(Edited) Background The following question comes from the order topology chapter from A First Course in Topology by Conover
Theorem 4.1 A subset A of R is open in the Metric topology induced by the absolute value metric iff each point x$\in$ A there exists real numbers a an b with a<x<b s.t x$\in$(a,b)$\subset$ A
He then mentions total order relation on a set X and an order topology on totally ordered set X
(3) There are open subsets of [0,$\Omega$) of cardinality $\aleph_0$ and subsets of [0,$\Omega$) of cardinality $\aleph_1$ ?
Question What does this question ask of me? I took it to imply to give an example or two as the book does
So for the first one. $\omega$+2=(0,1,2,3...,$\omega$,$\omega$+1),
So |$\omega$+2|=$\aleph_0$
How about $\omega_1$ +1, for the latter? Then |$\omega_1$+1 |=$\aleph_1$ Though I don’t know how to write the set .
Any help would be appreciated.l hope this is not a duplicate.
For convenience let $X=[0,\Omega)$. Then $[0,\alpha)$ is a countable open set for each $\alpha\in X\setminus\{0\}$, and $(\alpha,\Omega)$ is an open set of cardinality $\aleph_1$ for each $\alpha\in X$. That these sets are open is immediate from the definition of the order topology. That the sets $[0,\alpha)$ are countable follows at once from the fact that $\Omega$ is the smallest uncountable ordinal number: each $\alpha\in X$ is therefore countable, and $[0,\alpha)=\alpha$. Finally,
$$(\alpha,\Omega)=X\setminus[0,\alpha]=X\setminus[0,\alpha+1)\,,$$
$|X|=\aleph_1$, and $|[0,\alpha+1)|=\aleph_0$, so $|\alpha,\Omega)|=\aleph_1$.
Of course there are many other open subsets of $X$ of cardinality $\aleph_0$ and of cardinality $\aleph_1$. For instance, let $A\subseteq X$, and let $U_A=\{\alpha+1:\alpha\in A\}$; then
$$U_A=\bigcup_{\alpha\in A}(\alpha,\alpha+2)$$
is open, being a union of open intervals, and $|U_A|=|A|$, so we can start with any subset of $X$ of cardinality $\aleph_0$ or $\aleph_1$ to get an open subset of $X$ of the same cardinality. In fact, if $A$ is any countably infinite subset of $X$, then $U_A$ and $U_{X\setminus A}$ are disjoint open subsets of $X$ such that $|U_A|=\aleph_0$ and $|U_{X\setminus A}|=\aleph_1$