[QUOTE=davar55;248424]

A intersect B has 3 elements. Correct.
A union B has 5 elements. Correct.
(Not 8 as you now see.)[/QUOTE]

so in other words A union B = #A+#B - #(A intersect B)

[QUOTE=science_man_88;248418]Is {1, 2, 3, 4} equal to {1, 2, 3, 4, 1}?

my first guess is no, just on the basis that there's an extra 1 , which breaks my understanding as it violates the same number of elements. [/QUOTE]

Your understanding is wrong, but that's OK, we're getting there. Just as order doesn't matter, duplicates don't matter either.

"A = {1, 2, 3, 4}" is just a way of writing down five facts about A: 1[tex]\in[/tex]A, 2[tex]\in[/tex]A, 3[tex]\in[/tex]A, 4[tex]\in[/tex]A, nothing else [tex]\in[/tex]A. If you think about it, saying those things in a different order doesn't change anything. Neither does repeating any one of them.

[QUOTE=science_man_88;248425]So in other words:
# A union B = # A + # B - # (A intersect B)[/QUOTE]

Correct as edited. Very good.

[QUOTE=Mr. P-1;248428]Your understanding is wrong, but that's OK, we're getting there. Just as order doesn't matter, duplicates don't matter either.

"A = {1, 2, 3, 4}" is just a way of writing down five facts about A: 1[tex]\in[/tex]A, 2[tex]\in[/tex]A, 3[tex]\in[/tex]A, 4[tex]\in[/tex]A, nothing else [tex]\in[/tex]A. If you think about it, saying those things in a different order doesn't change anything. Neither does repeating any one of them.[/QUOTE]

True, except for the additional time it takes to say them or to unscramble
the order if there's an order. More later.

Redundancy, noops, you know about those from programming.
This will all get connected later.

[QUOTE=Mr. P-1;248428]duplicates don't matter either.[/QUOTE]

For sets. -fuples are different. "A = (1, 2, 3, 4)" is a way of saying "A has four elements. the first is 1; the second is 2; the third is 3; the fourth is 4".

[QUOTE=Mr. P-1;248431]For sets. -fuples are different. "A = (1, 2, 3, 4)" is a way of saying "A has four elements. the first is 1; the second is 2; the third is 3; the fourth is 4".[/QUOTE]

So set A = { 1,2,3,4 } = { 2,4,3,1 } = { 4,3,2,1 } = etc.

But 4-tuple (1,2,3,4) != (2,4,3,1) != (4,3,2,1) != etc.

I'd use the ne sign if I hadn't learned C and != and == and etc.

[QUOTE=science_man_88;248425]so in other words A union B = #A+#B - #(A intersect B)[/QUOTE]

I think you meant:

#(A union B) = #A+#B - #(A intersect B)

Can you show that this is always the case for finite sets?

[QUOTE=Mr. P-1;248435]I think you meant:

#(A union B) = #A+#B - #(A intersect B)

Can you show that this is always the case for finite sets?[/QUOTE]

well #A+#B is the maximum that it could be in a finite set, then to eliminate overlap (intersection) which wouldn't be in the union, you gain -#(A intersection B) if you sum these you get the equation above.

[QUOTE=davar55;248434]So set A = { 1,2,3,4 } = { 2,4,3,1 } = { 4,3,2,1 } = etc.

But 4-tuple (1,2,3,4) != (2,4,3,1) != (4,3,2,1) != etc.

I'd use the ne sign if I hadn't learned C and != and == and etc.[/QUOTE]

so in other words because you know a non-TEX symbol for it you don't use the TEX symbol ?

Sometimes. Depends on the complexity.

I'd write 2[sup]n[/sup] - 1 or 2^n - 1 depending on context,

but for the YJ-Conjecture I wrote

lim(n->infinity) M[sub]n[/sub] ^ (1/n) = 3/2 = 1.500000 exactly

because there's no infinity symbol in ascii (and there shouldn't be).

[QUOTE=davar55;248449]Sometimes. Depends on the complexity.

I'd write 2[sup]n[/sup] - 1 or 2^n - 1 depending on context,

but for the YJ-Conjecture I wrote

lim(n->infinity) M[sub]n[/sub] ^ (1/n) = 3/2 = 1.500000 exactly

because there's no infinity symbol in ascii (and there shouldn't be).[/QUOTE]"oo" has been used for many years.


Paul