I'm going through the earlier chapters in books and making sure I can do everything (and addressed many short-comings, like compactness) but I've come across something I can't do.
In "Introduction to Topological Manifolds" – by John Lee "Exercise 2.28" there is a question which I am stuck on, and confused about the wording.
It goes as follows:
Let $X$ be the half open interval $[0,1)\subset\mathbb{R}$ and let $\mathbb{S}^1$ be the unit circle in $\mathbb{C}$ – both with their Euclidian metric topologies.
Define a map $a:X\rightarrow\mathbb{S}^1$ by $a(s)=e^{2\pi js}$ (where $j=\sqrt{-1}$).
Show that $a$ is continuous and bijective but not a homeomorphism.
It includes a picture of [——) and a circle with -)[- at the "3-o'clock" position, with the [ facing vertically up. That's not really important here though.
I am (finally!) happy with the topology part. I know to implicitly consider the subspace topology. I can see it is clearly bijective (but I am not sure how to prove this, probably by saying "but there's only one solution on the interval $[0,2\pi)$ so I could do it)
But how do I show it is continuous and that its inverse isn't.
What have I tried?
I recently did http://www.maths.kisogo.com/index.php?title=Continuity_definitions_are_equivalent – this shows that definitions of continuity are the same. I could scrape together some proof by cutting the circle into little open intervals and looking at the pre-image. Then show they generate the topology but this isn't a very general method.
The "weakest" (as in most specific) continuity definition (the metric space one where $d(a,b)=|a-b|$) is something I can do. I have no experience with trying to show the most general one. Continuity in terms of open sets, and I want to know what methods I can employ to show continuity.
Quick addendum: other examples are welcome I'd like to practice this (proving continuity between topological spaces) so any other examples are welcome.
Best Answer
A continuous real-valued function defined on a compact set is bounded. You can see that semi-closed interval is not compact by defining an unbounded continuous function. Then you can conclude that no map from a compact set to this set could be a homeomorphism.