2
$\begingroup$
Suppose that $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous on $\mathbb{R}$ and that $f(r) = 0$ for every rational number $r$. Prove that $f(x) = 0$ $\forall$ $x \in \mathbb{R}$

Ok, here's what I've done:

Let $c \in \mathbb{Q}$.
Since, $f$ is continuous so $\forall \epsilon > 0$ $\exists \delta > 0$ such that $\forall x \in (c-\delta, c+\delta)$, we should have $|f(x)| < \epsilon$. Now, let $d \in (c-\delta, c+\delta) \cap \mathbb{R} \setminus \mathbb{Q}$, $|f(d)| < \epsilon$. Since $\epsilon$ is arbitrary, we have $f(d) =0$

Hence, $\forall c \in \mathbb{Q}$, $\exists \delta > 0$ such that $\forall$ $|x-c| < \delta$, $f(x) = 0$

From here, how do I deduce that $f(x) = 0$ $\forall x \in \mathbb{R}$?

(Just to mention : I can intuitively think that it should be true because like you can go on taking union of these neighbourhoods but I was unable to rigorously prove it)

$\endgroup$
8
  • $\begingroup$ An elegant way to prove the claim would be using that the rational numbers are dense in $\mathbb R$ $\endgroup$
    – Peter
    Nov 16 '16 at 14:15
  • $\begingroup$ Related (but this is far more general): math.stackexchange.com/questions/1084018/… $\endgroup$
    – Watson
    Nov 16 '16 at 14:16
  • $\begingroup$ This is not quite good enough. You pick your $d$ after being given your $\epsilon$. That means that if you choose a smaller $\epsilon$, then your $\delta$ might be smaller, which means that we no longer have $d \in (c-\delta, c+\delta)$, so you can't say that $|f(d)| < \epsilon$ while $\epsilon$ gets smaller and smaller. It's easier to make this work if you let let $c$ be irrational and show that $f(c) < \epsilon$ by finding rational $d\in (c-\delta, c+\delta)$ $\endgroup$
    – Arthur
    Nov 16 '16 at 14:16
  • $\begingroup$ @Peter Can you give me some hints on how to proceed with that idea? $\endgroup$
    – kishlaya
    Nov 16 '16 at 14:21
  • $\begingroup$ @Watson I came across that but like I am a beginner with analysis, so I don't know much about Hausdorff space! I'll try to read about them next. Thanks! $\endgroup$
    – kishlaya
    Nov 16 '16 at 14:25
3
$\begingroup$

The step following "since $\epsilon$ is arbitrary..." is not quite right.

You have shown that if $d \in (c-\delta, c+\delta) \cap \mathbb{R} \setminus \mathbb{Q}$, then $|f(d)|< \epsilon$. You want to say that you could just make epsilon smaller and smaller; the problem with this is that $\delta$ depends on $\epsilon$. If you want $|f(d)|< \frac{\epsilon}{2}$, you might need to be in a smaller interval $(c-\eta, c+\eta)$ for some $\eta< \delta$. So you cannot conclude that $|f(d)|<\epsilon$ for all $\epsilon>0$, just for the epsilon you chose.

The slickest way to prove this is to use the fact that $f$ is continuous if and only if for every sequence $(x_n)$ such that $\lim_{n \to \infty}x_n \to x_0$, we have $\lim_{n \to \infty}f(x_n) \to f(x_0)$. If you let $x_0 \in \mathbb{R} \setminus \mathbb{Q}$, can you choose a sequence $(x_n)$ so that $\lim_{n \to \infty} f(x_n)=0$? (This works if $f(x_n)=0$ for all $n$...)

If you want to go the delta-epsilon route, then given $c \in \mathbb{R}\setminus \mathbb{Q}$, you know for all $\epsilon>0$ there exists $\delta>0$ such that for all $x \in (c-\delta,c+\delta)$, $|f(x)-f(c)|< \epsilon$. In particular, for each $\epsilon$ and corresponding $\delta$, there will be an $x \in (c-\delta,c+\delta) \cap \mathbb{Q}$. What can you conclude from that?

$\endgroup$
2
  • $\begingroup$ Yep! Got it. Just choose a sequence of rational numbers which converges to some irrational and that works. And using delta-epsilon route, we can choose $c \in \mathbb{R}\setminus\mathbb{Q}$ so that $\forall \epsilon > 0$ $\exists \delta > 0$ such that $\forall x \in (c-\delta, c+\delta)\cap\mathbb{Q}$ we have $|f(x) - f(c)| = |-f(c)| = |f(c)| < \epsilon$. SInce we can always find a rational in the interval $(c-\delta, c+\delta)$, so $f(c) = 0$. Am I right here? $\endgroup$
    – kishlaya
    Nov 16 '16 at 16:00
  • $\begingroup$ Yep, you got it. $\endgroup$
    – kccu
    Nov 16 '16 at 16:11
2
$\begingroup$

Let $x\in\mathbb R$. By continuity, $f(x)=\lim_{y\to x}f(y)$. Since this limit exists, it is independent of the manner of approach $y\to x$; specifically, we may assume that $y\to x$ through rational values since $\mathbb Q$ is dense in $\mathbb R$. But then $f(x) = \lim_{y\to x} 0 = 0$.

$\endgroup$
2
$\begingroup$

A slightly more topological approach:

Theorem: Let $(X, \tau_X)$, $(Y, \tau_Y)$ be two topological spaces. Suppose $Y$ is Hausdorff and that $D \subset X$ is a dense subset in $X$. Let $f,g: X \to Y$ be two continuous functions such that $f_{|D}=g_{|D}$. Then $f=g$ on $X$.

The theorem is proven in this question.

Now, $f_{|\mathbb Q}=0$ and $\mathbb Q$ is dense in $\mathbb R$, which is Hausdorff. By the previous theorem, $$f=0$$

$\endgroup$
1
$\begingroup$

Only thing that remains to show is that $f(r)=0\forall r\in \Bbb Q^c$

Since $\Bbb Q$ is dense in $\Bbb R$ so given $r\in \Bbb Q^c;\exists r_n\in \Bbb Q$ such that $r_n\to r$

$f$ is continuous so $r_n\to r\implies f(r_n)\to f(r)\implies f(r)=0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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