Iterated suspensions of a finite discrete space If you take iterated suspensions of a one-point space, you get $n$-balls. If you start from a two-point space, you get $n$-spheres. 
Is there a natural interpretation to the iterated suspensions of a three-point space ? What about a general discrete finite space ?
 A: It's easier than it looks. Consider a finite discrete space $X_{n+1}$ with $n+1$ points. This is a wedge sum of $n$ copies of $S^0$, the two-points space, $X_{n+1} = \bigvee^n S^0$. And if $\Sigma$ denotes the reduced suspension, then $\Sigma(X \vee Y) \cong \Sigma X \vee \Sigma Y$. Thus by induction:
$$\Sigma^d X_{n+1} \cong \bigvee^n \Sigma^d(S^0) = \bigvee^n S^d$$
is a wedge sum of $n$ copies of the $d$-sphere.

If you use the unreduced suspension (since you say that the suspension of a point is a ball I guess this is actually the case), the result is a bit weirder
The first suspension of $X_{n+1}$ is $n$ circles, which you can imagine are all arranged in a row from left to right, and they are then glued as follows: the right hemicircle of the first one is glued to the left hemicircle of the second one, the right hemicircle of the second one is glued to the left hemicircle of the third one, and so on. This is of course homotopy equivalent to the wedge sum of $n$ circles, by collapsing the left hemisphere of the first circle.
The pattern continues: $S^d X_{n+1}$ consists of $n+1$ copies of $k$-spheres, glued in a similar manner: the right hemisphere of the first one is glued to the left hemisphere of the second one, the right hemisphere of the second one is glued to the left hemisphere of the third one, and so on. (This is still homotopy equivalent to a wedge sum of $n$ copies of $S^d$). If it helps for visualization, you can also imagine a homeomorphic space that is a $d$-dimensional "rectangular" shape divided in $n$ "chambers" (when $d=1$, this looks like a row of a chessboard).
