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?



"Namba forcing adds reals" independent of $ZFC + \neg CH$?
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 $x$. 


