# 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
6 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 $Z$!

- 6 years, 7 months ago

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

- 6 years, 7 months ago

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

- 6 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. :)

- 6 years, 8 months ago

2nd 1 is really difficult...

- 6 years, 8 months ago

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

- 6 years, 8 months ago