Infinitely many discontinuities for a bijective function from $[0,1)$ to $(0,1)$ Show that any bijection from $[0,1)$ to $(0,1)$ has infinitely many discontinuities.
I have thought about this question but I have no any idea. Any idea is valuable for me, thanks.
 A: HINT: Suppose that $f:[0,1)\to(0,1)$ is a bijection with only finitely many discontinuities. Let $F$ be the set of points at which $f$ is not continuous. Then $[0,1)\setminus F$ is a union of intervals, and $f$ is continuous on each of these intervals. Moreover, all of these intervals are open except possibly one of the form $[0,a)$, in the case in which $f$ is continuous at $0$. Let $\mathscr{I}$ be the set of these intervals.

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*Use the continuity of $f\upharpoonright I$ and the fact that $f$ is a bijection to show that $f[I]$ is an interval for each $I\in\mathscr{I}$. Moreover, $f[I]$ is an open interval if $I$ is.


*Show that if $0\in F$, so that each $I\in\mathscr{I}$ is an open interval, then on the one hand $|f[F]|=|\mathscr{I}|-1$, since $(0,1)\setminus f[F]$ is the union of $|\mathscr{I}|$ open intervals, but on the other hand $|F|=|\mathscr{I}|$. Conclude that $0\notin F$, so $f$ is continuous at $0$.
Let $F=\{a_1,\ldots,a_n\}$, where $0<a_1<\ldots<a_n<1$. Let $I_0=[0,a_1)$, for $k=1,\ldots,n-1$ let $I_k=(a_k,a_{k+1})$, and let $I_n=(a_n,1)$. Then $f$ is continuous on each of the intervals $I_k$ for $k=0,\ldots,n$.

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*Use the fact that $f[F]=F$ to show that $f$ must map each of the intervals $I_k$ to one of the intervals $(0,a_1)$, $(a_k,a_{k+1})$ for $k=1,\ldots,n-1$, or $(a_n,1)$.


*Get a contradiction by showing that $f[I_0]$ is a half-open interval, not an open interval.
A: Suppose that $f$ is discontinuous only at a finite set $A$, and let $B = A \cup \{0\}$.  Set $|B| = n$.
$[0,1)\setminus B$ is a disjoint union of $n$ open intervals $\{I_k\}$ on which $f$ is continuous, so they are mapped by $f$ to a disjoint union of open intervals $\{f(I_k)\}$ that must line up end-to-end (as their complement is finite).
But lining up $n$ open intervals end-to-end in $(0,1)$ gives us $n-1$ endpoints, which cannot be in bijection with $B$.
