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Exact(21)
71 (2009) 3794 3804]: in the process we also obtain a very useful fact that a random locally convex module with the countable concatenation property must have the same completeness under the two topologies.
Further, if E has the countable concatenation property, then (K_{alpha}) has the countable concatenation property.
Then (operatorname{epi}(f)) has the countable concatenation property.
By Lemma 3.7, A has the countable concatenation property.
Since E has the countable concatenation property, by Lemma 3.3 (K_{alpha}) has the countable concatenation property.
Further, since A has the countable concatenation property, it implies that (Acap[K_{alpha}+ x,r)]), (forall x,r in Etimes L^{0}(mathcal{F})) has the countable concatenation property.
Similar(39)
we denote a countable chain.
Since every countable level is itself countable (after all, there are only countably many possible defining formulas), and there are ω1 countable levels, there must be only ω1 real numbers.
= concatenation.
"A concatenation of circumstances. . .
Three-quarters have no countable assets.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com