I want to show that a function $f:\mathbb{R} \to \mathbb{R}$ that satisfies the Darboux property (Intermediate Value Theorem) and $f:\mathbb{R}\setminus\mathbb{Q} \to \mathbb{R}$ is inyective , is continuous.
I know that I somehow need to use Baire’s Category Theorem, but I don’t know how.
Please help!
Thanks!
Assume $f$ discontinuous at $x$ so $f$ has a limit point (could be finite or infinite) there $y \ne f(x)$ and wlog we can assume $y<f(x)$ and there is $x_n$ increasing to $x$ st $f(x_n) \to y$. If $y$ finite let $u=\frac{f(x)+y}{2}, v=\frac{2f(x)+y}{3}$ otherwise if $y=-\infty$ here take $u=f(x)/2, v=2f(x)/3$ so $y<u<v<f(x)$ and pick $x_1$ st $f(x_1) < u < f(x)$ (as $f(x_n) \to y<u$) so by Darboux there is $x_1<z_1<x$ with $f(z_1)=u$ and then by Darboux there is $z_1<w_1<x$ st $f(w_1)=v$ so also by Darboux we have $[u,v] \subset f[z_1,w_1]$
But now pick $w_1<x_2<x$ st $f(x_2)<u$ and repeat the above so find $x_2<z_2<w_2<x$ st $f(z_2)=u, f(w_2)=v$ so $[u,v] \subset f[z_2,w_2]$. By choice we have $[z_1,w_1]$ and $[z_2,w_2]$ disjoint.
But for every $c \in [u,v]$ there is $y_1(c) \in [z_1,w_1], y_2(c) \in [z_2,w_2], f(y_1(c))=f(y_2(c))=c$ while $y_1(c) < y_2(c)$ and of course $y_1(c) \ne y_1(c'), c \ne c'\in [u,v]$ and same for $y_2$
In particular the set $I$ given by $c \in [u,v]$ for which $y_1(c)$ is irrational is uncountable and then the set $J$ of $c \in I$ for which $y_2(c)$ is irrational is also uncountable hence non empty and this means that $f(y_1(c))=f(y_2(c))$ for two distinct irrationals ($y_1(c)<y_2(c)$) and that is a contradiction!
Hence $f$ is continuous.
We just need to establish that $f$ is injective on all of $\Bbb{R}$. From there, we can establish that $f$ is continuous.
Let us suppose that $f(a) = f(b)$ (obviously at least one of $a$ and $b$ must be rational, but this is not important!). Because $(a, b)$ contains at least two (or indeed, uncountably many) irrational points, there must be some $c \in (a, b)$ such that $f(c) \neq f(a) = f(b)$. So, for every $y$ strictly between $f(c)$ and $f(a) = f(b)$, since $f$ is Darboux, there must exist some $p_y \in (a, c)$ and $q_y \in (c, b)$ such that $f(p_y) = f(q_y) = y$.
Since $f|_{\Bbb{R} \setminus \Bbb{Q}}$ is injective, we know that $$(p_y, q_y) \in (\Bbb{Q} \times \Bbb{R}) \cup (\Bbb{R} \times \Bbb{Q}).$$ We can think of the above as the countable union of horizontal and vertical lines. Because there are uncountably many $y$ strictly between $f(c)$ and $f(a) = f(b)$, and our map $y \mapsto (p_y, q_y)$ is necessarily injective, we know, by pigeonhole principle, that there must exist distinct $y_1, y_2$ such that $(p_{y_1}, q_{y_1})$ and $(p_{y_2}, q_{y_2})$ lie on the same line, i.e. they lie on $\{r\} \times \Bbb{R}$ or $\Bbb{R} \times \{r\}$, for some fixed rational $r$.
In the former case, $f(p_{y_1}) = f(p_{y_2})$. In the latter case, $f(q_{y_1}) = f(q_{y_2})$. In either case, we get $y_1 = y_2$, which contradicts $y_1$ and $y_2$ being distinct. Thus, $f$ is indeed injective, and hence continuous.
