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So, start with some open cover, and we'll derive that there's a finite subcover. Firstly, one of the sets in the open cover contains omega. Include that set in the cover and let it be X^-U for some compact set U. The original open cover covers U so... I'll leave it there, I reckon you should be ok with it.

The other ones, I'll have to think about for a few minutes.

Yes, you only need to include one set like that in the finite subcover. Then, note that the cover that you start with has a finite subcover of U, so that the finite subcover of U and the set X^-U cover X^.

However, you've got to make sure that there is a finite subcover of U. To do that, you need to take the open subsets of X^ and turn them into subsets of X by ignoring omega if its an element. They're still open, because an set like that must be of the form X^-V where V is compact, so X-V is open in X.

I'll probably have some sort of anser to the h.e.p. question soonish too.

ok so...
let {Ua} be an open cover of X^
this open cover has an open set say Ub which contains {w}

if X-U (for any open U in X) is compact,
then {Ua}-Ub covers X - U, yes?
so um there's a finite subcover of X-U which is {Uai}-Ub?

and I should let Ub = X^ - C for a compact set C?


I can't quite figure the next step... am I on the right track here? confused

(incidentally, for the subspace part of that question,
can I say that since X c X^, open sets in X will intersect X^ => X is a subspace of X^ with the subspace topology?)

I'm sorry I'm being so stupid today, but could you give me a hint sweatdrop
I guess I just don't see intuitively how/why X^ would be compact at all

thank you so much,
I get it now.

Okay... not really knowing where you're starting off, I'll start with cells, then attaching and skeletons and then finally I'll define a cw pair. I'll use the annulus as an (imperfect) example to motivate and flesh out the discussion.

So, an n-cell is basically a compact blob of n dimensional space which has no holes. More precisely, it is the homeomorphic image of an n-simplex, or an n-ball of some finite radius (where the ball is considered as a subset of n-dimensional real space), though these are the same thing as an n-simplex is homeomorphic to an n-ball. But if you think about it, any set like this is a blob with no holes. Good example: the closed unit disc of radius 1 centred at the origin in R^2.

But what if you wanted to think about a closed annulus? That is, consider the set of points in R^2 which are such that their distance from the origin is not less than 1 and not greater than 2. It has a loop that you can't contract, the circle of radius 1 centred at the origin, so its not a cell. If it was a cell its homotopy groups would all be trivial ie all be just the group with one element. The annulus' pi_1 group is isomorphic to (Z,+) and is generated by the incontractable circle that was described, so it can't be a cell. Idea; what if we allowed collections of cells? Well, you don't want a bunch of pathological things happening. So how do you restrict the collections of cells to ones that aren't crazy? Also, we don't want the cells just floating around all higgeldy piggeldy. How do you make sure that you know how the cells are related to one another, ie how do you specify how the cells are all joined up? Is there a recipe for building up non pathological collections of cells?

To answer these questions you need two concepts: n-skeletons and attaching. I'll define them simultaneously (more or less). Start with an (n-1)-skeleton. To attach an n-cell to the n-1 skeleton we require a homeomorphism, call it f , from the boundary of the n-cell to the (n-1)-skeleton. The new space is the union of the points in the cells in the n-1 skeleton and the points in the cells of the n-cell that we're attaching, but we factor these points by the equivalence relation generated by the relation ~, where x~y iff f(x) = y. The points from the (n-1)-skeleton and the new cell form an n-skeleton.

This is good, but in general we might want to attach a whole bunch of cells at once, and we're still a little informal. So, formally, a 0-skeleton is a collection of points with the discrete topology. An n-skeleton is formed by attaching a collection of n-cells to an (n-1)-skeleton. You can do this attaching a finite or infinite number of times. That is, to get to an n-skeleton from an (n-1)-skeleton you can attach an infinite number of cells, and you don't have to stop at a finite dimension.

The union of the points in the skeletons (factored by the equivalence relations used to attach cells) is called a CW-complex and is topologised so that a sub set is open if and only if its intersection with any of the n-skeletons that makes it up is open.

