Need a hint: prove that $[0, 1]$ and $(0, 1)$ are not homeomorphic I need a hint: prove that $[0, 1]$ and $(0, 1)$ are not homeomorphic without referring to compactness. This is an exercise in a topology textbook, and it comes far earlier than compactness is discussed.
So far my only idea is to show that a homeomorphism would be monotonic, so it would define a poset isomorphism. But the can be no such isomorphism, because there are a minimal and a maximal elements in $[0, 1]$, but neither in $(0, 1)$. However, this doesn't seem like an elemenary proof the book must be asking for.
 A: Let $f:[0,1]\to (0,1)$ be a homeomorphism. Let us suppose, for a contradiction, that $f$ is not monotonic. In particular, either there exists:
(1) $a<b<c$ with $f(a)<f(b)$ and $f(b)>f(c)$ 
                    or

(2) $a<b<c$ with $f(a)>f(b)$ and $f(b)<f(c)$.
Let us consider case (1) without loss of generality.
Exercise 1: Prove that the image of $(a,c)$ under $f$ is not open in $(0,1)$. In particular, we have a contradiction and $f$ must be monotonic.
Let us assume without loss of generality that $f$ is monotonically increasing.
Exercise 2: Prove that $f$ is not surjective. In particular, we have a final contradiction.
Therefore, $f$ is not a homeomorphism.
The following exercises are relevant:
Exercise 3: Prove that there is a continuous surjection $f:(0,1)\to [0,1]$. Do you think that there is a continuous surjection $f:[0,1]\to (0,1)$?
Exercise 4: Does there exist a surjective open map (i.e., open sets are mapped to open sets) $f:[0,1]\to (0,1)$. Do you think that there is a surjective open map $f:(0,1)\to [0,1]$?
Exercise 5: Find an example of a continuous bijection between topological spaces that is not a homeomorphism. 
A: Here's a hint.  If $X$ and $Y$ are homeomorphic (via $f$) then $X \setminus \{x\}$ and $Y \setminus \{f(x)\}$ (subspace topologies) are homeomorphic via the restriction of $f$, for any $x \in X$.
A: $\text{id}_{(0,1)}$ has no maximum (easy)
Every continuous function $f$ on $[0,1]$ has a maximum, otherwise pick $x_n$ such that $f(x_n)>\text{sup}f([0,1])-\frac{1}{n}$ for all $n$. Then Bolzano-Weierstraß does it.   
A: There is no continuous and bijective function $f:(0,1) \rightarrow [0,1]$. In fact, if $f:(0,1) \rightarrow [0,1]$ is continuous and surjective, then $f$ is not injective, as proved in my answer in Continuous bijection from $(0,1)$ to $[0,1]$.  This is a consequence of the intermediate value theorem, which is a theorem about connectedness. Are you allowed to use that?
