Uncountable Set with this property

1. Aug 19, 2009

Dragonfall

Does there exist an uncountable set of positive reals such that every countable decreasing sequence from that set sums to a finite number?

2. Aug 19, 2009

Tac-Tics

I don't know how to prove it rigorously, but I'd think it reduces to a problem in measure theory.

If your uncountable set contains any non-empty interval [a, b], it fails to have the countable-sums property.

All non-empty intervals have a nonzero measure. I wouldn't be surprised if the converse was true for the real numbers. Let's assume it is for now. This means that if our uncountable set has a nonzero measure, it contains an interval, and fails to have our desired property. So, we conclude our set has zero measure.

Sets of zero measure are countable. This is our contradiction.

Again, this is just my intuition. I don't know much about measure theory, but if I did, this is what I would expect the proof to look like ;-)

EDIT: Or maybe the Cantor set is a counter example to the last assumption I made. It is uncountable with zero measure. (http://en.wikipedia.org/wiki/Measure_zero)

Last edited: Aug 19, 2009
3. Aug 19, 2009

CRGreathouse

Beat me to it.

4. Aug 19, 2009

Dragonfall

I'm thinking that any uncountable set has at least one accumulation point away from zero. The problem is that if the accumulation point is from a sequence of increasing numbers instead of decreasing. Then it wouldn't work.

5. Aug 19, 2009

VKint

Pretty sure this proof works:

Obviously, if a set $$S$$ is to have the desired property, then $$S \cap [a,b]$$ is finite for all positive $$a$$ and $$b$$. (Indeed, if $$S \cap [a,b]$$ has infinitely many elements, we can choose a countably infinite, decreasing sequence, which necessarily diverges, since all terms are $$\geq a$$.) Furthermore, if $$S$$ has this property, then any uncountable subset of $$S$$ also does; thus, WLOG, $$S \subseteq (0, \infty)$$ (if $$S \cap (0, \infty)$$ is countable, then $$S \cap (-\infty, 0)$$ is uncountable, and we may proceed in essentially the same way). Define $$\displaystyle U_i = \left[ \frac{1}{i + 1}, \frac{1}{i} \right]$$ for $$i \geq 1$$, and $$U_0 = [1, \infty]$$. By above, $$\displaystyle \bigcup_{i = 0}^{\infty} (S \cap U_i) = S$$ (since $$S \subseteq (0, \infty)$$); however, recall that each term $$S \cap U_i$$ has finite order. Thus, $$S$$ is a countable union of finite sets, and is therefore countable. This is a contradiction; hence, no such set can exist.

6. Aug 19, 2009

Hurkyl

Staff Emeritus
Actually, that's not guaranteed. For example, consider the infinite set containing
1, 1.9, 1.99, 1.999, 1.9999, 1.99999, ...​
Every decreasing sequence whose elements in this set is necessarily finite.

The thing we can prove with this argument is that $S \cap [a, \infty)$ is well-ordered by <.

7. Aug 19, 2009

VKint

Excellent point, Hurkyl; I guess that's what I get for trying to write up a post while on Vicodin. Wisdom teeth... :(

I think my proof can still be repaired, though. All I really need to show is that $$S \cap U_i$$ is countable for all $$i$$ (a countable union of countable sets is countable). We prove the following lemma:

Lemma: Let $$V$$ be a bounded set of real numbers. If, for all $$x \in V$$, the set $$\{ y \in V \; | \; y < x \}$$ is countable, then $$V$$ is countable.
Proof: Let $$q = \sup(V)$$. Clearly, if $$q \in V$$, we're done. Thus, WLOG, $$q \notin V$$. Let $$p$$ be a lower bound for $$V$$, and let $$d = q - p$$. Define $$\displaystyle T_i = \left[ q - \frac{d}{2^i}, q \right]$$. Since $$q$$ is the supremum of $$V$$, it is also an accumulation point, so $$V \cap T_i \neq \emptyset$$ for all $$i$$. Thus, using the axiom of choice if necessary, we can choose some non-decreasing sequence of points $$x_i \in V \cap T_i$$. Finally, define $$Q_i = \{ z \in V \; | \; z < x_i \}$$ (running out of letters...) and observe that $$\displaystyle V = \bigcup_{i = 0}^{\infty} Q_i$$, i.e., a countable union of countable sets (by hypothesis). This concludes the proof.

Now that we have this result, we can prove that $$S \cap U_i$$ is countable by contradiction. As before, the contradiction comes from assuming that $$S \cap U_i$$ is uncountable, and constructing a countable decreasing sequence of points in $$S \cap U_i$$. Indeed, if $$S \cap U_i$$ is uncountable, we can choose some $$x_0 \in S \cap U_i$$ such that $$\{ y \in S \cap U_i \; | \; y<x \}$$ is uncountable by the lemma. It should be easy to see now that we can construct a strictly decreasing sequence of points in $$S \cap U_i$$ inductively, which gives the same contradiction as before.

8. Aug 19, 2009

Dragonfall

Every well ordered (by <) set is countable. Then $S=\cup_a (S\cap [a,\infty))$ for a that are inverse powers of 2. S is therefore a countable union of countable sets, which is countable. Contradiction.

