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
['https://math.stackexchange.com/questions/3778961', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/813552/']
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 $f[I]$?
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.