Density with infinite cardinals Let κ ≤ µ infinite cardinals.
and lat D(µ, κ) = min{|D| : D ⊆ [µ]^κ ∧ (∀y ∈ [µ]^κ)(∃x ∈ D)(x ⊆ y)}
D(µ, κ) is called the density of κ-sets of µ.  
1) Suppose κ = cf(µ) < µ.  prove that D(µ, κ) > κ.
2) Suppose κ = cf(κ) < µ  and κ != cf(µ).  prove that D(µ, κ) = ΣD(λ, κ) for κ≤λ<µ.
 A: For question 1, you can do better: $D(\kappa,\mu)>\mu$. To prove it, suppose you're given $\mu$ (or fewer) subsets of $\mu$, each of size $\kappa$; you need to produce a subset $X$ of $\mu$, of size $\kappa$, that is not a superset of any of the given sets. Group the given sets into $\kappa$ families $F_\alpha$ (where $\alpha<\kappa$), each of size $<\mu$. Then choose the $\kappa$ elements $x_\alpha$ to form $X$, one at a time by induction on $\alpha$.  Choose $x_\alpha$ to be distinct from all the previously chosen $x_\beta$'s and to be outside all the sets in all the $F_\beta$'s for $\beta<\alpha$.  This can be done because all those $F_\beta$'s together contain fewer than $\mu$ sets, each of size $\leq\kappa$.  The resulting $X=\{x_\alpha:\alpha<\kappa\}$ is clearly not a superset of any set from any $F_\beta$.
For question 2, if $\kappa<\text{cf}(\mu)$, then just use that any $\kappa$-sized subset of $\mu$ is a subset of some ordinal $<\mu$.  If, on the other hand, $\kappa>\text{cf}(\mu)$, then every $\kappa$-sized subset of $\mu$ has $\kappa$ elements below some ordinal$\alpha<\mu$.
EDIT, in response to a comment-question from the OP:
The only difference between the two cases is that, in the first case, we get a stronger result and the proof of that stronger result is just the definition of cofinality. In the second case, we have a weaker result and the proof isn't quite so trivial (though it's still pretty easy). Namely, choose a cofinal subset $C$ of $\mu$ of size $\text{cf}(\mu)$. If a set $X\subseteq\mu$ has fewer than $\kappa$ elements below each element of $X$, then, by regularity of $\kappa$ and the case hypothesis $\kappa>\text{cf}(\mu)$, we get that $X$ has $<\kappa$ elements altogether.
Returning to question 2, and having the weaker result from case 2 available in both cases, we can proceed as follows. For each ordinal $\alpha<\mu$ (or just for cofinally many such $\alpha$), choose a dense subset of $[\alpha]^\kappa$ of cardinality $D(|\alpha|,\kappa)$. The union of these chosen sets will be dense in $[\mu]^\kappa$.
