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. :)

#GroupTheory #LinearAlgebra #MathProblem

Easy Math Editor

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Comments

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:

$Z$ is uncountable, since it has $Y$ as a subset,
$Z$ is closed under addition,
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 $Z$ !

Mark Hennings - 7 years, 7 months ago

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

Alex Meiburg - 7 years, 7 months ago

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

Surendra Ratha - 7 years, 7 months ago

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

Alex Meiburg - 7 years, 7 months ago

2nd 1 is really difficult...

Surendra Ratha - 7 years, 7 months ago

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

Surendra Ratha - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$