What does it mean to "identify" points of a topological space? I was recently reading about circle rotations (a basic example in dynamical systems) and got confused by some notation. It said consider the unit circle $S^{1} = [0,1]/{\sim}$, where $\sim$ indicates that $0$ and $1$ are identified. 
What does "identify" mean? and how is the set $[0,1]/{\sim}$ different from the set $[0,1]$? Thanks!
 A: It means first that $\sim$ is the following equivalence relation: for $x,y\in[0,1]$, $x\sim y$ iff $x=y$ or $\{x,y\}=\{0,1\}$. (It’s easy to check that this is an equivalence relation on $[0,1]$. As a set $[0,1]/{\sim}$ is the set of equivalence classes of $\sim$; these equivalence classes are the singletons $\{x\}$ for $x\in(0,1)$ and the doubleton $\{0,1\}$. Finally, it says that we’re endowing $[0,1]/{\sim}$ with the quotient topology, defined as follows: 

if $q:[0,1]\to[0,1]/{\sim}$ is the map that takes a point $x\in[0,1]$ to its $\sim$-equivalence class, then $U\subseteq[0,1]/{\sim}$ is open in $U\subseteq[0,1]/{\sim}$ iff $q^{-1}[U]$ is open in $[0,1]$.

It turns out that this has a rather simple intuitive meaning in this case: the effect of all this is to treat $0$ and $1$ as if they were the same point, a single point whose open nbhds are obtained by ‘gluing together’ open nbhds of $0$ and $1$ in $[0,1]$. That is, the construction effectively just bends the unit interval around into a circle and identifies the endpoints in the sense that it makes them identically the same point, and the result is the circle $S^1$.
A: Identified here means they are considered to be the same.  In your example, the unit interval has two ends.  If you identify $0$ and $1$, it has no ends, so looks like a circle.  In $[0,1]$, if you ask for points $\frac 18$ from $\frac {15}{16}$ there is only one-$\frac {13}{16}$.  In $[0,1]/$ ~ if you ask for points $\frac 18$ from $\frac {15}{16}$ there are two-$\frac {13}{16}$ and $\frac 1{16}$ 
