For all n, j > 0: (B^{j} is n-connected.
Indeed any convex set (including ∅) is n-connected.)
For all n, j > 0: S^{j} is n-connected just if n ≠ j+1.
“path connected” is 1-connected.
simply connected is (1-connected and 2-connected)
Some sets and their connectivity: The nth entry on the right (n>0) is whether the body described below is n-connected:
S^{0} | F T T T … |
S^{1} | T F T T … |
S^{2} | T T F T … |
B^{n} | T T T T … |
T^{2}: Torus | T F T T … |
ST^{2} = 3D volume bounded by T^{2}: Torus | T F T T … |
Q = ST^{2} omitting a small solid void: | T F F T … |
Two disjoint copies of B^{2} | F T T T … |
Two disjoint copies of ST^{2} | F F T T … |
Two disjoint copies of S^{2} | F T F T … |
Two disjoint copies of Q. | F F F T … |
If sets P' and Q' are base families (they each serve as a base for topologies P and Q respectively) then P'×Q' is a base family which serves as a base for R the product topology for P and Q. I think that for each positive integer n, R is n-connected just if both P' and Q' are both n-connected. I think this goes for infinite products too.
It seems the definition goes over smoothly to ω-connected by reasoning about Hilbert space over the reals and the unit ball there. For Hilbert space we must use balls or simplexes, not cubes.