Is the suspension of a countable collection of points in $\mathbb{R}$ a countable collection of circles? I am extremely new to topology and taking an algebraic topology course, and I need some help understanding the behavior of suspensions. The problem I am working on asks about the suspension of the subspace of $\mathbb{R}$ consisting of the sequence 1, 1/2, ... with the limit point 0 added. The main example I'm able to find of a suspension appears to turn a circle into the outer shell of a double-cone with the top and bottom points identified, but does that mean that a suspension generally "squeezes" the figure it starts with on its way up and down?
From what I can tell, the definition of a suspension implies to me that the suspension of a single point in $\mathbb{R}$ should be a circle, so the suspension of several or a countable collection of such points should then be a similarly-numbered collection of copies of $S^1$. Is this correct? And, if it is, would the circles in this case have 1, 1/2 ..., and 0 as their basepoints, or should they all somehow have the same basepoint somewhere?
 A: I'll try to address everything in your question :

The suspension of a point is not a circle. Let's go back to the definition. The suspension of a space $X$ is:
$$S X = X \times [0,1] / \sim$$
where $\sim$ is the equivalence relation generated by $(x,0) \sim (y,0)$ and $(x,1) \sim (y,1)$ (for all $x,y \in X$). This definition, applied to a point, gives $S \{pt\} = \{pt\} \times [0,1] / \sim$ where the equivalence relation is actually trivial, and $S \{pt\} \cong [0,1]$ is just a segment.

The suspension of a "collection" (more formally, a disjoint union) is not the collection of the suspensions. In other words, for example if $X = \bigsqcup_{i \in \mathbb{N}} X_n$ then $S X$ is not equal to $\bigsqcup S X_n$. As Mariano Suárez-Alvarez recommends, let's try to work out for example the collection of two points $X = \{a,b\}$. Then $S X = \{a,b\} \times [0,1] / \sim$ and in fact this is homeomorphic to two segments connected at their endpoints: the collapsed points $X \times \{0\}$ and $X \times \{1\}$.

More generally it's always the case that the suspension is path-connected (and therefore connected); indeed if $[x,t] \in S X$, there exists a path between $[x,t]$ and $[x,0] = [X \times \{0\}]$.

Finally the subspace $X = \{1, \frac{1}{2}, \frac{1}{3}, \dots\} \cup \{0\} \subset \mathbb{R}$ is not (topologically) the disjoint union of its components. Indeed, any neighborhood of $0$ has to contain an infinite number of points -- this wouldn't be the case if the topology was that of the disjoint union. Similarly, the suspension is more complicated than the suspension of a disjoint union of a countable number of points (which would be homotopically equivalent to a countably infinite wedge sum of circles). Based on what I wrote before, try to compute it now.

PS: Regarding base points: there is a priori no canonical base point on a suspension. There would be two candidates (the two collapsed subspaces). On the other hand, the reduced suspension $\Sigma X$ is homotopically equivalent to the suspension (in good cases only! eg. CW complexes. this isn't the case here), and it has a canonical base point.
