## Easy Direction

August 29, 2011 at 14:07 | Posted in Uncategorized | Leave a commentMy daughter was born last week, so this post has been written by someone for whom sleep is but a fond memory…

Anyway, picking up where we left off, let us define

and

We’ll dispose of the easy direction first, and show

Thus, let be a family of cardinality with the property that any set in is covered by a union of fewer than sets in .

For each , we define a function with domain by setting

whenever ; otherwise, set equal to .

Since elements of have cardinality , it is clear that each is an element of . We claim that the collection witnesses .

This should follow easily from the definitions involved via a proof by contradiction: if not, then there is a set of cardinality less than and a function such that cannot be majorized by taking the supremum of any collection of fewer than functions in .

By the definition of , we can find and such that

For each , let denote the function . Given , we know lies in for some , hence

Thus, for each there is an such that , and since each is in , we have a contradiction.

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