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Is true that the function $f:\mathbb{R}\rightarrow \mathbb{R}$

$f(t):=\begin{cases}t \: , \: t\in \mathbb{R}-\mathbb{Q} \\ 0\: , t\in \mathbb{Q} \end{cases}$

is continuous only in the point $0$ ?

If the answer is yes, how I can prove the assertion?

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  • $\begingroup$ Yeah, it is true. $\endgroup$ – Mankind Dec 23 '15 at 23:21
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It is pretty easy to show that $f$ is continuous at $0$. For any $x_0 \neq 0$, by Bolzano-Weierstrass any neighbourhood of $x_0$ has two limit points, so $f$ is not continuous anywhere else.

If you don't want to use Bolzano-Weierstrass theorem, assume that $f$ is continuous at $0 \neq x_0 \in \mathbb{Q}$. That is, $$ \forall \epsilon > 0, \exists \delta > 0\mid |x-x_0| < \delta \Rightarrow |f(x) - f(x_0)| = |f(x)| < \epsilon $$ However, inbetween every two real number there exists an irrational number, so another $x'\mid |x' - x_0| < \delta$ exists with $|f(x')| \geq \epsilon$ when $\epsilon$ is smaller than $x'$. Therefore $f$ is not continuous at any rational $x_0$.

Similarly $f$ is not continuous at any two irrational numbers since

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