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I'm working on a homework problem and I'm not sure I'm understanding how to construct the proof. I have done work on it, but it's not coming together. The problem, "Let $A \subseteq (0, 1)$ be a set of positive real numbers with the following property: For any finite subset $F \subseteq A$, the sum of the all of the numbers in $F$ is less than 1. Show that $A$ must be either finite or countable. [Hint: This is a problem about cardinality, but it shows that there is really no meaningful way to consider uncountable series. Show that for each $n \in N$, the set $A_n = A \cap ( \frac{1}{n+1} , \frac{1}{n}]$ must be finite.]"

What I have so far is something like this: Proof By assumption, any $F \subseteq A$ is finite. The definition of $A$ implies that $F \not\subset \mathbb{Z}$ and $F \not\subset \mathbb{N}$ because the open interval $(0,1)$ contains neither. By assumption, $F$ is finite and so $F \subseteq \mathbb{Q}$. (I know see this is faulty reasoning because F is defined to be finite not simply countable.) Since $\mathbb{Q}$ is countable and $F \subseteq \mathbb{Q}$, $F$ is also countable.

However, though I now see I wasn't on the right path, I'm perplexed by the HINT. How can I show that $A_n = A \cap (\frac{1}{n+1}, \frac {1}{n}]$ must be finite? I see that each $A_n$, $\inf A_n$ and $\max A_n$ exist. But since $A\subseteq\mathbb{R}$, how can it be finite?

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    $\begingroup$ Oh dear! Not every finite set is a subset of $\Bbb Q$. $\endgroup$
    – Hagen von Eitzen
    Commented Feb 11, 2017 at 18:49
  • $\begingroup$ That's why I recognized my reasoning to be faulty. Being finite doesn't mean it's in $\mathbb{Q}$. $\endgroup$
    – Andrew Falanga
    Commented Feb 11, 2017 at 18:51

2 Answers 2

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Improved hint: Show that $A_n$ contains at most $n$ elements.

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  • $\begingroup$ I'm sorry but I don't see what you're hinting at so I am misunderstanding something. Please correct me where I'm incorrect. I'm given $A_n = A \cap (\frac{1}{n+1},\frac{1}{n}]$. Since I'm working with subsets of $\mathbb{R}$, wouldn't that semi-closed interval still mean there's an infinite number of terms from $(\frac{1}{n+1}, \frac{1}{n}]$? How can $A_n$ have at most $n$ elements? I see that $\frac{1}{n+1}$ isn't in the set, but that doesn't rule out everything else does it? $\endgroup$
    – Andrew Falanga
    Commented Feb 12, 2017 at 3:32
  • $\begingroup$ @AndrewFalanga While $(\frac1{n+1},\frac1n]$ has uncountably many elements, the set $A\cap (\frac1{n+1},\frac1n]$ may have less. In fact, see Thomas Raberry's answer for what happens if we assume that it has even just $n+1$ elements. $\endgroup$
    – Hagen von Eitzen
    Commented Feb 12, 2017 at 13:00
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An error in your statement is that since $F$ is finite, $F$ must be composed of rational numbers. This is not the case; $\frac{\pi}{10},$ for instance, is irrational, and could comprise the entire set $A$ for all we know (all subsets of such a set are either empty or $\frac{\pi}{10}$ itself, and therefore have sum less than one).

To consider that $A_n = A \cap\big(\frac{1}{n+1},\frac{1}{n}\big]$ must be finite, assume that it is not finite. Then take any $n+1$ numbers from $A_n$; call this sequence $\{a_k\}_{k=1}^{n+1}$. Notice its sum is $$\sum_{k=1}^{n+1}a_k > \sum_{k=1}^{n+1} \frac{1}{n+1} =1,$$ contradicting our assumption about finite subsets of $A.$ Finally, consider what set $\cup_{n=1}^{\infty}A_n$ represents.

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  • $\begingroup$ Your $\ge$ can (and should) actually be a $>$. $\endgroup$
    – Hagen von Eitzen
    Commented Feb 12, 2017 at 13:01
  • $\begingroup$ Thomas, I'm considering your response now. Please forgive my ignorance but how does making a sequence $a_k$ in the manner you've done prove this? What's tripping me up is that, if I'm understanding, you're comparing the sum of whole numbers to the sum of fractions? Why is that valid? Honestly, what am I missing? $\endgroup$
    – Andrew Falanga
    Commented Feb 13, 2017 at 3:19
  • $\begingroup$ The sequence is chosen from $A_n=A \cap (\frac{1}{n+1},\frac{1}{n}),$ so the elements will not be integers by definition. $\endgroup$
    – Thomas Rasberry
    Commented Feb 13, 2017 at 19:31
  • $\begingroup$ Additionally, I am proving finiteness of $A_n$ (which are subsets of the indicated set) through assuming the contrary, i.e. that the subset is infinite in cardinality. Since the cardinality of this subset is assumed to be infinite for sake of argument, that means you are able to select $n+1$ distinct elements from any of the subsets $A_n.$ Since these elements are all greater than $\frac{1}{n+1}$ by definition of $A_n,$ their sum must equal a real number greater than $1$ for each $A_n.$ Therefore, a contradiction to the assumption is reached, so $A_n$ are finite. $\endgroup$
    – Thomas Rasberry
    Commented Feb 13, 2017 at 19:34
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    $\begingroup$ I'm marking this one as the "answer" because it is more verbose. Thank you @HagenvonEitzen and Thomas. Yesterday I met with my professor in order to clarify this topic and during that discussion, I realized I had a whopper of a misunderstanding. I did not understand that the definition of $F$ influenced what could be in $A$. I was seeing them as separate and distinct (though $F \subseteq A$). Understanding that helped a lot to understand your respective hints and explanations. $\endgroup$
    – Andrew Falanga
    Commented Feb 15, 2017 at 16:51

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