Thread: set theory ordering question

1. set theory ordering question

Using an alternative definition of ordered pairs, choose two different sets $\triangle$ and $\square$, define $\langle a, b \rangle = \lbrace \lbrace a, \square \rbrace, \lbrace b, \triangle \rbrace \rbrace$

State and prove an analogue of Theorem 1.2

so Theorem 1.2 states (a, b) = (a' , b') if and only if a = a' and b = b'.

now the only part I'm not sure about is $\triangle$ and $\square$, so would the statement be:

$\langle a, b \rangle = \langle a', b' \rangle
$
if and only if a = a' b = b', $\triangle = \triangle '$ and $\square = \square '$

or simply $\langle a, b \rangle = \langle a', b' \rangle
$
if and only if a = a' b = b'. where the triangle and square are sort of fixed?

2. Hi ||||| !

Originally Posted by lllll
or simply $\langle a, b \rangle = \langle a', b' \rangle
$
if and only if a = a' b = b'. where the triangle and square are sort of fixed?
This is what you have to do and do not forget $$\triangle \neq \square$$. It is important.

By the way, this definition of an ordered pair was proposed by Felix Hausdorff.

Best wishes,
Seppel