12
  • @André Nicolas So what if you were an assistant on the college and were in a process of valuating my test. Suppose also that I am a first year student and that the exam was in the calculus of functions of real variable, and, while reading my test, you observe that I solved some integral correctly but not with methods of real analysis but with methods of complex analysis, which is on the third year but I somehow know about methods used therein. Now if you need to decide whether to give me minimal 0 points because I used methods not teached in the class or maximal 20 points, what would you do? –  Mar 28 '13 at 16:33
  • @Thus: I think there is not a close analogy, since the point of the usual "Rudin"-style course is systematic development. – André Nicolas Mar 28 '13 at 16:59
  • @André Nicolas Well, it is not about analogy, I just wanted to know what would you do in the situation I described above? If you do not want to answer it, it is okay also. –  Mar 28 '13 at 17:02
  • On the calculus question, I have not seen the details of the hypothetical answer. As a guess, a pretty high mark. But if the question asked for a proof that $\sum \frac{1}{n^2}\lt 2$, and the answer used the $\frac{\pi^2}{6}$ result, low mark. – André Nicolas Mar 28 '13 at 17:18
  • @Thus: It isn't obvious to me. In fact, if you read the OP's first comment under Dominic's answer, you'll find that the OP wasn't even sure that you can define such a sequence. It never hurts to make sure. – Cameron Buie Mar 28 '13 at 17:41
  • @Cameron Buie Good example of mathematical blasphemy. I thought that OP (OP, did you know this?) knows that every irrational number can be expressed as a number with non-terminating decimal part and from that it is obviously obvious that such a number can be seen as a sequence of fractions which have that irrational number as a limit. –  Mar 28 '13 at 17:49
  • non-terminating fractional part –  Mar 28 '13 at 18:01
  • Related. – Cameron Buie Apr 11 '13 at 19:56

3 Answers3

5

Maybe you should be more explicit why it works (mentioning that every irrational number is a limit of a sequence of rationals (taking decimal digits will be the canonical way)).

Every irrational has a unique decimal expansion (if we avoid recurring (?) 9). So when $x$ is irrational we know $$x=\sum_{k=-m}^\infty a_k 10^{-k}$$ with $a_k \in \{ 0,1,2,3,4,5,6,7,8,9\}$, and $m \in \mathbb{N}$ When we take the sequence $$b_n=\sum_{k=-m}^n a_k 10^{-k} $$ with the same $a_k$ as in $x$ we see that $b_n$ converges to $x$. But every $b_n$ is an rational.

Yeah it does work, but it would be even easier using intermediate value theorem, as between two irrational is always an rational and vice versa all values must be zero.

Dominic Michaelis
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4

As long as you know that you can define such a sequence, then your argument is perfect.

Perhaps an easier way to go is to note that since $f$ is continuous, then the preimage of any closed set is closed. In particular, the preimage of $\{0\}$ is closed, and the only closed set of reals containing the rationals is the whole real line. Hence, $f\equiv 0$.

Cameron Buie
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3

Assume $f$ is not identically zero, so there must exist some irrational $s$ such that $f(s) \neq0$, By continuity there is some neighbourhood of $s$ where $f \neq0$, but since $\mathbb Q$ is dense in $\mathbb R$, this is not possible. And you have your result.

Mathematician
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