Continuous f(x)

We claim that the only point of continuity is $\frac{1}{2}$ . Indeed, we will prove that $f(x)$ is continuous at $\frac{1}{2}$ and discontinuous at any $a$ such that $a\neq \frac{1}{2}.$ Let us prove the continuity of $f(x)$ at $\frac{1}{2}.$ If $x\geq \frac{1}{2}$ then $1-x\leq f(x) \leq x,$ and if $x \leq \frac{1}{2}$ , then $x\leq f(x) \leq 1-x.$ Applying the Sandwich theorem we get that $\lim_{x\to \frac{1}{2}} f(x)=\frac{1}{2}=f(\frac{1}{2}).$ So the function $f(x)$ is continuous at $\frac{1}{2}$ .

Now let us consider any real number $a$ such that $a\neq \frac{1}{2}.$ We have to consider two cases: $a$ can be rational or irrational.

Case 1 . If $a$ is irrational, we can always construct a sequence of rational numbers $a_{n}$ that tends to $a,$ for example, $a_{n}=\frac{\left\lfloor{10^{n} a}\right\rfloor}{10^n}.$ Since $a_{n}$ is a sequence of rational numbers then $\lim_{n\to \infty} f(a_{n})= \lim_{n\to \infty} a_{n}=a \neq 1-a=f(a).$ Therefore, $f$ is discontinuous at $a$ .

Case 2 . When $a$ rational and different from $\frac{1}{2}.$ In this case, we can construct a sequence of irrational numbers $a_n$ that tends to $a,$ for example by making $a_n=a-\frac{\sqrt{2}}{n}.$ Then we have that $\lim_{n \to \infty}f(a_{n})=\lim_{n \to \infty}(1-a_n)=1-a \neq a=f(a).$ Therefore $f$ is discontinuous at $a$ also in this case.

So our proof is complete.