I know that, in the presence of $CH$, Namba forcing does not add reals. But when $CH$ fails, is it consistent that it still does not add reals?
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If CH fails, then Nm adds reals (or equivalently, new $\omega$-sequences of reals).
Proof: Let $f:\omega_2\to 2^\omega$ be 1-1. Let $\bar \alpha:=(\alpha_n:n\in \omega)$ be the name for the Namba sequence in $\omega_2$. Then $x:=(f(\alpha_n):n\in \omega)$ is the name of a new sequence of reals. The sequence $x$ is new because $\bar\alpha$ can be computed from $x$.
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$\begingroup$ How do you get a new real from the $\omega$-sequence? $\endgroup$– ZooradoCommented Mar 12, 2015 at 9:31
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2$\begingroup$ @Zoorado: Pick your favorite bijection in the ground model between $\omega$ and $\omega^2$, and use it to encode the sequence into a single real. $\endgroup$– Asaf Karagila ♦Commented Mar 12, 2015 at 10:19