I am supposed to solve these two exercises for a set theory course. I should have done the first one, but I do not have good ideas for the second one.
- Find the cardinality of the subsets of $\mathbb{Q}$ which are order isomorphic to $\mathbb{Q}$.
- The same for $\mathbb{R}$.
For the first one I think that the cardinality is $2^{\aleph_0}$. Of course the required cardinality is $\leq 2^{\aleph_0}$ because $\mathbb{Q}$ is countable. For the other inequality I defined the map which sends each $x>0$ irrational into $A_x=\{y\in \mathbb{Q}\,:\,y>x\}$: $A_x$ is countable, with no endpoints and dense, thus it is order isomorfphic to $\mathbb{Q}$; furthermore the map is clearly injective and hence the required cardinality is $\geq 2^{\aleph_0}$. Could anyone confirm this?
For the second one I know that a linear order which is complete, with no endpoints and which has a countable dense subset is order isomorphic to $\mathbb{R}$. Thus I am able to find $2^{\aleph_0}$ subsets of $\mathbb{R}$ wich are order isomorphic to $\mathbb{R}$ by simply subtracting a closed interval to $\mathbb{R}$. But now I am stucked and do not know how to proceed.
Your argument for $\mathbb{Q}$ is correct. A subset of $\mathbb{R}$ that is order-isomorphic to $\mathbb{R}$ is the image of a strictly increasing function $\mathbb{R}\to\mathbb{R}$. A strictly increasing function $\mathbb{R}\to\mathbb{R}$ can only have countably many points of discontinuity (because every discontinuity jumps over some rational value), and so such a function can be determined by specifying its values on a countable dense subset of $\mathbb{R}$ that contains all the points of discontinuity. There are only $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}$ ways to pick a countable dense subset of $\mathbb{R}$ and again $2^{\aleph_0}$ ways to map that countable set to $\mathbb{R}$, so there are at most $2^{\aleph_0}$ subsets of $\mathbb{R}$ that are order-isomorphic to $\mathbb{R}$.