Csdrhrt

We can check if some function $g(x) =...$ (some random function, irrelevant to this case) is continuous at a given point $x_{0}=c iff lim_{xto c^+} g(x) = g(c) quadlandquad lim_{xto c^-} g(x) = g(c).$

If both conditions are satisfied, then the function is continuous at such point $x_{0} = c$. I suppose so far everything is correct, I hope?

I need to prove that the function $f(x) = begin{cases}
x & text{if } x in mathbb{Q} \
0 & text{if } x in mathbb{R}/mathbb{Q}
end{cases} quad$ is continuous exactly at one point $x_{0}=0$. The domain is $D_{f}=mathbb{R}$.

I am struggling because this is a function that is "made out of two sub-functions".

It is clear to me that the reason why this function is continuous only at $x_{0}=0$ is because $0$ is a rational number, so we get $f(0)=0$ for both "sub-functions" - and that is the only point in the graph where both "sub functions" (how do I call it properly?) are in the same spot (touching each other). Thus, function is continuous only at that specific point.

How can I finish the proof in a good, math-fashioned style using limits?

I came up with: $$ x_{0}=0in mathbb{Q} rightarrow lim_{xto 0^+} x = 0 quad land quad lim_{xto 0^-} x = 0,$$hence function is continuous at $x_{0}=0$. But I am afraid it's not good enough to be a completed proof.

Hints, tips, advices to formalize this proof are appreciated. Thanks.

Let $x in mathbb R$, $x$ constant. If $(rho_n)$ be a sequence of rational numbers with $$lim rho_n = x Rightarrow lim f(rho_n) = lim rho_n = x$$
If $(a_n)$ is a sequence of irrational numbers with
$$lim a_n = x Rightarrow lim f(a_n) = 0$$
then, it is
$$lim f(rho_n)neqlim f(a_n) Leftrightarrow x neq 0$$
Thus, $f$ is not continuous at $mathbb R setminus {0}$.

We will now show that $f$ is continuous at $0$. It is $f(0) =0$. For all $|x| < 1$, we have :
$$|f(x) - f(0)| = |f(x)| leq max{|x|,0} = |x|$$
thus $lim_{xto 0} f(x) = f(0)$ which means that $f$ is continuous at $0$.

You need to deal with a point $ainmathbb{R} $ and consider the case $a=0$ and $aneq 0$.

The use of $epsilon, delta$ definition is particularly very easy here and hence we follow that approach. Let's first prove that $f$ is continuous at $a=0$. Clearly we have $f(a) =f(0)=0$. Let $epsilon >0$ be given and we choose $delta =epsilon $. If $0<|x-a|=|x|<delta$ then for irrational $x$ we have $$|f(x) - f(a) |=|f(x) |=|x|<delta =epsilon $$ and for rational $x$ we have $|f(x) - f(a) |=0<epsilon $ therefore by definition $lim_{xto a} f(x) =f(a) $ and $f$ is continuous at $a=0$.

For $aneq 0$ we need to consider the cases when $a$ is rational and when $a$ is irrational. I show here the case when $a$ is rational and $aneq 0$. Here $f(a) =0$ and $|f(x) - f(a) |=|f(x) |$. Let $epsilon=|a|/2$ and choose any arbitrary $delta>0$. Then there is an irrational number $x$ which satisfies $$0<|x-a|<min(epsilon,delta)$$ and we have $$|f(x) - f(a) |=|x|=|a-(a-x) |geq |a|-|a-x|>|a|-epsilon=epsilon$$ Thus the limit of $f(x) $ as $xto a$ is not $f(a) $ and $f$ is discontinuous at $a$. You should prove in similar manner that $f$ is discontinuous at $a$ if $a$ is irrational.