Can this result in cardinal arithmetic be established without using pcf theory? Suppose $\kappa\leq\mu$ are infinite cardinals.  Let us agree to call a family $\mathcal{P}\subseteq[\mu]^{<\mu}$ a countably generating family for $[\mu]^\kappa$ if every member of $[\mu]^\kappa$ can be written as a union of countably many elements of $\mathcal{P}$. Note that we can extend this in the obvious way to structures of the form $[\mu]^{<\kappa}$ as well.

Theorem: If $\mu$ is a strong limit singular cardinal of uncountable cofinality, then the minimum cardinality of a countably
generating family for $[\mu]^{{\rm cf}(\mu)}$ is the same as the
minimum cardinality of a countably generating family for
$[\mu]^{<\mu}$ (and hence, the same as the minimum cardinality of a
countably generating family for $[\mu]^\kappa$ for any $\kappa$ with
${\rm cf}(\mu)\leq\kappa<\mu$).

The only proof I have relies on heavy machinery from pcf theory, but I do not believe this should be required, and I think that a more direct proof would shed light on some related questions. The following vague question asks for such a proof in a simple special case:

Question : Suppose $\mu$ is a strong limit singular cardinal of cofinality $\omega_1$, and let $\mathcal{P}$ be a countably generating
family for $[\mu]^{\omega_1}$ (WLOG closed under subsets). Is there a reasonably constructive way
to build a countably generating family for $[\mu]^{\omega_2}$ (of the
same cardinality) from $\mathcal{P}$?

I don't have the language of Galois-Tukey connections
available in this context to formulate this question more precisely, but I want to avoid tricks like "let $M$ be the Skolem hull of $\mathcal{P}$ and the other parameters, and then $M\cap [\mu]^{<\mu}$ works because the theorem is true".
Roughly speaking, the pcf-theoretic proof relies on Shelah's "cov vs. pp Theorem" to convert things into an equivalent question about pseudopowers, and then solves the associated pseudopower question using other results from his book Cardinal Arithmetic.   Can we do better? I want to make sure I'm not just suffering from a blind spot and missing something easy.
 A: There IS an easy proof of this, but I just had to reframe the way I was thinking of the problem. The cardinals in question (and many of their relatives) turn out to be $2^{\mu}$ if $\mu$ is strong limit because we can do suitable coding:
Suppose $\mu$ is a singular strong limit cardinal of cofinality $\kappa$.  Since $\mu$ is a strong limit, we can enumerate the bounded subsets of $\mu$ as $\langle X_\alpha:\alpha<\mu\rangle$ in such a way that every such set is indexed cofinally often.
We also let $\langle \mu_\xi:\xi<\kappa\rangle$ be increasing and cofinal in $\mu$.
Given $X\subseteq\mu$, we code $X$ as an element of $[\mu]^\kappa$ in the natural way:  for each $\xi<\kappa$ we choose $\alpha(\xi)<\kappa$ such that $X\cap\mu_{\xi}= X_{\alpha(\xi)}$, and define the code of $X$, $cd(X)$ to be $\{\alpha(\xi):\xi<\kappa\}$.
Note that we can assume the sequence $\langle \alpha(\xi):\xi<\kappa\rangle$ is strictly increasing, so that without loss of generality $cd(X)$ has cardinality $\kappa$.
Now since the sequence $X\cap \mu_{\xi}$ increases with $\xi$, we can recover $X$ from any subset of $cd(X)$ of cardinality $\kappa$ as $X\cap\mu_\xi$ is indexed (in the enumeration of the bounded subsets of $\mu$) by the $\xi$th element in the increasing enumeration of $cd(X)$.  Thus,  given $Y\subseteq cd(X)$ of cardinality $\kappa$,
$$ X = \bigcup_{\alpha\in Y}X_\alpha.$$
Now if $\mathcal{D}$ is any dense subset of $[\mu]^{\kappa}$ (that is, $\mathcal{D}\subseteq [\mu]^{\kappa}$ and every set in $[\mu]^{\kappa}$ has a subset in $\mathcal{D}$, then the ``decoding map''
$$d:\mathcal{D}\rightarrow \mathcal{P}(\mu)$$
defined by
$$d(A) = \bigcup_{\alpha\in A} X_\alpha$$
maps $\mathcal{D}$ onto $\mathcal{P}(\mu)$.
Now turning to the actual question, if $\mu$ is a strong limit singular cardinal of cofinality $\kappa$, and $\mathcal{P}\subseteq[\mu]^{<\mu}$ has the property that every member of $[\mu]^{\kappa}$ is a union of fewer than $\kappa$ members of $\mathcal{P}$, then the set
$$\mathcal{D}=\mathcal{P}\cap [\mu]^{\kappa}$$
is dense in $[\mu]^\kappa$, and therefore $|\mathcal{P}|= 2^{\mu}$.
The pcf stuff I made reference to is then properly viewed as a relative of the above argument that holds true even when $\mu$ is a strong limit in a very weak sense.
