# How many subsets of $\mathbb{R}$ are order isomorphic to $\mathbb{Q}$?

• How many subsets of $\mathbb{R}$ are order isomorphic to $\mathbb{Q}$?
• How many subsets of the long line $\omega_1\times[0,1)$ are order isomorphic to $\mathbb{Q}$?

I can see that results in both cases are between $\mathfrak{c}$ and $\mathfrak{2^c}$.

• There are $\mathfrak c$ subsets of $\mathbb R$ (and of $\omega_1\times[0,1)$) of size $\mathfrak c$, so really there is only one possibility. – Andrés E. Caicedo May 26 '13 at 16:12
• @Andres Did you mean "of size $\aleph_0$"? – Hanna K. May 26 '13 at 16:30
• Note that both answers below work for the long line as well, since they're just cardinality arguments. – Noah Schweber May 26 '13 at 16:55
• (@HannaK. Yes, of course. Silly typo.) – Andrés E. Caicedo May 26 '13 at 17:35

## 2 Answers

There are continuum many countable subsets of the continuum (because $\mathfrak{c}^{\aleph_0}=2^{\aleph_0}$). Thus the answer is $\mathfrak{c}$. See this question.

There are $2^{\aleph_0}$ subsets of $\Bbb Q$ which are order isomorphic to $\Bbb Q$.

To see this, note that $\Bbb{Q\setminus N}$ is order isomorphic to $\Bbb Q$, and consider for every $A\subseteq\Bbb N$ the set $\Bbb{Q\setminus N}\cup A$.

Since there are no more than $2^{\aleph_0}$ countable subsets to $\Bbb R$ the answer has to be $2^{\aleph_0}$.

• Or you look at a bounded subset of the rationals, which you can't tell that Butch said that because he is using white font on white background for that part. Gerhard "Selecting Makes Grey On Grey" Paseman, 2013.05.26 – Gerhard Paseman May 26 '13 at 22:07