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VoNemo19
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Hi. The proof that I'm working is in the title. The Defition of "same number as" is given as follows: [TEX](1) [/TEX]Two sets [TEX]S[/TEX] and [TEX]T[/TEX] are said to have the same nuber of elements if each element of [TEX]S[/TEX] can be paired with precisely one element of[TEX] T[/TEX] in such a way that every element of[TEX] T[/TEX] is paired with precisely one element of [TEX]S[/TEX].
Notation: If [TEX]\pi[/TEX] is a pairing of the elements of [TEX]S [/TEX]with with those of Tand the element sof S is paired in [TEX]\pi[/TEX] to the element [TEX]t [/TEX]of T[TEX],[/TEX] we shall write [TEX]s\overbrace{\leftrightarrow}^{\pi}{t}[/TEX] (I don't know how to put the \pi above the \leftrightarrow without the \overbrace).
I don't really understand the book's proof: Since [TEX]S[/TEX] has the same number of elements as [TEX]T[/TEX], we can select a pairing between the elements of [TEX]S [/TEX]and [TEX]T[/TEX] in accordance with [TEX](1). [/TEX]We define a pairing as follows: If[TEX] s[/TEX] is paired with [TEX]t[/TEX] by the selected pairing [TEX]\pi[/TEX], then pair[TEX] t[/TEX] with [TEX]s[/TEX]. That is if [TEX]s\overbrace{\leftrightarrow}^{\pi}{t}[/TEX], then[TEX] t[/TEX] is paired with [TEX]s[/TEX] to form the pairing of the elements [TEX]T[/TEX] with those of [TEX]S[/TEX]. If the original pairing satisfied [TEX](1), [/TEX]then so will the new pairing. Specifically, since[TEX] \pi [/TEX]had each element[TEX] T[/TEX] paired with a unique element of [TEX]S[/TEX], then the second pairing also has this property. Therefore, [TEX]T[/TEX] has the same number of elements as[TEX] S[/TEX].
Please help me to understand why this proposition is not trivial, and also the procedure of the proof.
Notation: If [TEX]\pi[/TEX] is a pairing of the elements of [TEX]S [/TEX]with with those of Tand the element sof S is paired in [TEX]\pi[/TEX] to the element [TEX]t [/TEX]of T[TEX],[/TEX] we shall write [TEX]s\overbrace{\leftrightarrow}^{\pi}{t}[/TEX] (I don't know how to put the \pi above the \leftrightarrow without the \overbrace).
I don't really understand the book's proof: Since [TEX]S[/TEX] has the same number of elements as [TEX]T[/TEX], we can select a pairing between the elements of [TEX]S [/TEX]and [TEX]T[/TEX] in accordance with [TEX](1). [/TEX]We define a pairing as follows: If[TEX] s[/TEX] is paired with [TEX]t[/TEX] by the selected pairing [TEX]\pi[/TEX], then pair[TEX] t[/TEX] with [TEX]s[/TEX]. That is if [TEX]s\overbrace{\leftrightarrow}^{\pi}{t}[/TEX], then[TEX] t[/TEX] is paired with [TEX]s[/TEX] to form the pairing of the elements [TEX]T[/TEX] with those of [TEX]S[/TEX]. If the original pairing satisfied [TEX](1), [/TEX]then so will the new pairing. Specifically, since[TEX] \pi [/TEX]had each element[TEX] T[/TEX] paired with a unique element of [TEX]S[/TEX], then the second pairing also has this property. Therefore, [TEX]T[/TEX] has the same number of elements as[TEX] S[/TEX].
Please help me to understand why this proposition is not trivial, and also the procedure of the proof.