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Goldstern
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Each of the following implies that (the true) $\omega_1$ is inaccessible in $L$, and hence that there are only countably many constructible reals:

  • The proper forcing axiom
  • There is a Ramsey cardinal
  • $0^\#$ exists
  • All projective sets are Lebesgue measurable
  • All $\Sigma^1_3$-sets are Lebesgue measurable

The mere existence of a nonconstructible set, or even a nonconstructible real, does not imply that $\omega_1^L$ is countable. There are many forcing notions in $L$ which do not collapse $\omega_1$: adding one or many Cohen reals, destroying Souslin trees, etc. Each such forcing (over L) results in a model where $\omega_1=\omega_1^L$.

In fact, "Martin's axiom plus continuum is arbitrarily large" is consistent with $\omega_1^L=\omega_1$. (And also with $\omega_1^L<\omega_1$.)

Goldstern
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