$(0,1), [0,1), [0,1]$ are not homeomorphic I need to show $(0,1)$, $[0,1)$ and $[0,1]$ are not homeomorphic using  intermediate value theorem (without using connectedness).
I have already did proved that $(0,1)$, $[0,1]$ are not homeomorphic but I struggle with the 2 other couples.
My proof: assume there is an homeomorphism $f:(0,1)\rightarrow[0,1]$ and take $a,b$ such that $f(a)=0, f(b)=1$. So using intermediate value theorem we can say that $f([a,b]) = [0,1]$. so $f$ isn't injective.
Will appreciate any help
 A: Suppose $f:(0,1)\to [0,1)$ is a homeomorphism.
Then $\exists x\in(0,1)$ such that $f(x)=0$. By continuity for every  $\epsilon >0$, $ \exists y<x<z$ such that $\{f(y),f(z)\}\subset[0,\epsilon)$.
Without loss of generality assume that $f(y)<f(z)$.
Now by intermediate value property of $f$, we have $f$ on $(x,z)$ attains all the values in $(f(x),f(z))$. So there exists $z_0\in (x,z)$ such that $f(z_0)=f(y)$.
Contradicts the fact that $f$ is one-one.
A: You have to be a little more careful: you’ve written your argument on the assumption that $a<b$, but it could be that $b<a$, in which case it’s the interval $[b,a]$ that maps to $[0,1]$. You get the same contradiction, of course.
HINT: The same argument works if the domain of $f$ is $[0,1)$. For $(0,1)$ and $[0,1)$ you have to work a little harder, but you can still use the same idea. Suppose that $f:(0,1)\to[0,1)$ is a homeomorphism. There is an $a\in(0,1)$ such that $f(a)=0$, and for each $n\ge 2$ there is a $b_n\in(0,1)$ such that $f(b_n)=1-\frac1n$. Let $$I_n=\begin{cases}[a,b_n],&\text{if }a<b_n\\ [b_n,a],&\text{if }b_n<a\,;\end{cases}$$ clearly $f[I_n]\supseteq\left[0,1-\frac1n\right]$. The sequence $\langle b_n:n\ge 2\rangle$ has a convergent subsequence; let $c$ be the limit of this subsequence, and let
$$I=\begin{cases}
[a,c),&\text{if }a<c\\
(c,a],&\text{if }c<a\,.
\end{cases}$$
What can you say about $f[I]$?
A: $[0,1]$ is compact but the other two are not. Hence none of them are homeomorphic to $[0,1]$.
Now to show that  $(0,1)$ and $[0,1)$ are not homeomorphic.
Suppose $f:[0,1)\to(0,1)$ is a homomorphism.
Then $f(0)\in (0,1)$ so $(0,1)\setminus \{f(0)\}$ is a disconnected set.
But $f((0,1))=(0,1)\setminus \{f(0)\}$, which is not possible as the continuous image of a connected set is connected .
A: Observe that if $\;f:X\to Y\;$ is a homeomorphism between two toplogical spaces and $\;x\in X\;$ then
$$\;g:X\setminus\{x\}\to Y\setminus\{f(x)\}\;,\;\;g(a)=f(a)\;\;\forall\;a\in X\setminus\{x\}\;$$
is also a homeomorphism on the respective top. spaces with the induced topology, and thus: if
$$f:[0,1)\to(0,1)\;\;\text{is a homeomorphism, then}\;\;g:(0,1)\to[(0,1)\setminus\{f(0)\}$$
is also a homeomorhpism. but this can't be since $\;(0,1)\;$ is (path) connected whereas $\;[(0,1)\setminus\{f(0)\}\;$ isn't.
Complete the other pair now (there you have path connectedness if you don't want connectedness...I don't understand why, but whatever)
