So, I was wondering if one could find a continuous bijection from a non-compact set to a compact set. I had an exam today where one of the questions asked me to find a continuous bijection (if there is one) from $\mathbb{R}$ to $\left[0,1\right]$ which obviously does not exist because $f$ has to be monotonic. That generalization sort of popped up in my head, and at some point I thought I had a proof (I've tried to prove that for every compact $X$ in the codomain, $f^{-1}(X)$ is compact), which proved to be wrong. So the question is, is that fact even true, at least maybe by adding some more conditions?
General Topology – Continuous Bijection from Non-Compact to Compact Set
analysisgeneral-topology
Best Answer
Take the map $f : \mathbb{N} \to (\{\frac{1}{n} \mid n > 0\} \cup \{0\})$ defined by $f(0) = 0$ and $f(n + 1) = \frac{1}{n+1}$. Here we use the discrete topology on the domain, and the subspace topology inherited from $\mathbb{R}$ for the codomain.
The map is a continuous bijection, the domain is not compact, and the codomain is compact.