Continuous and Discontinuous Functions - Variation of Dirichlet function

Question

Be $f: R \longrightarrow R $ a real function where:

$ f(x) = \begin{cases} x + 1, & \text{if $x \in Q $} \\[2ex] 2, & \text{if $x \in R-Q$} \end{cases} $

Where the graph "jumps" among the two straight lines of the graph. There is just one point where both straight line are continuous, that is x = 1 and y = 2.

If we have an sequence $ (x_n) $ that converge to $\sigma $, then $(x_n) + 1 \longrightarrow \sigma + 1$ by rational values $(Q)$ and $(x_n) = 2 \longrightarrow 2 $ by irrational values $(R - Q)$. Then, the unique point of converge is $\sigma = 1$, where both straight lines are y = 2.

My question is for whats point's in the domain of $f$, $f$ is continious and discontinious?

Answer 1

Actually, $x=1$ is the only point of continuity. To see why, first let $x_0\in \Bbb Q^c$. Then there exist two sequences $$a_n=x_0+{1\over n}\to x_0\implies f(a_n)\to 2\\b_n={\lfloor nx_0\rfloor\over n}\to x_0\implies f(b_n)\to x_0+1$$since $x_0+1\ne 2$, continuity fails here.

Now let $x_0\in \Bbb Q-\{1\}$. Then by defining $$a_n=x_0+{1\over n}\\b_n=x_0+{\pi\over n}$$ we see again that $f(x)$ is not continuous in $x=x_0$ either in this case. So the conclusion is what we mention first.

More generally, the following function$$f(x)=\begin{cases}g(x)&,\quad x\in \Bbb Q\\h(x)&,\quad x\notin \Bbb Q\end{cases}$$is continuous in the roots of $g(x)=h(x)$.