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Consider the topological space $(X,\mathscr{U})$, where $X=\mathbb{R}^2$ and the topology $\mathscr{U}$ is generated by the collection of sets $\{(0,0)\}\cup \{I_a\}$ where $I_a$ are the open intervals on the rays departing from the origin.

We then make a quotient of this space, lets call it $Y$, by identifying the points on the closed disk $D^2$ centred at the origin of radius 1.

Intuitively I can see some closed subsets that are compact (essentially closed segments on the rays) and others which are not (any subset extending over a "continuity" of rays). But, being the question "Describe all the compacts", can you suggest me a way to find them all? Is there a procedure someone can follow in this or in similar cases?

Are there some results achievable by finding which separation axiom hold in this space?

Finally, what is the role of the identification in this particular problem? I can't find anything relevant.

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Quotient of $ X $ or $ U $? And what is a half line? –  user38268 Jul 17 '12 at 23:28
    
Thanks for raising the point. It should be clear now. –  Temitope.A Jul 17 '12 at 23:33
    
I think in the prequotient space you want the topology on $X$ generated by the set $U$. As it stands, $U$ is not a topology on $X$. –  J. Loreaux Jul 18 '12 at 0:06
    
I guess that would be it because as it currently stands the empty set is not in U –  user38268 Jul 18 '12 at 0:12
    
Not only that, but in general, unions of elements of $U$ are not elements of $U$. –  J. Loreaux Jul 18 '12 at 0:15

1 Answer 1

up vote 2 down vote accepted

The answer provided below should be sufficient to walk you through the proof. I've left some places for you to fill in the proof since this is homework.

I would start by describing the compact sets in the prequotient space. From here, images of compact sets in the prequotient space are compact in the quotient space (most of the time, these are not the only compact sets). In general, there is not a particularly good method of describing all the compact subsets of a given topological space. As BenjaLim pointed out, the space you described is Hausdorff, so any compact set must be closed.

Let $(\hat X,\mathscr{U})$ denote the topological space $\hat X=\mathbb{R}^2$ with $\mathscr{U}$ the topology generated by the set $U=\{(0,0)\}\cup\{I_a\}$, where $I_a$ are the open intervals on the rays departing from the origin. Let $(X,\mathscr{V})$ denote the topological space obtained by identifying all the points on the unit disk in the space $(X,\mathscr{U})$. Let $p:\hat X\to X$ denote the quotient map.

Now what is the space $(\hat X,\mathscr{U})$ like? Well, notice that each open ray from the origin is naturally homeomorphic to $\mathbb{R}$ with the standard topology. So, it is not difficult to see that $(\hat X,\mathscr{U})\cong \coprod_{a\in[0,1]} (X_a,U_a)$, where $\coprod$ denotes the disjoint union space with the disjoint union topology, and the topological spaces $(X_a,U_a)$ are all $\mathbb{R}$ with the standard topology, except $(X_0,U_0)$, which is the one point space. Finally, note that compact sets in the disjoint union topology are precisely those sets which are finite unions $\bigcup_{i=1}^n K_{a_i}$ where $K_{a_i}\subset X_{a_i}$ is compact in the $U_{a_i}$ topology (You should prove this, it's not particularly difficult). So, the compact sets in the prequotient are precisely those which are finite unions of "closed and bounded" (in the subspace topology, which is homeomorphic to $\mathbb{R}$) subsets of the open rays extending from the origin, possibly along with the point $(0,0)$. Hence, the images of these under the map $p$ are all compact by the continuity of $p$.

Now, what about the quotient space $(X,\mathscr{V})$? The purpose of this identification is to make it so that the space is no longer obviously a disjoint union of simpler spaces. Now, let $b=p((0,1))$, that is, $b$ is the image of the unit disk under $p$. Now, $p$ is a closed map (you could prove this), and so restricting it to the set on which it is injective we get a homeomorphism from $\mathbb{R}^2\setminus\overline{\mathbb{D}}$ (with the subspace topology inherited from $(\hat X,\mathscr{U})$) onto $X\setminus\{b\}$ (with its subspace topology). Thus, any set $K\subset X$ such that $a\notin K$, is compact in $X$ if and only if it is the image under $p$ of a compact subset of $\hat X$.

From the previous paragraph, we have reduced the problem to describing the compact sets of $X$ that contain the point $a$. Let $K\subset X$ be compact with $a\in K$. Suppose that there exist points $a_n\in K$ with $a_n\not=a$ such that $\hat{a}_n=p^{-1}(a_n)$ have the property that $\hat{a}_n$ and $\hat{a}_m$ lie on different rays if $m\not=n$. Now we form saturated open sets $\hat{V}_n$ in $\hat X$ such that $\hat{a}_k\in \hat{V}_n$ if and only if $k\le n$, and that $\bigcup \hat{V}_n = \hat X$. Then $K\subset\bigcup V_n$, but is not contained in any finite subcollection, which will contradict the compactness of $K$. Hence $K$ is the image of a set that is contained in finitely many of the rays extending from the origin. I leave it to you to construct some appropriate $\hat{V}_n$ sets (I know these exist, I have constructed them).

The previous paragraph proves that any compact set in $X$ must be the image of a compact set in $\hat{X}$. This completes the proof.

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I think that we are doing $X/A$ where $A$ is the closed unit disk and not the unit circle. I am not sure if this makes a difference to your answer above though. –  user38268 Jul 18 '12 at 1:10
    
Ah, my mistake, it doesn't change much though, the answer is still the same. –  J. Loreaux Jul 18 '12 at 1:11
    
I think you have some $X$ and $\hat{X}$'s mixed up too. But +1 anyway for this nice answer. –  user38268 Jul 18 '12 at 1:12
    
@J.Loreaux In paragraph 6 (line 4) did you mean "if and only if k = n"? If not K could be contained in finite subcollection, I Think (the saturated open sets looks like circle sectors with <=... Thanks for the answer, the notion I needed was the disjoint union topology. –  Temitope.A Jul 18 '12 at 6:36
    
@Temitope.A: No, I meant $k\le n$, but either one should work just fine. If $\hat{V}_{n_1},…,\hat{V}_{n_m}$ is any finite subcollection, then let $N=\max\{n_1,…,n_m\}$. Then $\hat{a}_k\in\bigcup_{i=1}^m \hat{V}_{n_i}$ if and only if $k\le N$. Thus no finite subcollection contains all of the $\hat{a}_k$, and so no finite subcollection covers $K$. –  J. Loreaux Jul 20 '12 at 18:48

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