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Statement[편집 | 원본 편집]
A group [math]\displaystyle{ P=( \left\{f: \mathbb{N} \to \mathbb{Z} \right\}, + ) \cong {\mathbb{Z}}^{\omega} }[/math] is not a free abelian group.
Proof[편집 | 원본 편집]
State a lemma related with freeness
Lemma. If F is free and C is a countable subgroup, then the quotient F/C is a direct sum of countable group and a free group.
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Recall Lemma : [math]\displaystyle{ \mathbb{Z}(B_0) }[/math] is countable if [math]\displaystyle{ B_0 }[/math] is countable.
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So the divisible group F/C should be in the countable summand of above lemma and thus countable.
Now suppose the group P is free. Take [math]\displaystyle{ X=\left\{f:\mathbb{N} \to \mathbb{Z} | |\{f(n) != 0 \}| \lt \infty \right\} }[/math] (A subset of P with only finite nonzero function values). Since X is countable, the divisible part of P/X would have countable numbers.
The cardinality of P is equal to [math]\displaystyle{ 2^{\aleph_0} }[/math], the cardinality of basis of P/X is also [math]\displaystyle{ 2^{\aleph_0} }[/math]
However, take a sequence [math]\displaystyle{ n \to n! \cdot a_n }[/math] for arbitrary sequence [math]\displaystyle{ (a_n ) \in \mathbb{Z}^{\aleph_0} }[/math]. Then [math]\displaystyle{ n! \cdot a_n }[/math] is divisible because for all [math]\displaystyle{ N \in \mathbb{N} }[/math], we can take [math]\displaystyle{ ( b_n ) }[/math] for [math]\displaystyle{ b_n = \begin{cases} 0 & n\lt N \\ n! a_n & n \geq N \end{cases} }[/math] satistying [math]\displaystyle{ b_n +C = n!\cdot a_n + C }[/math] and [math]\displaystyle{ (1/N) b_n \in P/C }[/math]. Also, [math]\displaystyle{ a_n \to n! a_n }[/math] is a bijective function of P so [math]\displaystyle{ \{ n! a_n \} +C }[/math] has cardinality [math]\displaystyle{ 2^{\aleph_0} }[/math], which contradicts the lemma.
Therefore, P is not a free abelian group.