Set theory problem - closed subsets of irrationals?

Definition: A set X is closed under addition if, for p and q both in X (with p and q not necessarily distinct!), p+q is always in X.

Easy question: Is there a non-empty subset of the irrationals that is closed under addition?

Tricky question? Is there an uncountably infinite non-empty subset of the irrationals that is closed under addition?

For example, the set {sqrt(2)} fails because it is not closed; taking p=sqrt(2), q=sqrt(2), we find that p+q=2*sqrt(2) is not in the set. The set of reals fails because it is not a subset of the irrationals.

I came up with this problem today, solved it, and I hope it makes you think. :)

Note by Alex Meiburg
5 years, 11 months ago

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The set of reals R\mathbb{R} is uncountably infinitely-dimensional as a vector space over Q\mathbb{Q}. Let XX be a Hamel basis for R\mathbb{R} which contains 11, and let Y=X\{1}Y = X \backslash \{1\} be everything in the Hamel basis except 11. Certainly YY is uncountable.

Let ZZ be the set of finite sums of elements of YY, so that ZZ is the set of real numbers of the form z  =  n1y1+n2y2++nmym z \; = \; n_1y_1 + n_2y_2 + \cdots + n_my_m where mNm \in \mathbb{N} and n1,n2,,nmNn_1,n_2,\ldots,n_m \in \mathbb{N}, while y1,y2,,ymYy_1,y_2,\ldots,y_m \in Y. It is clear that:

  1. ZZ is uncountable, since it has YY as a subset,

  2. ZZ is closed under addition,

  3. No element of ZZ is rational. If ZQZ \cap \mathbb{Q} \neq \varnothing, we could find m,n1,n2,,nmNm,n_1,n_2,\ldots,n_m \in \mathbb{N} and distinct y1,y2,,ymYy_1,y_2,\ldots,y_m \in Y and qQq \in \mathbb{Q} such that n1y1+n2y2++nmymq×1  =  0 n_1y_1 + n_2y_2 + \cdots + n_my_m - q\times1 \; = \; 0 But this tells us that there is a nontrivial rational linear dependence relation between the elements y1,y2,,ym,1y_1,y_2,\ldots,y_m,1 of XX, which is not possible.

Thus ZZ does the job. I would not want to attempt to write down a concrete example of a set ZZ!

Mark Hennings - 5 years, 11 months ago

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Nice! ^__^ And much better explained that I ever would have. :P

Alex Meiburg - 5 years, 11 months ago

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The set with elements multiples of root 2 (only positive multiples)

Surendra Ratha - 5 years, 11 months ago

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Yes, this satisfies the first one. :D It obviously does not satisfy the second one, though; I suggest trying that one too. :)

Alex Meiburg - 5 years, 11 months ago

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2nd 1 is really difficult...

Surendra Ratha - 5 years, 11 months ago

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I tried the 2nd one bt i could not, plz tell me the answer...

Surendra Ratha - 5 years, 11 months ago

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