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Exercise:

Let $f(x)$ be continuous for $0 \leq x \leq 1$. Suppose further that $f(x)$ assumes rational values only and that $f(x) = \frac{1}{2}$ when $x = \frac{1}{2}$. Prove that $f(x) = \frac{1}{2}$ everywhere.

Attempt:

Suppose $f(\frac{1}{2} + \delta_1) = p \neq \frac{1}{2}$ is rational.

There is a irrational number $q$ that is between $\frac{1}{2}$ and $p$.

For $f$ to be continuous, there must be a $f(\frac{1}{2} + \delta_2) = q$ where $\delta_2<\delta_1$.

But, by $f$'s definition, it contains no irrational numbers.

So, the only other option is that $f(\frac{1}{2} + \delta_1) = \frac{1}{2}$.

Request:

Is my proof correct? If not, where and why'd I go wrong?

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    You may want to spell out the name of the theorem you use (the Intermediate Value Theorem) – 2017-01-17
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    @HagenvonEitzen -- Regarding my second statement? – 2017-01-17
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    The shape of the proof looks good. How do you know that there is an irrational number between $\frac 12$ and $p$? For the proof to be tight you should justify each of the intermediate statements either by reference to previously established results, or by showing why they are true as part of the proof. – 2017-01-17
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    @MarkBennet -- Alright, thanks. I presume that's the same thing HagenvonEitzen pointed out, correct? – 2017-01-17
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    You should add that $\mathbb{R}$ is Archimedean to justify the existence of an irrational in your interval, and that you used the intermediate value theorem – 2017-01-17
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    @Canardini -- No idea what that means (but I bet you're right ). I'm a beginner in Analysis. – 2017-01-17

1 Answers 1

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Why do you say that $\delta_1<\delta_2$.

here is an other approach.

$f$ is continuous at $[0,1]$ thus $f([0,1])$ is an interval.

but $f([0,1]) \subset \Bbb Q$ so the only possibility is $f([0,1])=\{\frac{1}{2}\}$.