Is every countable ordinal homeomorphic to a subspace of $\mathbb R$? I know that every countable ordinal is isomorphic to some subset of $\mathbb R$ as ordered sets. Is it also the case that every countable ordinal (with the order topology) is homeomorphic to some subspace of $\mathbb R$?
 A: Yes. Every countable ordinal can be realized as a subspace of the rationals, and therefore as a subspace of the real numbers.
You can do this by transfinite induction, and note that both $\Bbb R$ and $\Bbb Q$ have the property that every open interval is homeomorphic, and order isomorphic, to the full space.
A: Every countable linearly ordered space with its order topology can be continuously embedded in $\Bbb Q$. Here’s a sketch of a proof. Let $\langle X,\preceq\rangle$ be a countable linear order. If $x,y\in X$, $x\prec y$, and $(x,y)=\varnothing$, write $y=x^+$ and $x=y^-$. Let $L=\{x\in X:x=y^-\text{ for some }y\in X\}$. For each $x\in L$ let $Q_x=\{x\}\times\Bbb Q$, and let
$$Y=X\cup\bigcup_{x\in L}Q_x\;.$$
Extend the $\preceq$ to $Y$ in the natural way, so that $x\prec\langle x,q\rangle\prec x^+$ for each $x\in L$. Then $\langle Y,\preceq\rangle$ is a countable dense order, so it’s order-isomorphic to one of the sets $\Bbb Q\cap[0,1]$, $\Bbb Q\cap[0,1)$, $\Bbb Q\cap(0,1]$, and $\Bbb Q\cap(0,1)$. Let $f$ be an order-isomorphism from $Y$ onto one of these sets; then $f$ is a homeomorphism from $Y$ with its order topology onto one of these sets with its usual (order) topology, and $f\upharpoonright X$ is a homeomorphic embedding of $X$ into $\Bbb Q$ (and hence into $\Bbb R$).
A: If all you want is a homeomorphic embedding of a countable ordinal $\alpha$ into $\mathbb R$, just take an injection $f:\alpha\to\omega$ and define $h:\alpha\to\mathbb R$ by setting
$$h(\xi)=\sum_{\beta\lt\xi}2^{-f(\beta)}.$$
