Why is this quotient space of $[-1,1]$ homeomorphic to $S^1$ and $[-1,1)$ is not?

Question

By using the Compact-to-Hausdorff property, I have shown that there is a homeomorphism between $[-1,1]\diagup\sim$ where $\sim$ is defined to be the equivalence relation $-1 \sim 1$.

Now I know that $[-1,1)$ is strictly not homeomorphic to $S^1$. Here is my confusion: in some sense, isn't $[-1,1)$ the same as $[-1,1]\diagup\sim$?

Another clarification would be useful to me: $[-1,1)$ is not homeomorphic to $S^1$ because, loosely, on the circle $-1$ and $1$ are very close, but in the interval they are not. Thus, the function $f(t)=(\cos(\pi t), \sin(\pi t))$ between $[-1,1)$ and $S^1$ cannot be continuous. I am not really sure on this reasoning - if we take an open set about the point $(-1,0)$, then is its inverse under $f$ not an open set in $[-1,1)$?

Any help would be appreciated!

Answer 1

Yes, there is an obvious bijection between $[-1,1)$ and $[-1,1]/\sim$ ... but ...

I'll say first that even from the perspective of set theory, having a bijection between two sets is not the same as saying that those two sets are "the same".

Even worse, if you are concerned not just with sets but with topological spaces, then having a bijection is a very, very far cry from having a homeomorphism. Starting with $[-1,1)$ and $[-1,1]/\sim$, there are many, many examples of two topological spaces which have a bijection between them but no homeomorphism. Here's just a couple more:

• The integers $\mathbb{Z}$ and the rational numbers $\mathbb{Q}$. All points of $\mathbb{Z}$ are isolated, no point of $\mathbb{Q}$ is isolated, and so they are not homeomorphic. But both sets are countable.
• The open square $(0,1) \times (0,1)$ and the open interval $(0,1)$. Removing any point of $(0,1)$ disconnects it, but there is no point of $(0,1) \times (0,1)$ whose removal disconnects it, and so they are not homeomorphism. But there is a bijection between these two sets.

Answer 2

One topological invariant is connectedness; so, no interval will be homeomorphic to a circle because if you remove any [non-end]point from the former it becomes disconnected, whereas any single point deletion does not disconnect the circle.