Help understanding how a topological cone is constructed. This is the definition of the cone from wikipedia:

I have trouble understanding why the cone has that particular shape. Based on what I understand about equivalence relations and quotients, shouldn't the cone consist of copies of X over the length $(0,1]$, together with the copy $X \times \{0\}$ collapsed to a point? The mental image of a cone that I have is that of a cylinder, together with just one point on top.
If we take the set $X$ to be $S^1$, for example, in what sense are the "middle" circle of the cone and the circle $S^1 \times \{0.5\}$ in the original cylinder equivalent?
Is the last remark in the pic supposed to explain this situation? If so please explain how that homeomorphism works. 
 A: The description of your mental image as a "cylinder, together with just one point on top" is misleading. 
For example, suppose we start with a "cylinder", namely $S^1 \times [0,1]$. And in order to match up with the picture in your post, let's imagine that the $[0,1]$ direction is upside down, with $0$ "on top" and $1$ on bottom. So, the top of this cylinder is $S^1 \times 0$.
Now, how shall we go about "collapsing $S^1 \times 0$" to a point? If you insist on keeping with the "cylinder" picture, the geometry will mislead you. Instead, you must somehow distort your imagination (and your cylinder) so that all of the infinitely many points in the set $S^1 \times 0$ somehow "become" or are "collapsed to" one point. 
One way to do this is to let your cylinder become distorted into a frustum (really I mean just the outer boundary of the frustrum, not the whole solid object). So far, the frustrum is still topologically the same as a cylinder. Now imagine the top circle of the frustrum gets smaller and smaller. As long as the top circle remains of positive radius, the frustrum is still topologically the same as the cylinder. But in the limit, when the radius of the circle becomes $0$, you get a cone.
Having said all of that, when topologists work formally with quotient spaces such as a cone, they do not bother with the "slow distortion" process. Instead, they learn the abstractions of the quotient topology and quotient maps, and they work formally with that. Nonetheless, the "slow distortion" process can still function as a useful intuition.
Addendum to address a question in the comments: Your question is best answered using the concept of a quotient map. In this case, consider the cylinder $S^1 \times [0,1]$ and the cone
$$C = \{(x,y,z) \in \mathbb R^3 \,\,\bigm|\,\, z = 1 - \sqrt{x^2 + y^2} , \,\, 0 \le z \le 1\}
$$
Let me use $A = (0,0,1)$ for the "apex" of the cone.
The quotient map for this example is the continuous, surjective function 
$$f : S^1 \times [0,1] \to C
$$
that is defined by the formula
$$f(x,y,z) = \left(x(1-z),y(1-z),z \right)
$$
Notice that the pre-image of the apex $f^{-1}(A)$ is equal to $S^1 \times \{1\}$, the "top" of the cone. Also, $f$ is one-to-one over the subset $C - \{A\}$, meaning that for each point $P \in C-\{A\}$ its preimage $f^{-1}(P)$ is a single point. Thus, the answer to your question regarding the "middle" circle of the cylinder is that $f$ maps $S^1 \times \{0.5\}$ homeomorphically onto the "middle" circle of the cone, namely the circle on the $z=0.5$ plane satisfying $\sqrt{x^2+y^2}=0.5$.
It might be instructive to pick up a topology book, read the definition of a quotient map, and verify that $f$ is a quotient map. That verification would be exactly how one rigorously checks the opening sentence of that wikipedia entry that you pasted into your question, by comparing that definition to the "ordinary" cone $C$. Furthermore, once that has been checked, one may quote a theorem about the "universal" nature of the quotient topology to conclude that the quotient space $CX$ is indeed homeomorphic to the "ordinary" cone $C$.
A: You might also want to think about paths. If you take the bottom and top circles  and connect two points. As the top circle gets deformed, that path also moves. So the path that was on the boundary of the cylinder will form on the boundary of the cone. 
