Prove that an open interval (0,1) and a closed interval [0,1] are not homeomorphic. Prove that an open interval (0.1) and a closed interval [0,1] are not homeomorphic.
I am trying to prove this statement but only vague ideas on how to start.
Not using connectedness properties.
Please help
 A: Suppose $f:[0,1]\to(0,1)$ is a homeomorphism. In particular, then, it is a continuous surjection. Since $f$ is a continuous real-valued function on the closed interval $[0,1],$ it has an absolute minimum value, call it $m$. Then $m\in(0,1),$ and $\frac m2\in(0,1).$ Since $f$ is surjective, there is some $x\in[0,1]$ such that $f(x)=\frac m2\lt m.$ This contradicts the fact that $m$ is the minimum value of $f.$
A: Consider the sequence $\left(\frac1n\right)_{n\in\mathbb N}$ in $(0,1)$. It has no subsequence which converges to an element of $(0,1)$. However, every sequence of elements of $[0,1]$ has a subsequence that converges to an element of $[0,1]$, by the Bolzano-Weierstrass theorem and because $[0,1]$ is closed. Therefore, $(0,1)$ and $[0,1]$ are not homeomorphic.
A: The open interval is not compact metric space,  but the closed is. 
A: Removing an endpoint (i.e. $0$ or $1$) from the closed interval $\left[ 0,1\right]$ leaves either $\left( 0,1\right]$ or $\left[ 0,1\right)$, both of which are connected spaces. On the other hand, removing any point from $(0,1)$ will leave a disconnected space (with two components). See the answer to this question. 
This formalizes the feeling that $0$ and $1$ are "special points" in the space $\left[0,1\right]$. This kind of argument is often useful when dealing with simple spaces. For another example, see my answer to this question.
