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Let be an infinite subset of such that Let (2.4).
Let ((X,f)) be a TDS and A be an infinite subset of X.
So, let A be an infinite subset of D. Assume that A has no accumulated point.
Let A be an infinite subset of X and (P f)) be dense in A. Then (P f cap A) is infinite.
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I is an infinite subset.
Thus either or is an infinite subset of.
there is an infinite subset I of ℕ so that the inequality (14) holds for all n ∈ I, or, there is an infinite subset J of ℕ so that the inequality (15) holds for all n ∈ J.
Let { a n } be a sequence in D. Since { a n } is an infinite subset of D, there exists an accumulated point a of { a n }.
Let N be the set of modules N i with i ∉ I, then N is an infinite subset of M and consists of pairwise non-isomorphic indecomposable modules.
Then the set $\{ n\in \omega: F(n)\in B\}$ is an infinite subset of $\omega$, hence countable, and so there is a bijection $G \omega \to \{ n\in \omega : F(n)\in B\}$.
This is equivalent to stating that either (i) there is an infinite subset I of ℕ so that the inequality (14) holds for all n ∈ I, or, (ii) there is an infinite subset J of ℕ so that the inequality (15) holds for all n ∈ J. .
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com