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Assume $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous on $\mathbb{R}$ and such that $f(r)=0$ for every rational number $r$. Show that $f(x)=0, \forall x\in\mathbb{R}$ using the $\varepsilon-\delta$ definition of continuity.

I am attempting to do this by contradiction. If we assume $f(y)\neq 0$ for some irrational number $y$, we should be able to come up with a contradiction to the definition of continuity at point $y$. I am not quite sure how to arrive at the contradiction, though.

Relevant facts:

  1. Let $A\subseteq\mathbb{R}, f:A\rightarrow\mathbb{R}$ and $c\in A$. Then $f$ is continuous at $c$ if $\forall\varepsilon>0, \exists\delta>0$ such that if $x\in A$ and $|x-c|<\delta$, then $|f(x)-f(x)|<\varepsilon$.

  2. If $x,y\in\mathbb{R}$ and $x<y, \exists r\in\mathbb{Q}$ such that $x<r<y$.

  3. If $x,y\in\mathbb{R}$ and $x<y, \exists z\in\mathbb{R\setminus Q}$ such that $x<z<y$.

The first is just the definition of continuity at a point. The second and third state that between any two real numbers you can find both a rational and irrational number, which I think may come in to play somewhere.

flubsy
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  • @NormalHuman This seems to be a similar question, but I want to prove this using the $\varepsilon-\delta$ definition of continuity specifically, something the answers in the other thread do not do. I added that information to the original question to clarify. – flubsy Nov 10 '15 at 03:46
  • Plenty of those, too: Show that $f(x) = 0$ for all $x \in \mathbb{R}$. There is not just one thread there, a dozen such questions are linked to it. –  Nov 10 '15 at 03:50
  • While you are doing this problem notice two things: (i) for an arbitrary function $f:\mathbb{R}\to\mathbb{R}$ the information about some of the values of $f(x)$ tells you absolutely nothing about the remaining values, and (ii) for a continuous function the information about most of the values does tell you exactly what the other values are. In fact most here is exactly the two conditions (2) and (3) described here. The rationals have these two conditions and any "big" subset of $\mathbb{R}$ that has the two properties (2) and (3) works too! – B. S. Thomson Nov 10 '15 at 04:03

2 Answers2

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Argue by contradiction:

Suppose there is some irrational $r$ such that $f(r) \neq 0$; let $d := |f(r)|$. If $f$ is continuous on $\Bbb{R}$, then there is some $\delta > 0$ such that $|x-r| < \delta$ only if $|f(x) - f(r)| < d/2$. But there is some rational $x$ such that $|x-r| < \delta$, implying that for that rational $x$ we have $|f(x) - f(r)| = |0 - f(r)| = d > d/2$, a contradiction.

Yes
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Since $f$ is continuous, given $\epsilon > 0$ there is some $\delta > 0$ so that $|f(x) -f(y)| < \epsilon$ whenever $|x-y|<\delta$. Let $x \in \mathbb{R}$. Then since $\mathbb{Q}$ is dense in $\mathbb{R}$ there is some $q \in \mathbb{Q}$ with $|x-q| < \delta$. Since $f(q) = 0$ and since $f$ is continuous, we have $|f(x) - f(q)| = |f(x)| < \epsilon$. Since this holds for every $\epsilon > 0$, we have $f(x) = 0$.

aquilegia.taylor
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