Homework Help: Metric Spaces and Open sets

1. Jun 23, 2010

Buri

I'm reading Analysis on Manifolds by Munkres and in the section Review of Topology Munkres states the following theorem without proof:

Let X be a metric space; let Y be a subspace. A subset A of Y is open in Y if and only if it has the form A = U ∩ Y where U is open in X.

All he has defined is a metric space, subspace, B(y;ε) = {x|d(x,y) < ε} which is the ε neighborhood of y. And he defined open sets to be: A subset U of X is said to be open in X if for each y ∈ U there is a corresponding ε > 0 such that B(y;ε) is contained in U.

I have never taken Topology or even read about it, so I wrote a proof for it which I'm not sure is correct.

My Proof

(⇒)

Since A is open then for x0 ∈ A there exists an ε > 0 such that B(x0;ε) is in A. Further, B(x0;ε) is open (I've proved this already). Now taking the union of all such ε neighborhoods of x's in A also produces and open set (I've also proved this). Therefore, letting U = U B(x; ε) proves this direction as clearly, A = U B(x;ε) ∩ Y.

(⇐)

A = U ∩ Y

This means that any x ∈ A is also in U and Y. Therefore, since x ∈ U, which is open, there exists B(x; ε) in U. But I want the open ball to be in A so letting ε' = ε and then B'(x;ε') = B(x;ε)∩Y. However since B(x;ε) is in U, then B'(x;ε') is in U∩Y which is A. □

My Questions

Am I on the right track? What if I have a metric X, and a closed set Y in X and then I choose U to be open and not entirely in Y but entirely in X. I define A to be U∩Y, but then A isn't open since it contains part of the boundary of Y. What am I not understanding? And if X is a metric space and Y a subspace, if A is open in Y, does it necessarily imply that A is open in X also?

Thanks for the helps!

2. Jun 23, 2010

Office_Shredder

Staff Emeritus
This half looks fine
This is a little sketchily worded but it looks like you have the right idea. B' here is the ball of radius epsilon inside of the set Y around the point x.

Keep in mind you only want it to be open in Y. If Y has boundary points, then the standard preconceptions of openness don't really apply. For example, in the set [0,1] the set [0,1/2) is open

3. Jun 23, 2010

Buri

Could you elaborate?

4. Jun 23, 2010

tmccullough

This is not really a theorem at all, but a definition, that's why you feel very unsure of your proof. If you have a topological space (set $$X$$ with associated topology $$\tau$$ definied on it), then for any $$Y \subset X$$, there is automatically a topology which can be associated with it - the same sets are open, just restrict them to $$Y$$. This is usually called the induced or subspace topology.

Your proof is fine, it's just a restatement of the fact that you started with a valid topology, then restricted to a subspace.

5. Jun 23, 2010

Office_Shredder

Staff Emeritus
Obviously any point in the set [0,1/2) that is not 0 has a neighborhood around it. So let's look at the point 0 now:

Let's pick the ball of radius 1/4 around 0. In the set [0,1], this is going to be [0,1/4). Notice you don't get any negative numbers because they just aren't in Y. [0,1/4) is contained in [0,1/2) so we've confirmed that [0,1/2) is in fact open

6. Jun 23, 2010

Buri

"Half-balls" are allowed? I never knew that. I find it confusing how open sets seem to be relative to the set in which they are contained in. Because according to the definition I'm given for epsilon-neighbourhoods its always open, however the ball you've chosen appears to be neither but its open in [0,1] and is not even a complete ball. If you could help me clear up this confusion I'd really appreciate it.

7. Jun 23, 2010

Buri

This is a bit confusing to me because it appears from Office Shredders posts that it wouldn't be true. So I'm probably not understanding the whole notion of a topology on X.

For example, I find it confusing that my text says that if $$A\subset Y\subset X$$ Then A is open in X if A is open in Y and Y is open in X. I don't see why we need to have Y open in X for this to be true. Though I guess Office Shredders example tends to be an example why, but then again I find it weird that open sets seem to be relative to the set in which it is contained. Once again, if you can help me understand this better I'd appreciate it.

8. Jun 23, 2010

Buri

Is this what restricting open sets means? For example let X = R, and choose Y = [0, 1]. Then (-1, 1/2) is open in X, but then when we 'restrict' this open set to Y we get the set [0, 1/2) and therefore open in Y?

Another question, if X were to have one topology and Y another, or they both have the same...how does this affect everything?

9. Jun 23, 2010

tmccullough

Without getting too long winded (remember that this is abstract and understanding will come with time) - yes, openness is always relative to your "universe".

You are inherently thinking of metric spaces as basically being the real line with the regular old $$|\cdot|$$ metric, but what if you restrict yourself to something smaller?

What do open sets (with this metric) look like if our universe is restricted to just the rational numbers? Lots of subsets are both open and closed - kind of different from the whole real line.

If the universe is restricted to the integers? Any subset is both open and closed.

Note that your book doesn't say that Y must be open in X for A to be open in X, but rather, if Y is open in X then A is open in X. If Y is not open, then no guarantees can be made - as with Office Shredder's example.

10. Jun 23, 2010

Office_Shredder

Staff Emeritus
It's not a half ball. The interval [0,1/4) contains every point in [0,1] within a distance of 1/4 from 0. The reason we don't get the set (-1/4,1/4) in this ball is because there are no negative numbers in our original set! So there's no way they lie within a distance of 1/4 from 0.

In general, the topology on a subspace barely resembles the topology of the original space, despite the fact that it seems they should be exactly the same (the more your subspace looks like your original space, the more the topologies will look the same)

11. Jun 24, 2010

Buri

My text says that the fact that the ball B(y;ε) = {x|d(x,y) < ε} is open follows from the triangle inequality, so it would appear that its 'openness' is completely dependent on the metric. Why is it that B(y;ε) appears to be so independent? Is this why we use it to prove a set is open?

12. Jun 24, 2010

Buri

Ahhh I see...I think I'm starting to understand this better now, but like tmccullough said, I'll need more time to fully grasp these ideas. Thanks a lot for your help! :)