TITLE: $[0, 1)$ and $S^1$ not homeomorphic?
QUESTION [6 upvotes]: Let $f:[0, 2\pi) \to S^1 = \{(x, y): x^2 + y^2 = 1\}$ be such that
$f(t) \to (\cos t, \sin t)$
$f$ is a continuous bijection but it is NOT a homeomorphism.
I suppose the only point of contention is $(1, 0)$ in $S^1$.
Is it because there is no open $U \in \mathbb{R^2}$ such that $U \bigcap S^1 =  f([0, 1))$? Because $f([0, 1))$ contains the point $(1, 0) \in \mathbb{R^2}$?

REPLY [8 votes]: Another way you can consider is that the numbers of connected components of two homeomorphic topological spaces have the same amount.
With that fact you can show that $[0,1)$ and $S^1$ are not homeomorphic:
If you remove $\frac{1}{2}$ from $[0,1)$ then we obtain $[0,\frac{1}{2}) \cup (\frac{1}{2},1)$ which is obviously not connected. 
But if you remove an arbitrary $z\in S^1$ then we obtain $S^1$\ $\{z\}$, which is connected.
So $[0,1)$ and $S^1$ can not be homeomorphic.