# 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
4 years, 8 months ago

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

Let $$Z$$ be the set of finite sums of elements of $$Y$$, so that $$Z$$ is the set of real numbers of the form $z \; = \; n_1y_1 + n_2y_2 + \cdots + n_my_m$ where $$m \in \mathbb{N}$$ and $$n_1,n_2,\ldots,n_m \in \mathbb{N}$$, while $$y_1,y_2,\ldots,y_m \in Y$$. It is clear that:

1. $$Z$$ is uncountable, since it has $$Y$$ as a subset,

2. $$Z$$ is closed under addition,

3. No element of $$Z$$ is rational. If $$Z \cap \mathbb{Q} \neq \varnothing$$, we could find $$m,n_1,n_2,\ldots,n_m \in \mathbb{N}$$ and distinct $$y_1,y_2,\ldots,y_m \in Y$$ and $$q \in \mathbb{Q}$$ such that $n_1y_1 + n_2y_2 + \cdots + n_my_m - q\times1 \; = \; 0$ But this tells us that there is a nontrivial rational linear dependence relation between the elements $$y_1,y_2,\ldots,y_m,1$$ of $$X$$, which is not possible.

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

2nd 1 is really difficult...

- 4 years, 8 months ago