Prove that an open interval $(a,b)$ and a closed interval $[c,d]$ are not homeomorphic.
I'm trying to prove this statement but have only vague ideas on how to start. How may I use the property of connectedness to show this?
[Math] Prove that an open interval and a closed interval are not homeomorphic
general-topology
Best Answer
Here's a proof not using connectedness properties.
Suppose $f: [a,b] \to (c, d)$ is a homeomorphism. Observe that $f(a)\not=f(b)$ as $f$ is injective and consider the point $f(a)+f(b)\over 2$ at half the distance between $f(a)$ and $f(b)$. As $f$ is surjective, $f(x) = {f(a)+f(b)\over 2 }$ for some $x \in [a,b]$.
Now let $\delta$ be enough for both the distances between $f(x)$ and $f(a)$, and between $f(x)$ and $f(b)$ to be less than $\delta$ and yet the open interval centered at $f(x)$ of radius $\delta$ not to cover the whole $(c,d)$.
As $f$ is continuous, there must be an open interval centered at $x$ large enough to contain both $a$ and $b$ and whose image under $f$ is within a distance of $\delta$ from $f(x)$. As such an interval must ecompass the whole $[a,b]$, the function $f$ cannot be surjective for $\delta$ is chosen in such a way that there're points in $(c,d)$ with a distance from $f(x)$ greater than $\delta$.