Could someone give me a hand on this problem?
Let such that for every there exists a neighborhood such that is at most countable. Prove that is at most countable.
1. is second countable (open balls centered at rational coordinates and having rational radii form a countable base).
2. Every subspace of a second countable space is second countable (the obvious proof works).
3. Every open cover of a second countable space has a countable subcover.
(Pf: Each set in an arbitrary open cover is some union of those basis sets, so the union of all the sets in the cover is the union of all of those associated basis sets. Now for each of those basis sets in that final union, pick one set of the original cover that it's a subset of. That picks out a countable subset of the original cover, since the basis is countable. The union of all those basis sets covers the space, and is also a subset of the union of that countable subset of the original cover. Thus that countable subset of the original cover is also a cover. This has shown how to produce a countable subcover for any open cover.)
With those onbservations in hand, the problem is straighforward to prove.
You must know that the countable union of countable sets is countable.
From the given, the collection is an open cover of
From #3 there is a countable subcover.
You can get as the countable union of countable sets that look like
Assume X is uncountable.
( is the open ball in with center and radius ).
If such that , then every ball about intersects in a countable set.
That would mean that is countable, and so is countable, which we've assumed is not the case.
Therefore . (This isn't necessary, but might help to avoid concern about the meaning of .)
For , define .
If were countable , then would be countable - but we're assuming it's not.
Therefore such that is uncountable. Let .
Now divide into a grid of solid n-dimensional cubes of side length
Do it so that the coordinates of the verticies of the cubes are like , where .
The cubes are solid, and so will intersect in their common faces. This won't matter. What matters is this:
The distance between any two points in a single cube is less than or equal to (a diagonal has length ),
and therefore, for each of those cubes , if , then .
The number of cubes is countable (its cardinality is less than or equal to the cardinality of all the verticies of all the cubes, which is countable).
Here are some thoughts on finishing the proof:
Consider how is defined. If , what does that say about ?
If , for some cube , what can you say?
So, combining those two observations: what can you about for those cubes ?
Now recall that is uncountable. Is it possible that is countable for ALL of those cubes ?
You should be able to produce a contradiction.
Thank you Plato and John. Today my professor showed us the proof of Lindelof's lemma. So, now I know how to prove this problem by applying the lemma. Let . Then is an open cover for . By the lemma, has a countable subcover that also covers . By hypothesis, is countable for every . Hence, is countable since it is a countable union of countable sets.
I really want to understand your geometric proof, John. However, I can't figure out the answers to the questions you asked. I spent more than an hour going over your steps, but I don't really know how to proceed. I would really appreciate it if you can guide me more on your geometric proof.
Let . Then if , then . I asked myself. Is ? If it is, then is countable. I am not able to connect the dots here. Also, I don't really know how to use the fact that the function is defined as the the supremum of the set of radii of the balls.Consider how is defined. If , what does that say about ?
Then since for any .If , for some cube , what can you say?
There are 4 established facts that, with a little bit of reasoning, will complete the proof.
By the way each was defined: if , then is countable for all .
In particular, (Fact1) if , then is countable.
By the way the size of those cubes were choosen, it's been established that:
(Fact2) if , then
Also, by construction, (Fact3) those cubes, countable in number, cover all of .
Finally, the specific was found so that (Fact4) is uncountable.
I envisioned the proof proceeding this way:
Pick any one of those cubes .
If , then , and so is countable (Fact1).
If , then , and so (Fact2).
Thus, if , then , so .
Thus, if , then is countable.
(*) Incorporating that the empty set is countable proves: for every one of those cubes C, is countable.
(I use "countable" to mean finite or countably infinite, which means at most countably infinite. Sometimes people use "countable" only for exactly the cardinality of the integers, and "denumerable" for cardinality less than or equal to the cardinality of the integers. Since the premises and conclusion of the theorem are about "at most countable", the choice of terminology doesn't make any difference, because on one definition you're using "at most countable" everywhere, which is the same as "denumerable", and on the other definition - my definition - you're using "countable" everywhere. Note that the meaning of uncountable is the same for all of these definitions.)
Now, since (Fact3), it follows that .
Because a countable union of countable sets is countable, and because is uncountable (Fact4), it follows that there's at least one cube such that is uncountable.
But that's a contradiction with (*). Therefore the assumption that is uncountable is false.
Therefore is countable.
Sorry if attempting it was frustrating. Since I already knew where the proof was going, it was hard for me to gauge exactly how difficult it would be to complete from any given stopping point. I had hoped to leave it requiring some work and thought, but not so much that it was frustrating.
Another thing. It was unforntunate for me to include that function f defined as a supremum. It's not necessary, and adds undue complication. It's a legacy from how I reasoned out the proof. I apologize for including it and not realizing that it was no longer needed.
If you're curious about the origin & purpose of that f, I came to the above proof by reasoning:
"What's the biggest r can be so that B(x,r) intersect X remains countable? It can't be infinite, or X would be countable. OK - what happens at its surpemum, f(r)? What happens is that any bigger ball than that will intersect X uncountably. That means that between the sphere of radius f(r), and any larger ball centered at x containing it, there are an uncountable number of X. So the closed n-dimensional annulus with inner radius f(r) and outer radius f(r)+epsilon has an uncountable number of X for all epsilon>0. That's a compact set. (Perfect, I thought, since this looks like a problem that wants to use compactness - it's about finiteness and gives something about the open sets around each point of the space in question - so surely it's about compactness.) That annulus is a compact set where each of those x's is bounded away from any uncountable portion of X. But that's impossible - you've an uncountable number, so they must be smushed up against one another closely *somewhere*. THEN it dawned on me that the compactness doesn't matter to that - ant uncountable set in must have some uncountable portion of itself smushed up against itself closely *somewhere*. Now matter how finely I slice up , if X is uncountable, then some arbitrarily tiny slice of will always contain an uncountable number of X elements." And that got me to this proof. So that's where the f(x) came from - as a legacy of my trying to prove this using a compact annulus centered at x for each x in X.