real analysis - The point of discontinuity of a bijective map

Prove/Disprove:The set of point of discontinuity of a bijective map from $(0,1)$ to $[0,1]$ is always countably infinite.

It's a standard exercise to show that any bijective function from $(0,1)$ to $[0,1]$ can't have finitely many discontinuity.Next,in order to find a explicit map I found that the map can be constructed so that it has countably infinite discontinuity.So, I wonder whether such map always have countably many discontinuity.

Answer

The claim, as stated, is false. Consider the map $f:(0,1)\to[0,1]$ as below$$f(x)=\begin{cases}0&x=\frac12\\\frac1{2n}&x=\frac1{2n+2},\;n\in\mathbb N\\\frac1{2n-1}&x=\frac1{2n+1},\;n\in\mathbb N\\1-x&x\notin\mathbb Q\\x&\text{otherwise}\end{cases}$$

Note that even if I weren't able to find such a map, the impetus lies on you to show such a map doesn't exist, not the other way around.

Proof that the function is injective: Consider $x,y\in(0,1)$. Note that rationals are only mapped to rationals, and irrationals only to irrationals, so we can assume that both are either rational or irrational.

If $x,y$ are irrational, and $x\neq y$, then clearly $1-x\neq1-y$. If $x,y\in\mathbb Q$, and neither is of the form $\frac1n$, then $f(x)=x$ is clearly a bijection.

If $x=\frac1n$, note that $x$ is mapped to either $1$, or another element of the form $\frac1m$. Hence, if $y$ is not of form $\frac1m$, then $f(x)\neq f(y)$.

Finally, if both $x,y$ are of the form $\frac1n$, then clearly $f(x)=f(y)\to x=y$, so the function is injective.

Proof that the function is surjective: Clearly, given any irrational $r$ in $[0,1]$, $1-r$ is irrational, and also in $[0,1]$. Hence, $f$ maps to all irrationals in $[0,1]$. Also, $f$ also maps to $0$ and all rationals of the form $\frac1n$. And since all rationals not of the form $\frac1n$ are mapped to themselves, all rationals not of this form are mapped to. So the function is surjective.

Proof that the function is nowhere continuous:

Let $x\notin\mathbb Q$. Clearly, $x\neq\frac12$. Consider a sequence of rationals $\{q_i\}_{i\in\mathbb N}$ that converge to $x$ such that none of the rationals are of the form $\frac1n$. Clearly, $\{f(q_i)\}_{i\in\mathbb N}$ converges to $x$. However, $f(x)=1-x\neq x$, which implies that the function is not continuous at irrational $x$.

Let $x\in\mathbb Q$, and let $x\neq\frac12,\frac13,\frac14$. Consider a sequence of irrationals $\{r_i\}_{i\in\mathbb N}$ converging to $x$. Note that $\{f(r_i)\}_{i\in\mathbb N}$ converges to $1-x$. But, $f(x)=x\neq1-x$, or both $x,f(x)<\frac12$, which means that $f(x)\neq1-x$. So, the function is not continuous at all rationals except $\frac12,\frac13\frac14$.

Consider $x=\frac12,\;\frac13,\text{ or }\frac14$, and consider a sequence of irrationals $\{r_i\}_{i\in\mathbb N}$ converging to $x$. Note that $\{f(r_i)\}_{i\in\mathbb N}$ converges to $\frac12,\frac23,\frac34$ respectively, which $f(x)=0,1,\frac12$ respectively. This implies that the function is not continuous at these three values.