# If S has the same number of elements as T, then T has the same number as S...

#### VoNemo19

##### New member
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.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
(I don't know how to put the \pi above the \leftrightarrow without the \overbrace).
\stackrel{\pi}{\leftrightarrow} or \overset{\pi}{\leftrightarrow} gives $\overset{\pi}{\leftrightarrow}$.

It is trivial to people because we have a great deal of intuition concerning the number of elements in a set and one-to-one correspondences between sets. Some of these intuitions probably come from visual images. If one has to construct a formal proof, for example, in order to convince a computer, which does not have these intuitions, then the task becomes more challenging.

For example, it is an obvious corollary of the pigeonhole principle that if f : {1, ..., m} -> {1, ..., n} and m > n, then there exist i1 and i2 such that 1 ≤ i1 < i2 ≤ m and f(i1) = f(i2). But if we have a formal proof of this fact, we can do more. We can mechanically extract a program that, given f, would compute such i1 and i2. Writing such program from scratch, though not hard, does not seem so trivial, especially if we are not allowed to use arrays. This implies to me that in considering the original fact obvious, we are cheating and coming up with some arguments that are different from a formal mathematical proof.

This problem is still simple enough, and it means that we are expected to provide a rather detailed proof.

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]
This just means that there exists a function $\pi:S\to T$...

in such a way that every element of[TEX] T[/TEX] is paired with precisely one element of [TEX]S[/TEX].
Formally, this means

$\forall y\in T\,\exists! x\in S\,\pi(x)=y~(1)$

Here $\exists!x$ means that there exists a unique x. If, given y, we denote the x whose existence is guaranteed in (1) by $\pi'(y)$, then (1) expresses the following two facts.

$\forall y\in T\, \pi(\pi'(y))=y~(2)$
$\forall y\in T\,\forall x'\in S.\,\pi x'=y\to x'=\pi'(y)~(3)$

Now we need to show that $\pi'$ establishes a pairing that shows that T has the same number of elements as S. We already have that $\pi'$ is a function from T to S, so we need to check

$\forall x\in S\,\exists! y\in T\,\pi'(y)=x~(4)$

Of course, given x, we let $y = \pi(x)$. Now the unique existential quantifier results in the following two facts to prove.

$\forall x\in S\,\pi'(\pi(x))=x~(5)$
$\forall x\in S\,\forall y'\in T.\,\pi'(y')=x\to y'=\pi(x)~(6)$

Hmm, the first attempt to prove (5) would be to use (2), but it has functions in the wrong order. It does not seem so trivial now, does it? Hint: (5) is proved using (3) and (6) is proved using (2).

• Jameson and Sudharaka