Equality of Ordered Pairs

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Theorem

Two ordered pairs are equal iff corresponding elements are equal:

$\left({a, b}\right) = \left({c, d}\right) \iff a = c \land b = d$


It follows directly that:

$\left({a, b}\right) = \left({b, a}\right) \iff a = b$

or, equivalently, that:

$a \ne b \iff \left({a, b}\right) \ne \left({b, a}\right)$


Proof

Let $\left({a, b}\right) = \left({c, d}\right)$.

From the Kuratowski formalization:

$\left\{{\left\{{a}\right\}, \left\{{a, b}\right\}}\right\} = \left\{{\left\{{c}\right\}, \left\{{c, d}\right\}}\right\}$


Suppose $a = b$.

Then:

$\left\{{\left\{{a}\right\}, \left\{{a, b}\right\}}\right\} = \left\{{\left\{{a}\right\}, \left\{{a}\right\}}\right\} = \left\{{\left\{{a}\right\}}\right\}$

Thus $\left\{{\left\{{c}\right\}, \left\{{c, d}\right\}}\right\}$ has only one element.

Thus $\left\{{c}\right\} = \left\{{c, d}\right\}$ and so $c = d$.

So:

$\left\{{\left\{{c}\right\}, \left\{{c, d}\right\}}\right\} = \left\{{\left\{{a}\right\}}\right\}$

and so $a = c$ and $b = d$.

Thus the result holds.


Now suppose $a \ne b$. By the same argument it follows that $c \ne d$.

So that means that either $\left\{{a}\right\} = \left\{{c}\right\}$ or $\left\{{a}\right\} = \left\{{c, d}\right\}$.


Since $\left\{{c, d}\right\}$ has distinct elements, $\left\{{a}\right\} \ne \left\{{c, d}\right\}$.

Thus:

$\left\{{a}\right\} = \left\{{c}\right\}$

and so $a = c$.


Then:

$\left\{{a, b}\right\} = \left\{{c, d}\right\}$

and so $b = d$.


Now suppose $a = c$ and $b = d$.

Then:

$\left\{{a}\right\} = \left\{{c}\right\}$ and $\left\{{a, b}\right\} = \left\{{c, d}\right\}$

Thus:

$\left\{{\left\{{a}\right\}, \left\{{a, b}\right\}}\right\} = \left\{{\left\{{c}\right\}, \left\{{c, d}\right\}}\right\}$

$\blacksquare$


Sources

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