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If $f:[a,b] \rightarrow R$ be continuous , let $f(x) =0$ when $x$ is rational. Prove $f(x) =0$ for all $x$ that is an element of $[a,b]$.

Thanks for the help I can't seem to find the f[a,b] notation in my notes and it is throwing me off.

Kaj Hansen
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Chris Millett
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    If you're asking about the notation $f([a,b])$, it is simply shorthand for the set ${f(x):x\in[a,b]}$. It's the image of the set $[a,b]$ under $f$. – Cameron Williams Mar 24 '15 at 03:56
  • $f : [a, b] \rightarrow \mathbb R$ means that $f$ is a function which maps reals in the interval $[a,b]$ to reals. – dalastboss Mar 24 '15 at 03:56
  • @CameronWilliams thanks for that! Now it's a matter of solving it. That's the main part. – Chris Millett Mar 24 '15 at 04:01
  • @KajHansen thanks for the kind words. Any help on actually solving it. I've been playing around with a delta epsilon proof and I'm getting stuck and getting lost in my work – Chris Millett Mar 24 '15 at 04:05
  • You should include details on what you are having trouble with, it could help people answer the question in a way that is helpful to you. –  Mar 24 '15 at 04:51

1 Answers1

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Go back to the definition of continuity. If such a function is continuous, then for any given $\varepsilon > 0$ and $x \in [a, b]$, there exists a $\delta > 0$ such that $|x-y| < \delta \implies |f(x) - f(y)|< \varepsilon$.

So what problems arise if there is an $x \in [a, b] \setminus \mathbb{Q}$ such that $f(x) \neq 0$? Well, we can always find a $y \in \mathbb{Q}$ that is as close to $x$ as we'd like (in math jargon, $\mathbb{Q}$ is dense in $\mathbb{R}$). As a hint, let $\varepsilon$ be the value of $f(x)$.

Can you fill in the details?

Kaj Hansen
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  • I'm sure I'll get it in a minute or so. I'll comment back if I get stuck again. Thanks! – Chris Millett Mar 24 '15 at 04:11
  • Yup officially lost in my proof. Any other hints? I understand that y is an element of the rationales and it has to be close to x but now I'm lost. – Chris Millett Mar 24 '15 at 04:21
  • So we have chosen a $\varepsilon > 0$ and $x$ such that $f(x) \neq 0$, and we are trying to show that no $\delta$ exists. To do this, suppose we did have a $\delta$, and remember we can always find a $y \in \mathbb{Q}$ such that $y$ is within $\delta$ distance of $x$. What goes wrong? Unfortunately it's hard to say much more without giving it away, so I'll let you think about it for a little longer. (Math is about struggling with the concepts, after all) – Kaj Hansen Mar 24 '15 at 04:25
  • Yea sorry I have nothing. |x-delta, x+delta| < y? I'm officially lost sorry for keeping you up with this. – Chris Millett Mar 24 '15 at 04:31
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    Yeah, so suppose $y \in [x-\delta, x+\delta]$ is rational. Now try computing the difference of the function values at $x$ and $y$. – Kaj Hansen Mar 24 '15 at 04:36
  • So y = 0 which is less than delta and therefore a contradiction? – Chris Millett Mar 24 '15 at 04:39
  • @ChrisMillett it is possible that $0\notin [x-\delta,x+\delta]$ for specific choices of $x$ and $\delta$, so you cannot choose $y=0$. But as Kaj has mentioned, you can choose $y$ to have a specific Qualitative property, which should tell you something about $f(y)$. – JMoravitz Mar 24 '15 at 04:48
  • @JMoravitz so then f(y) can not =0 and therefore f(y) > 0 <delta? – Chris Millett Mar 24 '15 at 04:53
  • @ChrisMillett Last comment for me, then I'm heading to bed. Although we cannot gaurantee that $y$ can equal zero, according to our definition of $f$, $f$ of any rational number is equal to zero. $x$ and $y$ are in the Domain and we compare them to $\delta$ in our epsilon-delta proofs. Meanwhile $f(x)$ and $f(y)$ are in our range and we compare to $\epsilon$ in our proofs. We do not under any circumstances try to compare $f(y)$ to $\delta$. I strongly recommend looking over definitions of continuity you used when first learning calculus. – JMoravitz Mar 24 '15 at 05:00