I want to prove that the metric space $([0,\infty), |\cdot|)$ is not homeomorphic to $(\mathbb{R},|\cdot|)$ (or $((0,\infty),|\cdot|)$, whichever is easier) without using the notion of connectedness.
I have only been given the definition of a homeomorphism in the question (a continuous bijection between topologies/metric spaces whose inverse is also continuous). I am also allowed to use the following definition of continuity:
$f:X\to Y$ is continuous if and only if the preimage of every open set is open. I.e $U\subseteq Y$ is open if and only if $f^{-1}(U)=\{x\in X: f(x)\in U\}\subseteq X$ is open.
Part 1 of the question asks us to prove that for homeomorphic spaces, the image of every open set is open. I am not sure if this is helpful. Can someone give some pointers or an outline of a solution?
Does the intermediate value theorem count as using connectedness?
If $f\colon [0,\infty) \to \mathbb{R}$ is a homeomorphism, then it is in particular a continuous bijection. By surjectivity, there must be $y,z>0$ such that $f(y) = f(0) - 1$ and $f(z) = f(0) + 1$. Use the intermediate value theorem to deduce that there is some $x>0$ such that $f(x)=f(0)$, contradicting injectivity of $f$.