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Let us have a sequence $a_n$ of positive rational numbers, for which $\sum a_n = R \in \mathbb R$. Now suppose $b_n$ is a subsequence of $a_n$ such that $\sum b_n = r < R$. The question is "Can there be an uncountable number of subsequences of $a_n$ whose sum is $r$, for some sequence of positive rationals $a_n$ and some $r\in\mathbb R$?"

To be clear on definitions, a sequence and all its subsequences must be mappings from $\mathbb N\to X$, where in this case $X$ is the set of positive rational numbers.

psmears
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Benji Altman
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3 Answers3

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Yes, consider this series:

$$1+\frac12+\frac12+1+\frac14+\frac18+\frac18+\frac14+\frac1{16}+\frac1{32}+\frac1{32}+\frac1{16}+\cdots=4$$

There are uncountably many subsequences that sum to $2$. From each quadruplet of the form $2a+a+a+2a$, we may select either $2a+a$ or $a+2a$.

Chris Culter
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The series $$\frac3{5^1}+\frac2{5^1}+\frac1{5^1}+\frac3{5^2}+\frac2{5^2}+\frac1{5^2}+\frac3{5^3}+\frac2{5^3}+\frac1{5^3}+\cdots=\frac32$$ has continuum many subseries converging to $\frac34.$

bof
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Hmm... Why so complicated? $ \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

$\dfrac12 + \dfrac12 + \dfrac14 + \dfrac14 + \dfrac18 + \dfrac18 + \cdots = 2$.

$\dfrac12 + \dfrac14 + \dfrac18 + \cdots = 1 < 2$. (Just pick one of each pair.)

Better still, for every positive real $r < 2$, there are uncountably many subseries that converge to it! Here is a sketch of the proof. First pick some subseries converging to $r$. If it is finite, replace the last term $\lfrac1{2^k}$ by $\lfrac1{2^{k+1}}+\lfrac1{2^{k+2}}+\lfrac1{2^{k+3}}+\cdots$. So we can assume the subseries has infinitely many terms. Now if the subseries omits some term from infinitely many pairs of the original series, then we are clearly done. If not, the subseries includes every term from some point onwards, but must omit some term before that, and hence we can change the subseries to use the last omitted term in place of one of every pair after that. We now have a subseries with infinitely many terms that do not come in pairs, and hence as before we are done!

If you want a sequence with distinct terms, that is easy to obtain from the above. Simply replace each pair $(\lfrac1{2^k}+\lfrac1{2^k})$ by $(\lfrac1{2^k}+\lfrac1{2^k·7}+\lfrac6{2^k·7})$. By basic number theory, no two rationals of the form $\lfrac{6^a}{2^k·7^b}$ can be the same unless $k,a,b$ are the same. Then use the same argument as above, treating $(\lfrac1{2^k·7}+\lfrac6{2^k·7})$ the same as $\lfrac1{2^k}$ in the original.

user21820
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    That works, provided that subseries that select different indices are considered different. I made my answer more complicated in order to produce different functions $\mathbb N\to X$, since it seemed like that's what OP wanted. – Chris Culter Dec 06 '17 at 17:32
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    @ChrisCulter: Technically, the asker didn't stipulate that requirement. Anyway, I've just updated my answer to satisfy that, and yet maintain the property that every positive real number less than the original limit has uncountably many subsequences converging to it, which is much stronger than what the other answers have shown. – user21820 Dec 07 '17 at 04:33