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Hi so I'm practicing for a exam and I need help to figure this proof out,

Suppose $A\subseteq \mathbb R^+$, $b\in\mathbb R^+$, and for every list $a_1,a_2,\ldots,a_n$ of finitely many distinct elements of $A$, $a_1+a_2+\ldots+a_n \le b$. Prove that $A$ is countable. (Hint: For each positive integer $n$, let $A_n=\left\{\,x\in A\mid x\ge\tfrac 1n\,\right\}$. What can you say about the number of elements in $A_n$?)

I have no idea how to go about this.

Note: Countable means finite or denumerable in this context and $\mathbb R^+$ are the positive real numbers.

Asaf Karagila
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ChemDude
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3 Answers3

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Follow the hint. If you pick $k$ elements from $A_n$ then their sum is at least $\frac kn$. As this sum must be $\le b$, there can be at most $nb$ elements in $A_n$, so $A_n$ is finite. On the other hand, for any $x \in A$ we have (because $x>0$) that $x\ge \frac1n$ if $n$ is large enough. That is, $A$ is the union of the countably many finite sets $A_n$, $n\in\mathbb N$, hence countable.

Hagen von Eitzen
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The hint is valuable. Note that $b$ is some fixed number. Then since $$ b \geq \sum_{k|a_k\in A_n} a_k \geq \sum_{k|a_k\in A_n} \frac{1}{n} = \frac{N(A_n)}{n}$$ where $N(A_n)$ is the number of elements in $A_n$, $$N(A_n) \leq nb$$

So each $A_n$ is finite, and $A$ is the union of all the $A_n$ but there are only a countably many $A_n$ as they can be indexed by $n \in \Bbb{N}$. And there is a theorem that the union of a countable set of finite sets is itself countable.

Mark Fischler
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  • Could you write the formal proof for me please? I really need it. Thanks. – ChemDude Apr 21 '15 at 20:48
  • @MathHelpNeeded That looks like a pretty good proof to me. Check out this thread for a proof for the last statement: http://math.stackexchange.com/questions/603456/prove-that-the-union-of-countably-many-countable-sets-is-countable (the theorem that the union of a countable set of finite sets is itself countable) – Tyler Hilton Apr 21 '15 at 23:19
  • I'm wondering if you need the Axiom of Choice to prove the union of a countable collection of finite sets is countable. (You do need it to prove a countable collection of countable sets is countable, but that is not what we have here.) However, what we need to prove the statement is that the union of a countable number of finite sets of real numbers is countable. This can be proven without the axiom of choice, because you can algorithmically select a sequence of particular numbers from each set. – Mark Fischler Apr 22 '15 at 18:21
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The idea is:

If $|A|$ is finite the proof is obvious.

If $|A|\geq \aleph_0$:

You know that

$$A=\bigcup_{i\in\mathbb N} (A_n )$$

If $A$ is more than countable, then it must exists $k$ such that

$$ |A_k|>\aleph_0 $$

Because a countable union of countable sets is countable.

So you could consider more than $j=\lceil \frac{b}{k} \rceil$ elements $h_1,...,h_j \in A$ such that $$\sum_{i=1,..,j} h_i >b.$$ Absurd.

Ludox
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