Show that there is no invertible continuous function $f:[0,1) \to (0,1)$ This is a problem I came across recently and have been trying to figure out, but I'm not having a lot of luck!

By considering the restriction of $f$ to $(0,1)$ or otherwise, show that there is no invertible continuous function $f:[0,1) \to (0,1)$.

Hint: there are bijections which map $[0,1)$ onto $(0,1)$. Can you construct one?
The main idea is that such a function has to be always increasing or always decreasing. Assume that there exist three points $a&lt;b&lt;c$ such that $f(a) \leq f(b)$ and $f(b) \geq f(c)$.

If either $f(a)=f(b)$ or $f(b)=f(c)$ we can conclude that the function is not injective hence cannot be invertible. Now we can assume $f(a) &lt; f(b)$ and $f(b) &gt; f(c)$ Then $\exists p \in [f(a),f(b)]\cap[f(c),f(b)]$. 

Then by IVP $\exists m_{1},m_{2}$ such that $a&lt;m_{1}&lt;b$ and $b&lt;m_{2}&lt;c$ with $f(m_{1})=f(m_{2})=p$. Thus $f$ is not injective hence not invertible. This proves that f has to be always increasing or always decreasing. 

Assume $f$ is always increasing and $f(0)=t\in (0,1)$ what maps to the rest of the values in $(0,t)$? This shows that f cannot be bijective (because it is not surjective) hence not invertible.
a theorem states: if f:U->V is a continuous function, and A is an open set in V, then f^-1(A) is also open. the same for closed set. (c.f. Baby Rudin Chap.4 theorem 4.8.)now (0,1) is obviously open and [0,1) is obviously not open.

Edit:Note:This answer is incorrect(see the comments).