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we have $f(x) = x$ if $x$ is rational number

$f(x) = 1-x$ if $x$ is irrational number.

By definition of continuity: $|x - 1/2| <\delta => |f(x) - 1/2| <\epsilon$

a) if we will pick any rational number x within $\delta$ and $1/2$: $|x - 1/2|<\epsilon =\delta$

b)a) if we will pick any irrational number x within $\delta$ and $1/2$:$|1 - x - 1/2|<\epsilon =\delta$

Then we can use the fact, that by continuity of f at x:

$f(x) = f(\lim x_n) = \lim f(x_n) = x$ , if x is rational

$f(x) = f(\lim x_n) = \lim f(x_n) = 1 - x$ , if x is irrational

the only possibility of choice is $1/2$

But my question is: how can we prove last two statements by epsilon-delta definition?

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    I do not get it. The proof you are looking for is in your first 5 lines. – 2017-02-16
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    @MiguelAtencia : I mean instead of sequence representation of the problem(last two lines) is there equivalent proof usinng only epsilon-delta definition? – 2017-02-16

1 Answers 1

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If $ x \in \mathbb Q $: $|f(x)-1/2|=|x-1/2|$

If $ x \notin \mathbb Q $: $|f(x)-1/2|=|1-x-1/2|=|x-1/2|.$

Consequence: $|f(x)-1/2|=|x-1/2|$ for all(!) $ x \in \mathbb R$.

Your turn !