In the c.c.c. space $2^{\omega_1}$, there is a sequence of open sets such that whenever $\alpha<\beta$, $U_\alpha$ is a proper subset of $U_\beta$.

Question

Show that in the c.c.c. space $2^{\omega_1}$, there is a sequence of open sets, $\langle U_\alpha | \alpha<\omega_1\rangle$ such that whenever $\alpha<\beta$, $U_\alpha$ is a proper subset of $U_\beta$.

This is an exercise form Kunen's Set theory. Here is my attempt at solving it:

Define $U_0=\{0\}\times\prod\limits_{0<\gamma<\omega_1}\{0,1\}$ and $U_\alpha=\left(\prod\limits_{\gamma<\alpha}\{0,1\}\right)\times\{0\}\times\left(\prod\limits_{\alpha<\gamma<\omega_1}\{0,1\}\right)\cup\bigcup\limits_{\delta<\alpha}U_\delta$. Then $\langle U_\alpha\rangle$ is an increasing(with respect to inclusion) sequence of sets. Also, each $U_\alpha$ is open as a union of basis elements.

Now let $\alpha<\beta$. Then define a function: $$f(\gamma) = \begin{cases} 1, \text{ if } \gamma<\beta, \\ 0, \text{ otherwise. } \end{cases} $$ Then, we have $f\in U_\beta$ and $f\notin U_\alpha$. So $U_\alpha$ is a proper subset of $U_\beta$.

Is this attempt correct? It seems a little strange to me that I did not use anywhere that the space is c.c.c..

Answer 1

This works, but notationwise I would just write $V_\alpha = \pi_\alpha^{-1}[\{0\}]$ which are (sub)basic open. And then define $U_\alpha = \bigcup_{\beta \le \alpha} V_\alpha$. The proof of properness is fine.

Note that the existence of such a chain implies (is in fact equivalent too) the fact that $2^{\omega_1}$ is not hereditarily Lindelöf (the subspace $A = \{x_\alpha : \alpha < \omega_1\}$, where we pick each $x_\alpha \in U_\alpha \setminus \cup_{\beta < \alpha} U_\beta$ is not Lindelöf). This contrasts with classes of spaces (like metric or more generally monotonically normal spaces) where ccc is equivalent to being hereditarily Lindelöf. Maybe that is why Kunen mentions the ccc here?