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I'm going through the set theory material in the appendix of Knapp's Basic Algebra. I want to make sure that I understand what he says is the set theoretic notion of the indexed cartesian product, [itex]\prod_{x\in S}A_{x}[/itex].
He says that this can be thought of as the set of all functions [itex]f:S\rightarrow \bigcup_{x\in S}A_{x}[/itex] such that [itex]f(x)\in A_{x}[/itex] for all [itex]x\in S[/itex]. Let's call this set [itex]F[/itex].
So as an example, let [itex]S = \left\{1,2\right\}[/itex], [itex]A_{1} = \left\{3,4\right\}[/itex], and [itex]A_{2} = \left\{5,6\right\}[/itex]. Then [itex]\bigcup_{x\in S}A_{x} = \left\{3,4,5,6\right\}[/itex] and the functions f, being subsets of [itex]S\times \bigcup_{x\in S}A_{x}[/itex], are [itex]f_{1} = \left\{(1,3), (2,5)\right\}[/itex], [itex]f_{2} = \left\{(1,3), (2,6)\right\}[/itex], [itex]f_{3} = \left\{(1,4), (2,5)\right\}[/itex], [itex]f_{4} = \left\{(1,4), (2,6)\right\}[/itex].
Then the set [itex]F[/itex] is [itex]\left\{\left\{(1,3), (2,5)\right\},\left\{(1,3), (2,6)\right\},\left\{(1,4), (2,5)\right\},\left\{(1,4), (2,6)\right\}\right\}[/itex].
Since we can form a bijection [itex]g[/itex] from [itex]F[/itex] to [itex]A_{1}\times A_{2}[/itex] with [itex]g:F\rightarrow A_{1}\times A_{2}[/itex] such that [itex]g(f) = (f(1), f(2))[/itex] for all [itex]f\in F[/itex], we can say that F is isomorphic to [itex]A_{1}\times A_{2}[/itex] and thus they are the same set. Is my understanding correct?
He says that this can be thought of as the set of all functions [itex]f:S\rightarrow \bigcup_{x\in S}A_{x}[/itex] such that [itex]f(x)\in A_{x}[/itex] for all [itex]x\in S[/itex]. Let's call this set [itex]F[/itex].
So as an example, let [itex]S = \left\{1,2\right\}[/itex], [itex]A_{1} = \left\{3,4\right\}[/itex], and [itex]A_{2} = \left\{5,6\right\}[/itex]. Then [itex]\bigcup_{x\in S}A_{x} = \left\{3,4,5,6\right\}[/itex] and the functions f, being subsets of [itex]S\times \bigcup_{x\in S}A_{x}[/itex], are [itex]f_{1} = \left\{(1,3), (2,5)\right\}[/itex], [itex]f_{2} = \left\{(1,3), (2,6)\right\}[/itex], [itex]f_{3} = \left\{(1,4), (2,5)\right\}[/itex], [itex]f_{4} = \left\{(1,4), (2,6)\right\}[/itex].
Then the set [itex]F[/itex] is [itex]\left\{\left\{(1,3), (2,5)\right\},\left\{(1,3), (2,6)\right\},\left\{(1,4), (2,5)\right\},\left\{(1,4), (2,6)\right\}\right\}[/itex].
Since we can form a bijection [itex]g[/itex] from [itex]F[/itex] to [itex]A_{1}\times A_{2}[/itex] with [itex]g:F\rightarrow A_{1}\times A_{2}[/itex] such that [itex]g(f) = (f(1), f(2))[/itex] for all [itex]f\in F[/itex], we can say that F is isomorphic to [itex]A_{1}\times A_{2}[/itex] and thus they are the same set. Is my understanding correct?