EDIT: I think you went overboard there with your lemma.

9. Aug 19, 2009

VKint

Not if you allow the axiom of choice...I think.

10. Aug 19, 2009

Dragonfall

But then an uncountable well-ordered set would be a counterexample to your contradiction, it would have the property we're looking for!

11. Aug 19, 2009

VKint

Not at all. You'll notice that in the statement of my lemma, I specifically noted that the result only applied to set of real numbers, implying the the usual ordering on the reals must be used for the result to remain valid. In other words, the lemma proved that no uncountable set is well-ordered under the usual ordering of $$\mathbb{R}$$, which does not preclude the existence of some other well-ordering.

EDIT: I just realize that this was what you were getting at all along when you said "well ordered (by <)," in which case you were correct. My lemma was essentially just a proof of your statement that "every well ordered (by <) set is countable." Apologies for the misunderstanding.

Last edited: Aug 20, 2009
12. Aug 19, 2009

VKint

Incidentally, where did this problem come from?

13. Aug 20, 2009

Dragonfall

I thought it up when I was playing around with the idea of uncountable sums (a previous post).

14. Aug 20, 2009

Elucidus

If one permits the Axiom of Choice, the reals can be well ordered.

(In fact the Well Ordering Theorem, the Axiom of Choice, and Zorn's Lemma (among others) are all equivalent.)

--Elucidus

15. Aug 20, 2009

Hurkyl

Staff Emeritus
So trying to rewrite what you've written in a more generally useful form, I think you've shown the following.

Define an uncountable accumulation point of a set X to be a point such that every neighborhood of that point contains uncountably many points of X.
(I doubt this is a standard term)

Every uncountable subset of R has an uncountable accumulation point​

and a slight modification will allow

Every uncountable subset of R has at least countably infinitely many uncountable accumulation points​

I haven't worked out whether or not it must have uncountably many uncountably accumulation points....

16. Aug 20, 2009

VKint

I don't see how this follows directly from what I proved. However, it's fairly easy to prove this by using the second-countability of $$\mathbb{R}$$. Indeed, if $$S \subseteq \mathbb{R}$$ is uncountable but has no uncountable accumulation point, then for all $$x \in S$$ there exists an open set $$U_x$$ with $$x \in U_x$$ and $$S \cap U_x$$ countable. Open sets are closed even under uncountable unions, so the set $$\displaystyle T = \bigcup_{x \in S} U_x$$ is open. However, since $$\mathbb{R}$$ is second-countable, each $$U_x$$ may be written as a union of base open sets $$B_{x,i}$$ for $$i \in I_x$$ (i.e., some countable index set), each of which has $$S \cap B_{x,i}$$ countable. We then have $$\displaystyle T = \bigcup_{x \in S, \; i \in I_x} B_{x,i}$$. However, since the base is countable, this is at most a countable union. The set $$S$$ is then a countable union of countable sets, so it is countable.

Your next proposition follows fairly trivially from the first (by simply removing successive uncountable accumulation points). I also think I have a proof that any uncountable $$S \in \mathbb{R}$$ must have uncountably many uncountable accumulation points. Indeed, suppose not. Let $$X \subseteq S$$ be the set of all uncountable accumulation points. Then, since $$X$$ is countable, $$S \setminus X$$ is uncountable. Thus, $$S \setminus X$$ must contain an uncountable accumulation point, contradicting assumption.

Thoughts?

17. Aug 21, 2009

Hurkyl

Staff Emeritus
I don't remember precisely what I was thinking -- but I had convinced myself my first proposition was either an easy consequence of what you had proved, or that your proof of your lemma contained most/all of the bits needed to prove my proposition.

The proof you give of my second and third propositions is flawed -- accumulation points of S don't actually have to be elements of S, so you don't get the contradiction you were relying on.

18. Aug 21, 2009

g_edgar

"condensation point"

19. Aug 21, 2009

VKint

I must have misunderstood what you intended. The proof I gave in #16 assumed that uncountable accumulation points ("condensation points") must lie in the uncountable set itself, and proved that every uncountable subset of $$\mathbb{R}$$ must have at least one such point. If you wanted to prove the existence of a condensation point not necessarily in the target set, you could do it by an inductive method similar to that used to prove the Bolzano-Weierstrass theorem--dividing a bounded region containing an uncountable set into finitely many pieces at each step, and picking one point from a piece with uncountably many points. In the end, you get a convergent sequence of points in the set, which must converge to a condensation point.

20. Aug 21, 2009

Preno

For any a>0, there must be a right neighbourhood of a containing no members of S (otherwise we could create a decreasing sequence approaching a), i.e. a number b>a such that there are only countably many members of S on the interval $\langle a; b\rangle$. Assuming that there are uncountably many members of S on the interval $\langle a;+\infty)$ yields a contradiction: just take the supremum* of all c>a s.t. there are only countably many members of S on the interval $\langle a; c \rangle$ and extend the interval a little. Therefore, for any a>0, there is only a countable number of members of S greater than a. But this proves the proposition, as S is the countable union of sets of the form $S \cap \langle \frac{1}{n} ; \infty)$.
edit: * if no such number exists, then $S \cap \langle a; \infty)$ is countable, being a countable union of countable sets.