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Question: Consider the following construction: take any uncountable set of elements and form a chain ($x < y < z <\dots$ ) of the elements (in particular, I think one can well-order the reals using the AoC, and then take this ordering) and now consider these elements to be 0-cells. Now attach a 1-cell between two 0-cells $x, y$ if there is no $z$ between $x$ and $y$. [Intuitively, this kind of looks like a "really long" real line.] The claim is that this is not a CW-complex. But why?

Motivation: On a final, I was asked a question which sparked my interest in this finite vs infinite distinction in CW complexes. In particular, I thought that this "long line" construction, though somewhat artificial, should qualify as a "nice space" since it is contractible. When I asked about it, I was told that this was not a CW complex because the topology restricts us to putting a finite number of cells in each dimension. I'm not quite sure why this is true.

Can anyone shed some light on this for me?

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    This space shouldn't be contractible; it's too long to contract (see http://en.wikipedia.org/wiki/Long_line_(topology) ). But I don't see any reason why this shouldn't be a CW-complex as long as you put the right topology on it.2011-04-20
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    Precisely what topology are you placing on your set? If you topologize it as a CW-complex, it's not a single really long line, it falls apart as a disjoint union of continuum-many copies of $[0,\infty).$2011-04-20
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    @Qiaochu: the long line has all homotopy groups trivial but is not contractible, hence is not homotopy equivalent to a CW complex by Whitehead's theorem.2011-04-20
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    @Chris: interesting. Could you elaborate on the second sentence of your first comment?2011-04-20
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    @Qiaochu: suppose for definiteness we start with some uncountable limit ordinal with the discrete topology (since the 0-cells always have the discrete topology) and for each ordinal $\alpha$ we glue in a 1-cell between $\alpha$ and $\alpha+1$. It's not hard to see that in the resulting quotient space, for each limit $\lambda$, the set of ordinals of the form $\lambda+n$, along with the 1-cells between them, form a clopen set homeomorphic to $[0, \infty)$, and the whole space is just the topological disjoint union of these pieces.2011-04-20
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    In the long line proper, instead of starting with the discrete topology and then passing to a quotient topology, we start with the order topology on the ordinals and pass to the order topology on the copies of $[0,1)$ that have replaced the ordinals. This gives a much coarser topology: for example, $\omega$ is in the closure of the finite ordinals in the order topology, but not in the discrete topology.2011-04-20

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To elaborate on Chris Eagle's point, if you really only add edges between elements $x$ and their immediate successors $S(x)$, then the resulting space is not connected. For example consider the well-ordered set $\{1,2\}\times \mathbb Z_+$ in the dictionary order. Then your construction fills in the edges in the two vertical copies of $\mathbb Z_+$ giving you a disjoint union of two copies of $[0,+\infty)$. Things are even worse when you look at an uncountable well-ordered set, since you will have uncountably many points with no immediate predecessor.

On the other hand, if you glue together all of the copies of $[0,+\infty)$ you get the "long line," which is not a CW complex. In each case, the underlying set is the same, but the CW topology is different.

In general, CW complexes are allowed to have infinitely many cells in each dimension. The topology is given by the so-called weak topology. A subset is open if and only if its intersection with each cell is open in that cell.