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I was reading this book called "discrete maths by susanna epp"
The topic of representing ordered pair by sets came in which it was done by
$$ \{ \{a\}, \{a, b\} \} = (a, b)$$
In a set order doesn't matter
My question is isn't this also valid? $$\{ \{b, a\}, \{a\} \}=\{ \{a\}, \{a, b\} \}? $$ if yes, then $\{ \{b, a\}, \{a\} \}$, $b$ comes first, then $a$ so how come $ \{ \{a\}, \{a, b\} \} = (a, b)$ be valid?

Mauro ALLEGRANZA
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Rambal heart remo
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  • You are right that in "a set order doesn't matter." So look carefully at the definition (and presumably the explanation given in that book on discrete math) and try to see how the first element of the ordered pair might be determined without resort to the order (arrangement) in the set notation of elements. – hardmath Sep 09 '22 at 04:21
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    The "decoding" of this ordered pair has been discussed here before. It helps to know that it is commonly called Kuratowski's ordered pair. – hardmath Sep 09 '22 at 04:26
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    The first element is not whatever you write first. The first element is the unique element of $\cap(a,b)$. – Arturo Magidin Sep 09 '22 at 05:02
  • ${a,b}$ and ${b,a}$ are the same thing. – DanielWainfleet Sep 09 '22 at 07:56

1 Answers1

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You are correct that the order you write the elements of the set doesn't matter. So instead, consider that in both cases one of the element of the set is a subset of the other, which does create an ordering of the elements. In both the case of $\{\{a\}, \{a, b\}\}$ and $\{\{a, b\}, \{a\}\}$, you can pick the "first" element as being $\{a\}$ since $\{a\} \subseteq \{a, b\}$ but $\{a, b\} \not\subseteq \{a\}$ unless $a = b$.

ConMan
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