Equality of Ordered Pairs/Lemma
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Theorem
Let $\set {a, b}$ and $\set {a, d}$ be doubletons such that $\set {a, b} = \set {a, d}$.
Then:
- $b = d$
Proof
We have that:
- $b \in \set {a, b}$
and so by definition of set equality:
- $b \in \set {a, d}$
So:
- $(1): \quad$ either $b = a$ or $b = d$.
First suppose that $b = a$.
Then:
- $\set {a, b} = \set {a, a} = \set a$
We have that:
- $d \in \set {a, d}$
and so by definition of set equality:
- $d \in \set {a, b}$
and so as $\set {a, b} = \set a$ it follows that:
- $d = a$
We have $b = a$ and $d = a$ and so:
- $b = d$
and so $b = d$.
$\Box$
Next suppose that $b \ne a$.
We have from $(1)$ that either $b = a$ or $b = d$.
As $b \ne a$ it must be the case that $b = d$.
$\Box$
So in either case we see that:
- $b = d$
$\blacksquare$
Sources
- 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): Exercise $1.1: \ 12$
- 2010: Raymond M. Smullyan and Melvin Fitting: Set Theory and the Continuum Problem (revised ed.) ... (previous) ... (next): Chapter $2$: Some Basics of Class-Set Theory: $\S 4$ The pairing axiom: Lemma $4.3$.