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/']
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.
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)