Proof that $(a,b)\not\cong[a,b]$ How to prove that
$$(a,b)\not\cong[a,b]$$
(not homeomorphic) as subsets of real line?
Is it true that in some topology $(a,b)$ is closed?
Thanks a lot!
 A: To show that two spaces are not homeomorphic you can find a property that holds for one and fails for the other, and is invariant under homeomorphism.
For example $[a,b]$ is compact, whereas $(a,b)$ is not. The continuous image of a compact set is compact, so if $f\colon[a,b]\to(a,b)$ were a homeomorphism you would have that $(a,b)$ is the continuous image of a compact set and therefore compact. Contradiction.
As for the second question, you can always declare a set is closed. Namely, $A\subseteq\mathbb R$ in the topology generated by $\mathbb R\setminus A$ you have that $A$ is closed.

Here is a slightly more hands-on approach:


*

*Show that if $f\colon[a,b]\to(a,b)$ is continuous and injective then it is either strictly increasing or strictly decreasing. Assuming without loss of generality that $f$ is increasing.

*Denote by $b'=f(b)$, then we have to have that $f(x)\leq b'$ for all $x\in[a,b]$ by the above argument.

*Since $b'<b$ we have to have some point $x\in(b',b)$ which cannot be in the range of $f$.
A: This is more of a continuation of Asaf's answer, but there is an even stronger sense that $(0,1)$ can be closed in a topology on $\mathbb{R}$.  One can actually construct a metric $d$ on $\mathbb{R}$ satisfying the following properties:


*

*every open set in the usual topology on $\mathbb{R}$ is open in the $d$-metric topology;

*the metric $d$ is complete (all Cauchy sequences with respect to this new metric converge);

*$\mathbb{R}$ is separable (has a countable dense subset) with respect to the new topology; and 

*$(0,1)$ is a clopen (closed and open) set in the new topology.

A: Another way similar to the answer already given:
Let $f\colon[a,b]\to(a,b)$ is a continuous map then $[a,b]\setminus\{b\}=[a,b)$ is connected set but $f([a,b))$ is not connected as we have deleted $f(b)$ from $(a,b)$.
A: Assume that $f: (a,b) \rightarrow [a,b]$ is a homeomorphism, i.e. in particular it's continuous and bijective (one-to-one and onto). Then there are $u,v \in (a,b)$ with $f(u)=a$, $f(v)=b$. From the intermediate value theorem, it follows that $f([u,v]) \supset [a,b]$. Since $\text{rng } f = [a,b]$, it cannot be an actual superset, so it further follows that $f([u,v]) = [a,b]$. Now, since $f$ is bijective, it must be that $(a,b) \setminus [u,v] = \emptyset$, since there's no place to map any remaining points to without destroying bijectivity. You obviously also have $[u,v] \subset (a,b)$, and together this shows $[u,v] = (a,b)$, which is impossible.
There is thus no homeomorphism from $(a,b)$ to $[a,b]$, hence the two sets are not homeomorphic.
