real analysis - If f1 and f2 are 2 continous functions then f = { f1(x), x is rational and f2(x) , x is irational is continous in x0 if only if f1(x0)=f2(x0)

I have to prove that.

I know that i have to take a string with rational number and a string with irational numbers but i don;t know how to do next.

Please help me !

Answer

Hint (just one side)

Assume $x_0$ irrational.

Being the rational numbers dense in $\mathbb{R}$ one could take a succesion $x_n \in \mathbb{Q}$ $x_n \to x_0$
By continuity of $f_1$ and $f$ if we take $\epsilon>0$ $\exists \delta>0$:
$$|x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon$$
$$|x-x_0|<\delta \Rightarrow |f_1(x)-f_1(x_0)|<\epsilon$$

Take $N \ \forall n>N: |x_n-x_0|<\delta$

Then by continuity:

$$|x_n-x_0|<\delta \Rightarrow |f(x_n)-f(x_0)|<\epsilon$$
$$|x_n-x_0|<\delta \Rightarrow |f_1(x_n)-f_1(x_0)|<\epsilon$$

How $x_n$ is rational, and $x_0$ is irrational:

$$|x_n-x_0|<\delta \Rightarrow |f_1(x_n)-f_2(x_0)|<\epsilon$$

So:

$$|f_1(x_0)-f_2(x_0)|\leq |f_1(x_0)-f_1(x_n)+f_1(x_n)-f_2(x_0)|$$
$$\leq |f_1(x_0)-f_1(x_n)|+|f_1(x_n)-f_2(x_0)|\leq2\epsilon$$

Then the difference $|f_1(x_0)-f_2(x_0)|$ is arbitrary small then are equal.

When $x_0$ is rational is analogous using the density of the irrational numbers.