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Remark Let $\lambda $ be an infinite cardinal. Then the exponential cofinality $\mathrm{ecf}(\lambda )$ satisfies $\aleph _0 \leq \mathrm{ecf}(\lambda ) \leq \mathrm{cf}(\lambda )$. In particular, we have $\mathrm{ecf}(\lambda ) \leq \lambda $. The inequality $\aleph _0 \leq \mathrm{ecf}(\lambda )$ is a reformulation of the fact that the collection of $\lambda $-small sets is closed under finite products (Proposition To prove the other inequality, choose a set $S$ of cardinality $\mathrm{cf}(\lambda )$ and a collection of $\lambda $-small sets $\{ T_ s \} _{s \in S}$ for which the coproduct $T = {\coprod }_{s \in S} T_{s}$ is not $\lambda $-small. We now observe that $T$ can be identified with a subset of the product ${\prod }_{s \in S} ( T_ s \coprod \{ s\} )$. Since each of the sets $T_{s} \coprod \{ s\} $ is also $\lambda $-small, we obtain $\mathrm{ecf}(\lambda ) \leq \mathrm{cf}(\lambda )$.