# Can a cell-complex have no zero cell?

My question is very simple, but I wasn't able to find an answer in various sources. Cell-complexes are commonly presented using an inductive construction where $$n$$-cells are attached to $$(n-1)$$-cells, starting with the data of a collection $$X^0$$ of $$0$$-cells.

But sometimes, one will define a $$2$$-cell for instance, as being a sub-cell complex of something, raising the question: can a cell complex $$0$$-skeleton be empty? And more generally, can a cell-complex have empty skeletons until a dimension $$k$$?

It doesn't seem absurd to me that the answer should be yes, but it worries me to never see the case taken into account in the inductive construction.

• The empty space is a CW-complex with no $0$-cells. – Lord Shark the Unknown Jan 2 at 7:41
• What do you mean by "cell complex"? Are you using that as a synonym for CW-complex? – Eric Wofsey Jan 2 at 7:44
• No, I mean a cellular complex, which is basically the same as a CW complex with no closure-finiteness or weak-topology condition (see link ) – TryingToGetOut Jan 2 at 7:49

Suppose $$X$$ is a nonempty cell complex and let $$n$$ be minimal such that $$X$$ has an $$n$$-cell. If $$n>0$$, then this $$n$$-cell has an attaching map $$S^{n-1}\to X^{n-1}$$ where $$X^{n-1}$$ is the $$(n-1)$$-skeleton of $$X$$. But by minimality of $$n$$, $$X^{n-1}=\emptyset$$. Since $$S^{n-1}$$ is nonempty, there are no maps $$S^{n-1}\to\emptyset$$, so this is a contradiction.
So, if $$X$$ is any nonempty CW-complex, it must have a $$0$$-cell. (Of course, the empty space is a CW-complex with no cells at all!)
• In this game, the boundary of the $0$-disc is empty so it attaches perfectly well to the empty $-1$-skeleton :-) @TryingToGetOut – Lord Shark the Unknown Jan 2 at 8:01