"Foreman`s maximality principle" states that every non-trivial forcing notion either adds a real or collapses some cardinals. This principle has many consequences including:
1) $GCH$ fails everywhere,
2) there are no inaccessible cardinals,
3) there are no $\kappa-$Souslin trees,
4) Any non-trivial $c.c.c.$ forcing adds a real,
5) Any non-trivial $\kappa^+-$closed forcing notion collapses some cardinals.
Consistency of (1) is proved by Foreman-Woodin, (2) clearly can be consistent and the consistency of (4) is shown in "Forcing with c.c.c forcing notions, Cohen reals and Random reals".
My interest is in the consistency of (5). Let's consider the case $\kappa=\omega.$
Question 1. Is it consistent that any non-trivial $\aleph_1-$closed forcing notion collapses some cardinals?
The above question seems very difficult, and it is not difficult to show that for its consistency we need some very large cardinals. But maybe the following is simpler:
Question 2. Is it consistent that any non-trivial $\aleph_1-$closed forcing notion of size continuum collapses some cardinals? Does its consistency imply the existence of large cardinals?
Best Answer
The answer to the second question is yes, without any large cardinals assumptions.
Claim: if $2^{\aleph_0}$ is singular then every non-trivial $\sigma$-closed forcing of size $2^{\aleph_0}$ collapses the continuum.
The reason is that such forcing must add a new set of ordinals of size $\lambda < 2^{\aleph_0}$, $\tau$ (since the minimal $\lambda$ must be regular). I want to code $2^\omega$ into $\tau$.
If we choose $\lambda$ to be minimal, every initial segment of $\tau$ is in $V$. Now, we can define a function $F\in V$ from all possible initial segments of $\tau$ onto the reals, such that for every $p\in P$ and every $x\in 2^\omega$ there is $q\leq p$ and $\beta < \lambda$ such that $q\Vdash \tau \restriction \beta = \check{a}$ and $F(a)=x$:
Let $\langle (p_i, x_i) | i < 2^{\aleph_0}\rangle$ enumerate $P\times 2^\omega$. For every $\alpha < 2^{\aleph_0}$, we use the $\sigma$-closure of $P$ and the fact that $\tau \notin V$ in order to find $q \leq p_\alpha$ such that $q \neq p_i$ for every $i < \alpha$, $q\Vdash \tau \restriction \beta = \check{a}$ (for some $\beta$) and $F(a)$ is not determined yet, and set $F(a)=x_\alpha$:
Let $p = p_\alpha \in P$. We start by building a tree of $2^{<\omega}$ incompatible conditions $q_s,\, s\in 2^{<\omega}$ such that $q_\emptyset = p$ and for every $s\in 2^{<\omega}$, $q_{s\frown (0)}, q_{s\frown (1)} \leq q_s$, there is $\beta_s < \lambda$ in which $q_{s\frown (i)} \Vdash \tau \restriction \beta_s = a_{s\frown (i)}$ for $i\in\{0,1\}$, $a_{s\frown (0)} \neq a_{s \frown (1)}$. This is possible since $\tau \notin V$ but every initial segment of it is in $V$.
For every $f\in 2^\omega$, let $q_f \in P,\,\forall n\, q_f \leq q_{f\restriction n}$ (by the closure of $P$). Without loss of generality, $\forall f \in 2^\omega\,q_f\Vdash \tau \restriction \beta = \check{a_f}$, and for every $f\neq f^\prime$, $a_f \neq a_{f^\prime}$ (take $\beta = \sup_{s\in 2^{<\omega}} \beta_s$). Since we picked already only $|\alpha |<2^{\aleph_0}$ values, there must be some $f\in 2^\omega$ such that $F(a_f )$ is not determined.
By density arguments, in $V[G]$, $\{F(\tau \restriction \beta) | \beta < \lambda\} = (2^\omega)^V$.