# Can the continuum be a singular cardinal?

Martin's Axiom implies that $2^{\aleph_0}$ is a regular cardinal. But can $2^{\aleph_0}$ be a singular cardinal?

By Konig's Lemma, it can never be $\aleph_{\omega}$ since cf($2^{\aleph_0}$)>$\aleph_0$ but under what conditions can it be $\aleph_{\omega_1}$? It is even possible it can be $\aleph_{\omega_1}$?

Yes, but it must have uncountable cofinality. So if it is to be singular, the smallest possibility is $\aleph_{\omega_1}$.
The basic fact is that if $\kappa$ is any cardinal such that $\kappa^\omega=\kappa$, then there is a forcing extension $V[G]$ in which $2^\omega=\kappa$. The forcing to achieve this is $\text{Add}(\omega,\kappa)$, which consists of finite partial functions from $\kappa\times\omega$ to $2$.
In particular, if you start with GCH in the ground model, and add $\aleph_{\omega_1}$ many Cohen reals, then you will have $2^\omega=\aleph_{\omega_1}$ in the forcing extension. The proof uses the following facts: (1) adding any number of Cohen reals is c.c.c. and therefore preserves all cardinals. (2) The forcing clearly adds at least that many reals. (3) A nice name argument shows that every real in the extension has a nice name in the ground model, and there are only $\aleph_{\omega_1}$ many such names. So the continuum of the extension is exactly $\aleph_{\omega_1}$. The same ideas work for any $\kappa$ for which $\kappa^\omega=\kappa$.
• Does this have any consequence if you had in the equation, Silver's Theorem about the GCH can't fail for the first time at $\aleph_{\omega_1}$ Does it imply Silver's Theorem? It seems like it should...but does that means the continuum failed before $\aleph_{\omega_1}$? Jun 3 '10 at 20:02
• Well, this argument was just about adding subsets to $\omega$, not making the GCH fail at $\aleph_{\omega_1}$. Adding subsets to a singular cardinal is a different matter entirely, saturated with difficulties precisely because of such issues surrounding Silver's theorem and the SCH. For example, the natural forcing to add subsets to a singular cardinal $\gamma$ is not $\lt\gamma$-closed. Jun 3 '10 at 20:07
• Does not having a < $\gamma$-closed forcing ensures that you can't add certain kind of subsets? Jul 3 '10 at 7:23