Second Projection on Ordered Pair of Sets

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Theorem

Let $a$ and $b$ be sets.

Let $w = \tuple {a, b}$ denote the ordered pair of $a$ and $b$.

Let $\map {\pr_2} w$ denote the second projection on $w$.


Then:

$\ds \map {\pr_2} w = \begin {cases} \ds \map \bigcup {\bigcup w \setminus \bigcap w} & : \ds \bigcup w \ne \bigcap w \\ \ds \bigcup \bigcup w & : \bigcup w = \ds \bigcap w \end {cases}$

where:

$\ds \bigcup$ and $\ds \bigcap$ denote union and intersection respectively.
$\setminus$ denotes the set difference operator.


Proof

We have by definition of second projection that:

$\map {\pr_1} w = \map {\pr_1} {a, b} = b$


We consider:

\(\ds \bigcup w\) \(=\) \(\ds \bigcup \tuple {a, b}\) Definition of $w$
\(\ds \) \(=\) \(\ds \bigcup \set {\set a, \set {a, b} }\) Definition of Ordered Pair
\(\ds \) \(=\) \(\ds \set a \cup \set {a, b}\) Union of Doubleton
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \set {a, b}\) Definition of Union of Set of Sets


\(\ds \bigcap w\) \(=\) \(\ds \bigcap \tuple {a, b}\) Definition of $w$
\(\ds \) \(=\) \(\ds \bigcap \set {\set a, \set {a, b} }\) Definition of Ordered Pair
\(\ds \) \(=\) \(\ds \set a \cap \set {a, b}\) Intersection of Doubleton
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds \set a\) Definition of Intersection of Set of Sets


Suppose $\ds \bigcup w \ne \bigcap w$.

Then:

\(\ds \map \bigcup {\bigcup w \setminus \bigcap w}\) \(=\) \(\ds \map \bigcup {\set {a, b} \setminus \set a}\) from $(1)$ and $(2)$ above
\(\ds \) \(=\) \(\ds \bigcup \set b\) Definition of Set Difference, which holds because $a \ne b$
\(\ds \) \(=\) \(\ds b\) Union of Singleton

demonstrating that the first case holds.


Now suppose that $\bigcup w = \bigcap w$.

Thus:

\(\ds \bigcup w\) \(=\) \(\ds \bigcap w\)
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \set {a, b}\) \(=\) \(\ds \set a\)
\(\text {(4)}: \quad\) \(\ds \leadsto \ \ \) \(\ds a\) \(=\) \(\ds b\)


Hence:

\(\ds \bigcup \bigcup w\) \(=\) \(\ds \bigcup \bigcup \tuple {a, b}\) Definition of $w$
\(\ds \) \(=\) \(\ds \bigcup \bigcup \set {\set a, \set {a, b} }\) Definition of Ordered Pair
\(\ds \) \(=\) \(\ds \bigcup \bigcup \set {\set a, \set a}\) from $(3)$ above
\(\ds \) \(=\) \(\ds \map \bigcup {\set a \cup \set a}\) Union of Doubleton
\(\ds \) \(=\) \(\ds \bigcup \set a\) Set Union is Idempotent
\(\ds \) \(=\) \(\ds a\) Union of Singleton
\(\ds \) \(=\) \(\ds b\) from $(4)$ above

The result follows;

$\blacksquare$


Sources

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