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let $B \subset \mathbb{R}^+ $ with property that summing together any finite subset of elements of $B$ sums to $2$ or less

Show $B$ has to be finite or countable

Intuitively,

$B \subset (0,1]$ since if there is $x \in B, x > 1$ then $\{1,x\} \subset_{finite} B$ and $1+x > 2$

If we take a finite subset $C $ of $B$ then it sums to less than or equal to $2$ and we remove $C$, that is take $B \backslash C$ then this will still have the property of $B$ since it is itself a subset of $B$. So it must be if this is repeated then that is what will give that $B$ is finite or countable.

Seems using the density of $\mathbb{Q}$ in $\mathbb{R}$ might be useful since if we took the union of every disjoint interval inside $B$ it will be countable.

Also noting $\displaystyle \sum_{n=1}^{n} \frac{1}{n}$ diverges makes it clear that $B$ can't be uncountable..

So assuming the $B$ has the property and $\neg (B: \text{finite or countable}) $

Then, every $C \subset_{finite} B$ is such that $\forall c \in C \,\, \text{and some N} \in \mathbb{N}, \sum_{i = 1}^{N} c_{i} \le 2$. Now taking all finite disjoint subsets $C $ of $B$, $\bigcap_{n=1}^{\infty} C_{n} = \emptyset$ and $\bigcup_{n=1}^{\infty} C_{n} = B$. So then $\displaystyle \sum_{n=1}^{\infty} \bigg (\sum_{i=1}^{N} c_i \bigg) \le 2$ $\rightarrow\leftarrow$ Since if $B$ is not countable then $\displaystyle \sum_{n=1}^{\infty} \bigg(\sum_{i=1}^{N} c_i \bigg) \,\,\text{diverges}$

I think this might be the right idea but not executed quite right.

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  • $\begingroup$ I think you missed part of the problem statement (you've introduced the property that $B$ defines but haven't said what you want to prove about it). $\endgroup$
    – Ian
    Sep 27 '15 at 1:14
  • $\begingroup$ @Ian sorry, made the edit to make it more clear. $\endgroup$ Sep 27 '15 at 1:17
  • $\begingroup$ @mfl I know it can be infinite, countable implies that part. but if it can be countable then it can clearly be finite as well. Just trying to show that it can't be that B is uncountably infinite and have the property. I saw where I had inifinite not uncountable in there, my mistake. $\endgroup$ Sep 27 '15 at 1:18
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    $\begingroup$ Hint: notice that any member of $B$ is in $(1/(n+1),1/n]$ for some $n \in \mathbb{N}$. (This is actually not quite right; $B$ could have a single element larger than $1$, but this is not really a problem.) Since these are countably many sets, if $B$ is uncountable then uncountably many elements of $B$ must be in one of these sets. What happens then? $\endgroup$
    – Ian
    Sep 27 '15 at 1:23
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Hint: You know already that $B\subset [0, 1]$. Now write

$$B_n = B \cap [ \frac{1}{2^{n}}, \frac{1}{2^{n-1}}), n = 1, 2, \cdots $$

Then if $B$ is uncountable, there is $B_k$ such that it has infinitely many elements.

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