A simple problem in Real Analysis

[rumjum]

Homework Statement

I am having somewhat a difficult time just understanding a simple concept. I am trying to prove that every open subset G of a separable metric space X is the union of a sub collection {Vi} such that for all x belongs to G, x belongs to some Vi (subset of G).

I am thinking of this problem for the sake of simplicity in just R. Let (a,b) be any open set G in the R. Then I need to show that (a,b) can be shown as union of countable collection of neighborhood of Q ( set of rational numbers).

Homework Equations

The Attempt at a Solution

I need to show that (a,b) is a union of countable collection of {Vis} where each Vi is a neighborhood (qi-e, qi+e) , where qi belongs to Q. Let 's say 0<a <b

I take any point d in Q, such that d belongs to (a,b). This is possible because rational numbers are dense in (a,b). Now, I have some Vi = (d-ri, d+ri) for some b-d>ri >0 and ri belongs to Q. In other words, Vi is within (a,b). Now, I can keep on finding another rational number rj,rk, until d+rj > b and d-rk <a. Let's say d+ rj = b + h' and d-rk = a-h (where h,h' >0).

But, d +rj belongs to Vj and d - rk belongs to Vk. And so (a,b) can be represented as
Vi U ... Vk U Vj. This is countable.

Is my solution ok? Is there an elegant way of proving the same?

Please help. I am stuck. Because I need to extend this to any metric space X (not necessarily R). Thanks

 

[morphism]

You have to remember that not every open set in R is an interval (a,b). What is true is that every open set is a union of intervals -- this is precisely what it means for the collection of open intervals to be a basis. Now of courses this collection is uncountable (it has the same cardinality as R), but we can easily limit the number of open intervals in it: simply take those open intervals with rational end points. This gives us a countable basis.

As in your previous thread, I don't think you understand what this question is asking. You're supposed to give us a countable collection {V_i} that you claim is a basis. This means that any open subset of your metric space can be written as a union of elements of {V_i}. For example, if our metric space is R then as I mentioned above, {(a,b) : a,b in R} works as a basis, as does {(a,b) : a,b in Q}. The latter just happens to be countable.

So now let's try to generalize this to any separable metric space X (whose countable dense subset we'll call D). We know that the collection of balls {B(x, r) : x in X and r>0} forms a basis for X. But of course this isn't necessarily countable. So how about we use the collection {B(d,p) : d in D and p is a positive rational number} instead? This is certainly countable. So all we need to do now is prove that it's a basis. Try to visualize the situation (draw a picture!).

Given a nonempty open set G in X, let's pick a point x in G. We know that there exists an r such that B(x,r) sits inside G. Now as D is dense in X, we can find a d sitting in D$\cap$B(x,r/3) (notice that I shrunk the original ball's radius). We can also find a rational number p such that r/3<p<r/2. So B(d, p) sits inside B(x,r) and contains x.

Can you take it from here?

 

[rumjum]

Thank you very much to start with. I think I now understand the problem better.

Just before you posted your approach, I had mine as follows. I think a part of my approach is similar to your. But yours is much more cleaner.
I will appreciate if you can read the following and understand my line of thought. Learning this course on my own has been difficult and so your proofreading shall be useful.

My method:
______
<So Far>
Consider the dense subset A of the X. Let there be any point "d" in the A,
then let us construct balls of radius (1/n), where n belongs to N.
Let Vn = B(d,1/n), i.e., a sphere with center d and radius 1/n. Now the
collection of such Vn be called as V.

V = {Vn}. This collection is countable as rational numbers are countable.

Let G be any open set in X , such that G is a subset of X.I need to
prove that G is union of a subcollection of V , call it V*.

(Here is my further attempt).

Let there be any open space G subset of X such that there exists
some element x of G that does not belong to V.
Let us have a sphere of radius e, with x as center point. Since x belongs to X, the neighborhood B(x,e) will have points that belong to A (because A is the countable dense subset of X).
Hence some point "q" that belongs to A shall belong to B(x,e).
Now, let us take "q" as center and draw a rational radius 2e> r > e. Hence, the new
sphere B(q,r) shall have x as its point. But B(q,r) shall belong to V ( as it is centered around a point in dense subset and has rational radius).

Also, B(q,r) belongs to V.

Hence, this contradicts the assumption that any open sub set G of X can
have element that does not belong to V. In other words, if x belongs to G,
then x belongs to V.

Besides , B(q,r) can be chosen as a subset of G (take radius "r" such that the
r < d(y,q) , where y is some limit point G. (I need to polish this part).

Hence, for all points x in G we can find some B(q,r) such that B(q,r) is a subset of G. Hence, G can be shown as the union of such B(q,r)s. But, the collection of B(q,r) belongs to V and so this subcollection of B(q,r) is countable. Hence, the collection of V
form the basis of X.

I feel better about this except that I need to polish some ports. But is this clear enough?
Thanks once again

 

[morphism]

I'll have a look later this evening. Right now I have to go.

 

[rumjum]

No problem. Have a good day!

 

[morphism]

I think you have the right idea, but the write-up is kind of messy. For example, your definition of V, as stated, doesn't work. You've defined it to be the collection of balls of radius 1/n centered at a particular point d in A, while we want to do this for all d in A.

Another thing worth commenting about is that you're saying "x is in V", when this is actually nonsense because the elements of V are sets. You probably meant to say that x is in some set that belongs to V.

And the major issue with your solution is the following:

Besides , B(q,r) can be chosen as a subset of G (take radius "r" such that the
r < d(y,q) , where y is some limit point G. (I need to polish this part).

You really need to clean this up because it is the crux of the problem. If you look at my post, you can find out how to do this. (By the way, when you started your proof you used balls of the form B(q,1/n), while here you're using B(q,r), which is sort of inconsistent, but I guess in the long run it doesn't matter.)

Next:

Hence, for all points x in G we can find some B(q,r) such that B(q,r) is a subset of G. Hence, G can be shown as the union of such B(q,r)s.

The first sentence should be: "for all x in G, we can find some B(q,r) such that B(q,r) is a subset of G containing x" (the last bit is important, for otherwise your second sentence wouldn't follow).

Finally:

But, the collection of B(q,r) belongs to V and so this subcollection of B(q,r) is countable.

I have no idea what the second half of this sentence is trying to say, or why it's relevant at all.

 

[rumjum]

Thanks for the detailed response. Really appreciate it.