Show that $S^1$ is not homeomorphic to $[0,1)$ 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.
 A: $S^1$ 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.
A: Now I understand your confusion, actually it is more like a matter of logic:
You were assuming the statement $P$ (that they are homeomorphic) at the very beginning, your aim is to deduce that $\neg P$ (that they are not homeomorphic) but you obtain by another way that the  second argument that they are homeomorphic (that $[0,1)\setminus\{0\}$...), this does NOT show that it is in fact $P$ because you have had assumed at the beginning, this is not a way of showing a statement by assuming at the very beginning.
