# Thread: Open sets

1. ## Open sets

I need to show that an open set K \subseteq S^2 is compact in S^2 if and only if it is also compact in R^3..

I seem to be banging my head in a wall here, but for it to be compact, it need to be closed and limited right? And obviously K \subseteq S^2 is closed and limited, hence its compact. But how do i show <=>?

2. What is S^2?

3. Sorry, missed that.

S^2 is the unitsphere. S^2 = {(x,y,z)∈R^3 | x^2+y^2+z^2=1}

4. Does the phrase "compact in s^2" have some fancy topological meaning, or does it just mean what it normally means: that any open cover has a finite subcover?

5. Yes, what it normally means, sorry if my english is a bit sloppy

6. No, I don't think your English is necessarily sloppy - only my understanding is.

I guess what I'm really asking is this: what is an open set in S^2? If you take the whole space, the whole sphere, it most certainly is not open in R^3, but you might be considering an open set in the context of, say, the topological space of S^2 (a topological space has the null set and the whole set being both open and closed).

See what I'm asking?

7. Ok, quoting a bit from my book:

"The compact subsets in R^k are the closed and limited subsets in R^k."

And i need to show that if any open set in S^2 is compact, then it is also compact in R^3. And if it isn't compact in S^2, then it isnt compact in R^3 either.

8. You mean, any subset of S^2, not an open set in S^2, right?

Ah, but I didn't ask what are open sets in R^3. I asked what is an open set in S^2. Because, technically, any subset of S^2 is not open in R^3. Think about it: if I take any subset of S^2, and I construct any neighborhood around any point in that subset, I'm going to get points in that neighborhood that are not in the subset (just move in a direction directly away from the origin, and you'll get off the sphere). Hence, any subset of S^2 is not open in R^3. And, since compactness is defined using open sets, what constitutes an open set is a very important question.

9. /headdesk, you're absolutely right. I misread it, it should just be any SET in S^2 is compact, then it is also compact in R^3.

So => if a set K \subseteq S^2 is compact in S^2 it naturally is also compact in R^3 since S^2 \subseteq R^3. and <= If a set K is compact in R^3 and K \subseteq S^2 it is also compact in S^2? Is that it? Seems so simple now, but not sure if im right, thanx for the help

10. No, that proof doesn't work, because it's what we call "proof by blatant assertion". In other words, you're simply stating the result and claiming that you've proven it.

The reason I'm not sure it works is because, although compact sets in R^3 are the closed and bounded sets (Heine-Borel theorem), I don't know yet what constitutes a compact set in S^2, because I don't know what an open set is, and compactness is defined in terms of open sets.

So, like I said before, you need to define open sets in S^2. Somewhere in your book, they should do that.

11. Originally Posted by Ackbeet

So, like I said before, you need to define open sets in S^2. Somewhere in your book, they should do that.
Sorry if i wasn't clear enough, i misread the question, and it should be "I need to show that a set K \subseteq S^2 is compact in S^2 if and only if it is also compact in R^3.."

12. Ok, the wording "open set" seems to throw you off. Let me say this differently:

Compact sets in R^3 are, by the Heine-Borel theorem, precisely the closed and bounded sets, as we've said before.
Question: are the compact sets in S^2 the same as the closed and bounded sets? They may not be. How do you know what the compact sets in S^2 are?

I, for one, don't know right now what the compact sets in S^2 are. If I don't know that, then there's no way I can prove that a subset of S^2 is compact in S^2 if and only if it is compact in R^3, which is your desired goal.

So you tell me, what constitutes a compact set in S^2?

13. I'm gonna have to try to read up a bit more on this, and then come back to you one of these days. I'm probably mixing up some terms as well. Ah heck, it's not meant to be easy

14. Ok, sounds good. Let us know if you get stuck again.

15. In general, if you are given a subset of a topological space, which is itself referred to as a topological space, then it is assumed to have the subspace topology. Any subset of a topological space inherits the ambiant topology in a natural way. Let $\iota : S^2 \to \mathbb{R}^3$ be the inclusion map. The topology on $S^2$ is the topology generated by the open sets of the form $\iota^{-1}(U)$, where $U$ is open in $\mathbb{R}^3$.

Now it is easy to solve the problem. Show that any open cover of a set in $S^2$ lifts to an open cover of the same set, considered as a subset of $\mathbb{R}^3$. Conversely, any open cover of some subset of $S^2$ (considered as a subset of $\mathbb{R}^3$) can be considered as an open cover of the same set as a set of $S^2$.

By the way, the notation $S^n$ for the $n$-sphere is universal!

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