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There is a similar question to this already on MSE, but I would like to ask about a different approach:

The full question is, "Let $B$ be a set of positive real numbers with the property that adding together any finite subset of elements from $B$ always give a sum of $2$ or less. Show that $B$ must be finite or countable."

My approach was:

For each $n \in \mathbb{N}$, $B_n = ${$x \in B \vert x \gt \frac{2}{n}$}; $B = \bigcup_{n \in \mathbb{N}}B_n$ by assumption. Let's make the assumptive claim that $B$ can have at most $n-1$ elements. Let's see what happens when $B$ has $n$ elements:

$B = ${$b_1, b_2,...,b_n$}

However, this means the sum of $B$'s elements is $\frac{2}{n} * n = 2$, which can't be possible based on how we defined $B$.

Hence, $B$ cannot be infinite and is therefore finite.

Is this valid or did my reasoning fault somewhere? I am quite certain it's wrong, since my logic resulted in the determination that $B$ is strictly finite, rather than "finite or countable" like the question asked.

Any help is appreciated! Thank you!

robert lewison
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    I think you have a pretty devastating typo right at the start: I'm guessing it should say "Show that B must be finite or countable", as in your last paragraph, rather than uncountable. – Izaak van Dongen Feb 06 '24 at 03:12
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    In that case is should just be saying "must be countable". – Max Meijer Feb 06 '24 at 03:15
  • Assuming that's what you meant, your reasoning doesn't make sense. It doesn't follow from $B = {b_1, \dotsc, b_n}$ that $b_i$ all lie in $B_n$, which seems to be your reasoning. It merely follows that there is some $m$ such that $b_i$ all lie in $B_m$, but $m$ may be much larger than $n$. – Izaak van Dongen Feb 06 '24 at 03:16
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    A correct argument is given eg here. @Max, "countable" is not a universally standardised term, sadly. To some people it means "bijects with $\Bbb N$", while to some it means "injects into $\Bbb N$" (see eg Wikipedia). The most common unambiguous phrasing would be "must be at most countable". – Izaak van Dongen Feb 06 '24 at 03:24
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    A tangent, but I'm not sure why the question has "at most 2" as the sum when "at most 1" has the same proof, but is a much more relevant claim! It is used in probability, to show that any probability space can have at most countably many outcomes $\omega$ such that the probability of the singleton event ${\omega}$ is positive. – Misha Lavrov Feb 06 '24 at 18:22

2 Answers2

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Your question must be wrong (perhaps it got lost in translation?). Counterexample: Consider the set $B = \{1/2,1/4,1/8,1/16,...\}$. Note that it is countably infinite and any finite subset of elements has a sum less than 2.

You can also use this counterexample to discover where your reasoning goes wrong.

Max Meijer
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  • You're right—I accidentally used the word "uncountable" instead of "countable" in the original question. Therefore, your $B$ is a great example of an example of a set that does in fact work. However, how can it be generalized and formalized? – robert lewison Feb 06 '24 at 03:39
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    Your first step is good but assuming that B has at most n-1 elements cannot be a good approach if you take into account the counterexample. You have already decomposed B into a countable number of finite sets that are easier to reason about. Your conclusion that then B is finite is wrong, but the countable union of finite sets is in fact (at most) countable. This is a well-known set-theoretic result with a well-known proof based on diagonals: https://math.stackexchange.com/questions/603456/prove-that-the-union-of-countably-many-countable-sets-is-countable – Max Meijer Feb 06 '24 at 04:17
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You have confused $B$ and $B_n$. You have shown that all the $B_n$ are finite.

$B$ is countable because it is the union of countable many $B_n$

Etoplay
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