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Recently I've had some questions about cardinality and the real numbers:

For any infinite subset $S \subseteq \mathbb R$ , can we find a set $H$ such that $S \subseteq H \subseteq \mathbb R$ , such that $H$ is closed under addition, and $|H| = |S|$ ? Can we require $H$ to be closed under multiplication as well and still have this result?

I have no experience in these sorts of problems, but I have a feeling these statements are true. Could results like this be generalized to apply to infinite subsets of an arbitrary set closed under some (possibly infinite) set of binary operations? Thanks a lot!

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The answers to all your questions are yes. The general construction proceeds by setting $S_0 = S$ and inductively: $$S_n = \{s+t : s , t \in S_{n-1}\}$$ and letting $H = \bigcup_{i\geq 0} S_i$. It is easy to check that $H$ is closed under addition. Further, we have$$|S_n| \leq |S_{n-1}\times S_{n-1}| \leq |S_{n-1}| + \aleph_0.$$ In the case that $S_{n-1}$ is infinite, we do not need the $\aleph_0$ but you can do this construction in the finite case too. Either way, we have $|H| = |S| + \aleph_0$. This probably requires the axiom of choice or some variant.

You can generalize this to countably many $n$-ary operations where $n < \omega$ in the obvious way.

EDIT: Note that $S$ does not need to lie in some set where the binary operation exists to carry out this construction, we can carry out something like this by defining $S_n' = \{(s,t) : s,t \in S_{n-1}$ and define $s+t = (s,t)$ and quotient out by the axioms of the binary relation at each stage.

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  • $\begingroup$ By $\mathcal N_0$ I think you mean $\aleph_0$, which is \aleph_0 in TeX. But I'm not entirely clear on what it does in that formula. $\endgroup$ – Henning Makholm Oct 25 '15 at 1:35
  • $\begingroup$ That was quick, thanks! Could you clarify on $|S_{n-1} * S_{n-1}| = |S_{n-1}| + \aleph_0$, as couldn't $S$ be larger than $\aleph_0$? Or am I misreading your symbol? I don't want to assume the continuum hypothesis. $\endgroup$ – B Gunsolus Oct 25 '15 at 1:36
  • $\begingroup$ For a finite $S_{n-1}$ you get $\le$ instead of $=$ in the second displayed equation, though. $\endgroup$ – Henning Makholm Oct 25 '15 at 1:40
  • $\begingroup$ Thank you. I hope there aren't any more mistakes :( $\endgroup$ – Asvin Oct 25 '15 at 1:41
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    $\begingroup$ This can be strengthened drastically: en.wikipedia.org/wiki/L%C3%B6wenheim%E2%80%93Skolem_theorem $\endgroup$ – Noah Schweber Oct 25 '15 at 1:46

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