Good day,
To be honest this semester I give an exercise class on Analysis I and currently I am sitting before a question and I can't get this proof. Let me first give you the task.
Let $f: \mathbb{R} \to \mathbb{R}$, $x \mapsto \begin{cases} 1, &x \in \mathbb{Q} \\ 0, &x \in \mathbb{R} \backslash \mathbb{Q} \end{cases}$ and show that $f$ is discontinuous for all $x \in \mathbb{R}$. Use the $\epsilon$-$\delta$-criterion.
And now the proof.
Let $c \in \mathbb{Q}$, $\varepsilon:=\frac{1}{2}$. Suppose that $\exists \delta>0 \forall x \in \mathbb{R}$ with $|x-c|<\delta$ : $|f(x)-f(c)|<\varepsilon=\frac{1}{2}$. Now there exists a sequence $(x_n) \in \mathbb{R} \backslash \mathbb{Q}$ such that $\lim_{n \to \infty} x_n = c$. Therefore $\exists x_n \in \mathbb{R} \backslash \mathbb{Q}: |x_n -c|<\delta$.
But: $|f(x_n)-f(c)|=|0-1|=1 > \varepsilon=\frac{1}{2}$ and therefore $f$ is not continuous in $c$. Further if we take a $c \in \mathbb{R} \backslash \mathbb{Q}$ this proof follows analogously.
I don't see why there exists this sequence in $\mathbb{R} \backslash \mathbb{Q}$ that goes to our arbitrary chosen $c \in \mathbb{Q}$. Wouldn't this mean that $\mathbb{R} \backslash \mathbb{Q}$ is dense in $\mathbb{Q}$? This doesn't make sense to me. Can someone please clarify this for me? Thanks a lot.
We can define a sequence $(x_n)\to c$ only with irrational terms. By example pick
$$x_n=\frac{c}{\sqrt{\frac{n+1}{n}}}$$
Now observe that $\sqrt{\frac{n+1}{n}}\in\Bbb R\setminus\Bbb Q$ because doesnt exists two consecutive squares for $n\in\Bbb N$, i.e. if $n$ is a square then $n+1$ isnt a square, and viceversa.
But, in general, the point is that $\Bbb R\setminus \Bbb Q$ is dense in $\Bbb R$. This mean that in any interval of reals there are irrational numbers. By example: in the intervals $(\frac{c\cdot n}{n+1},\frac{c(n+1)}{(n+2)})$ there exists irrationals in each one, then picking one irrational from any of these intervals we have a sequence that converge to $c$.