Cone over $X$, equivalence relation I have a question about the definition of the cone over $X$.

If $X$ is a space, define an equivalence relation $X\times [0,1]$ by $(x,t)\sim (x',t')$ if $t=t'=1$. Denote the equivalence class of $(x,t)$ by $[x,t]$. The cone over $X$, denoted by $CX$, is the quotient space $X\times [0,1]/\sim$.

I acutally do not get this definition of the equivalence relation....
Sure, two points $(x,t), (x',t')$ are equivalent if $t=t'=1$. 
But which points are equivalent to (for example) $(x,\tfrac12)$? The relation does not tell anything about the cases when $t\neq 1$ or $t\neq t'$, which feels incomplete. But I think I am doing a horrible mistake here.
How does this relation include every pair $(x,t)\in X\times [0,1]$, when the relation is only defined for $t=1$? 
I am currently studying "Introduction to algebraic topology" by Joseph J. Rotman.
An exercise goes as follows:

For fixed $t$ with $0\leq t<1$, prove that $x\mapsto [x,t]$ defines a homeomorphism from a space $X$ to a subspace of $CX$.

Which revealed my misunderstanding.
So I am not understanding which equivalence classes there are.
For every point $(x,t)$ with $t\neq 0$, the equivalcence class should just contain this one point.
Can you elaborate more?
Thanks in advance.

The short answer to your question "which points are equivalent to $(x,1/2)$?": Only $(x,1/2)$.
The cone should really look like the geometric cone you know. For a generic example, start with a cylinder and identify the points at the top face. This is the motivation for the name cone for $CX$. Up to pathological examples, the other examples look similar. You start with a topological space $X$, make a cylinder with $X$ as a face, so we get $X \times [0,1]$. Then, we identify the points of one face.

