If f(x) = x if x is rational , f(x) = 1-x if x is irrational,prove that function is continious at 1/2

Question

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?

Answer 1

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 !