Back to the annulus. Can we describe it as a CW-complex? Yes. How do you prove it? You contruct the skeleton. Lets sort of work backwards a bit. We'll need two dimensional cells and no three dimensional cells, that's for sure. But if you look at how attaching happens, you'll see we'll need lower dimensional cells too, or else the map from the boundary of the 2d cells won't be definable. In particular, the circles of radius 1 and 2 centred at 0 will have to be in the boundary of the 2d cells. We'll also need to start with a zero skeleton, which is only points. So, we start with points, then add curves (that aren't loops, because a loop has a hole), then areas. Each time we add stuff, it has to be compact. Working out the skeletons is a little tricky, but this example is relevant to the assignment questions asked.

So, the 0-skeleton will be the union of the points

P_1=(1,0), P_2=(2, 0), P_3=(-1,0) and P_4=(-2,0).

These have the discrete topology and we let X_0= U P_i. Next, we'll join these up with some lines.

L_1={(t,0): 1≤ t ≤ 2},
L_2={(t,0): -2≤ t ≤ -1},
L_3={(x,y): x^2 +y^2=1, x≥0},
L_4={(x,y): x^2 +y^2=1, x≤0},
L_5={(x,y): x^2 +y^2=4, x≥0},
L_6={(x,y): x^2 +y^2=4, x≤0}.

So, if we let D be the boundary operator, we have that

DL_1= {P_1, P_2},
DL_2= {P_3, P_4},
DL_3= {P_1, P_3},
DL_4= {P_1, P_3},
DL_5= {P_2, P_4},
DL_6= {P_2, P_4}.

Importantly, the boudary of every line iss two points: we can't have a degenerate boundary, we can't let the endpoints of the line touch, or else there's no chance of the relevant homeomorphisms.

Finally, the areas we want to add are

A_1= {(x,y): 1≤x^2+y^2≤2, x≥0} and
A_2= {(x,y): 1≤x^2+y^2≤2, x≤0}.

You can check the boundary stuff out yourself. The annulus is X=A_1 U A_2. I've kinda glossed over the fact that you aren't really attaching subsets of R^2 but rather homeomorphic images of the canonical n-cell (the n-simplex or n-sphere typically, though the n-cube is also possible, as are a bunch of other things). You probably shouldn't gloss over this.

Now, finally, say I have a CW complex X. If A is a subset which is a union of cells in the complex, then that (X,A) (or (A,X) depending on how you like your bread buttered) is a CW-pair.

Back to the annulus. If X is the annulus and C is the circle of radius 1 centred at the origin, then (X, C) is a CW pair, because X is a CW complex (as I just demonstrated with the skeletons) and C is a union of cells, that is C= L_3 U L_4.

Something that is important is that what is a CW pair is relative to the way that you express the large set as skeletons. So, basically, to do your problem you need to show that (R^n, S^(n-1)) is a CW pair, because CW pairs have the hep (which should have been shown in class, if not find a text that proves it and cite the fact, or you can use this very CW specific text which you can download free). What's a little nice about the question asked is that when you show that its true for n, you build an n-skeleton and you can then use this as a part of the n-skeleton needed for the n+1 case. You might have shown that R^n is a CW-complex in class. If not, think about generalised annuluses. Also, you should have shown that the boundary of an n-cell is the union of two (n-1)-cells. Have a look at that stuff in your notes if its there. It will help.

In the end, just writing the homotopy out is probably quicker, but it takes a lot of thinking to find it. In an exam situation, finding the CW complex proof is probably more doable because its more mechanical, ie you can just turn the handle in your head to get the skeletons (once you've defined a few skeletons you've defined em all).

The modern approach to homotopy requires CW complexes, so if you're going to need this stuff for later work, you might want to read some of that text I linked to.

Ok well I got question2 done eventually, but I've nooo idea about questions 1 or 3. I mean I know kinda how to test them, I just can't figure out what explicit retract would make the thing work. Any hints? sweatdrop