In Spivak's Calculus, he provides a """rigorous construction""" of the ordered pair.
To do so, he defines (a,b) as {{a},{a,b}}.
Can one of you explain how this is significant or rigorous. The definition and accompanying theorem seem arbitrary to me.
In Spivak's Calculus, he provides a """rigorous construction""" of the ordered pair
Brody Allen
Angel Jenkins
Set inclusion/difference gives the ordering.
Carson Parker
Could you explain this further? I'm completely lost.
Asher Nelson
>Can one of you explain how this is significant or rigorous.
He is showing that if you have 2 ordered pairs which are equal to each other they are made up out of the same members, it is written right there on the page.
It is obviously important to prove this, else the definition wouldn't make much sense.
>rigorous
It obviously is rigorous, there are no uncertainties and the definition is clear and unambiguous.
>significant
You need ordered pairs to construct the rational numbers and other things but the necessity for such a thing should be obvious.
Leo Moore
The fuck? What's so hard to get about [math] \{a\} \subset \{a,b\} [/math]
Colton Green
One more hint: [math] \{a,b\} \eq \{b,a\} [/math].
You should figure the rest out by yourself (if you can't, I suggest you study something else.)
Camden Wood
can this be extended to triplets and n-tuples
with (a,b,c)->{{a},{a,b},{a,b,c}}
(a1,a2,...,an)->{{a1},{a1,a2}...,{a1,a2,...,an}}
using induction on the same argument?
Nathaniel Carter
No, it's rigorous in the sense that it is set theoretical, i.e. it is based on more general primitive notions.
Parker Hall
What do you think?
Tyler Fisher
I think that you're a bad teacher.