# 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 $U_\beta$.

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..

• Perhaps the author is mentioning the ccc just because he thinks it interesting that a space with the ccc can have a long chain of open sets, i.e., the result of this exercise is supposed to be an interesting example. – bof Jun 2 at 5:45
• I think the point is that despite the name "countable chain condition" that does not mean all chains are countable, which is why ccc is pointed out explicitely – Alessandro Codenotti Jun 2 at 10:31
• @AlessandroCodenotti "countable (downward) anti-chain property" is a more correct name, but ccc is too settled already so that's not gonna change. – Henno Brandsma Jun 2 at 10:52
• @HennoBrandsma someone once explained to me that there are historical reasons for the name, because it was introduced in some context in which antichains conditions could be turned into chain conditions and viceversa but I don't remember any detail – Alessandro Codenotti Jun 2 at 11:24

## 1 Answer

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?