user193319

Oh, okay. So my argument for that part was correct.

Hmm...Okay. I don't see either one. I'll have to think about it, although I'm covering the class in like an hour, so maybe I'll just have to talk about something else.

@BenSteffan Are you saying that if $e \subseteq \cup X_i$, then $e$ must actually be one of the cells in the union? I guess I don't see why that's true. Can't a cell be a proper subset of another cell?

I know that $\overline{e}$ must intersect at most finitely many cells, being compact. I think from here I could say that $\overline{e}$ and therefore $e$ is contained in finitely many of the $X_i$'s, but I don't see how to argue it...Well, it could be contained in infinitely many of the $X_i$'s, but somehow we can throw out the subcomplexes to get finitely many.

Yeah, I can see that $\cup_{i} X_i$ must be a union of cells. But assuming $e \subseteq \cup X_i$, I don't know how to conclude $\overline{e} \subseteq \cup X_i$.

@BenSteffan Yeah, I tried using his definition initially (then I moved to the definition where a subcomplex is closed, but that didn't work either). But I don't know how to conclude the closure of a cell must be contained in the union.

Sure, yeah. This is the book I was referencing: lib.ysu.am/disciplines_bk/97a93f6fead36e46864819c339b339ad.pdf

Doesn't feel rigorous enough to me, and sometimes it seems like it matters.

Yeah, I don't like it though.

$\bigsqcup_{i \in I} A_i = \cup_{i \in I} \{(x,i) : x \in A_i\}$.

x and $\varphi_{\alpha}(x)$ are not actually elements in the disjoint union.

Literally in the definition. A disjoint union is actually a collection of ordered pairs.

I don't like all the identifications Hatcher does, and he doesn't use disjoint unions properly.

Okay. I'll have to translate things a bit. Because I'm actually using Massey's book for the definition of a CW complex (he doesn't talk about characteristic maps). He defines CW complexes in a more "internal" way which I thought was clearer.

@BenSteffan The only characterization I'm seeing is that a set $A$ is closed iff $A \cap X^n$ is closed in $X^n$ for all $n$. Is that what you had in mind?

Yes, that's the whole crux of it.

So, is it true that an arbitrary union of subcomplexes is a subcomplex? I don't know how to prove it...

Hmm...yeah, that makes sense.

Yes, that one.

Yeah, that was pretty much what I was attempting.

@BenSteffan anakElip's example is a correct example showing that arbitrary union of subcomplexes might not be a subcomplex, right?

@BenSteffan Is your comment in reference to me trying to find a counterexample using $\{1/n : n \in \mathbb{N} \}$? Yeah, I can see it won't work for a variety of reasons.

I'm still trying to parse the definitions.

Hmm...I don't know...You might be right.

Okay. Good to know. And in the example, I guess you don't even need to glue in spheres. You can glue in loops at each point on the interior of the 1-cell. That would be enough.

Ah, meaning that isn't enough, right? You either need to add that it is closed, or that if $e$ lies in it, so does $\overline{e}$...Yeah, I think so.

Okay...Yeah, I think this checks out.

The 0-skeleton has to have the discrete topology.

Wouldn't you have to attach the spheres onto the 0-skeleton or something?

Wait...is that a cw complex? Are you treating every point in (0,1) as a 0-cell?

I was trying to do something similar using $\{1/n : n \in \mathbb{N}\}$ as the one skeleton, and then gluing in intervals. But I couldn't get it to work.

Sounds like it to me. Yeah, I don't know what that other guy was thinking when he said that arbitrary unions of subcomplexes are subcomplexes.

I know I'm just being a knucklehead as usual.

Yeah...and I don't think that the proof I was proposing is entirely correct either...hmm..

Hmm...something still seems off...hmm...

In this case, I think the easiest definition to work with is that a subcomplex is a closed subspace that is a union of closed cells.

Yeah, I don't think that $e$ will necessarily be in a single $A_i$. But I think what I wrote gives us the conclusion that $\overline{e}$ is in the entire union.

And then my prove of the finite case implies that $\overline{e}$ is in the union of those and hence in the entire union.

Oh, so that would imply that $e$ is contained in finitely many of the $A_i$'s, right?

Hmm...Maybe that's what I am worried about, but I am not sure. I know that the closure of any cell is contained in finitely many cells (because it is compact), so that would imply $e$ itself is contained in finitely many cells.

@anankElpis Why does $e$ have to be contained in a single subcomplex $A_i$?

Maybe...I don't see it yet though.

So, assume $e \subseteq \bigcup_{i \in I} A_i$. Why does it follow that $\overline{e} \subseteq \bigcup_{i \in I} A_i$?

To be a subcomplex, you have to be a union of (closed) cells, and if a cell $e$ is contained in the subspace, then $\overline{e}$ is also contained in the subspace.

@Jakobian Yes, that part of the definition is trivial. But verifying the other condition isn't obvious--namely, that if $e$ is a cell contained in the union, then its closure is also in the union.

@jak

I was trying to fiddle around with $\{\frac{1}{n} : n \in \mathbb{N}\}$ as being my the $0$-skeleton and "gluing" in intervals, but $\bigcup_{n \in \mathbb{N}} \{1/n\}$ is a union of subcomplexes but it isn't closed (in $[0,1])$...But this doesn't quite work.

But, yeah, I don't see why an arbitrary union of subcomplexes is a subcomplex. But @BenSteffan says it is true.

Yes, I think those two definitions are equivalent.

Sweet! Thanks for the confirmation. The proofs are now trivial!