I want to show that $S^1$ is not homeomorphic to $[0,1)$.
When $X$ is homeomorphic to $Y$ then $X-\{a\}$, where $a \in X$ is homeomorphic to $Y-\{f(a)\}$. Assuming $S^1$ and $[0,1)$ are homeomorphic, I remove a point $a$ other than $0$. We see that $[0,1)-\{a\}$ is homeomorphic to $S^1-\{f(a)\}$. But $S^1-\{f(a)\}$ is connected while $[0,1)-\{a\}$ is not. This is a contradiction which proves the claim.
However when I remove $0$, We see that $ [0,1)-\{0\}$ is homeomorphic to $S^1-\{f(0)\}$ which is in fact true. This shows $S^1$ is indeed homeomorphic to $[0,1)$ opposite of what I am trying to show. So I am not sure what has gone wrong in these reasonings.
Best Answer
$S^1$ is not homeomorphic to $[0,1)$. Indeed, if you remove any point from $S^1$, you get a connected space while if you remove any point that is not equal to $0$ from $[0,1)$, you get a disconnected space. Thus, $S^1$ and $[0,1)$ cannot be homeomorphic.
In details, assume that there exists a homeomorphism $f \colon S^1 \rightarrow [0,1)$, show that $f|_{S^1 \setminus \{ f^{-1}(1/2) \}}$ must be a homeomorphism between ${S^1 \setminus \{ f^{-1}(1/2)}\}$ and $[0,1/2) \cup (1/2,1)$ and show that this is not possible.
Your mistake is that if they are homeomorphic, then connectedness should be preserved regardless the point you remove, which must be arbitrary.