On Wed, 14 May 2008, Oxnard wrote:
> E = is element of s
> N = integers >= 0
>
> {y | y E N ^ y = y+1}
A = { x in N | y = y + 1 } = nulset
> The books answer is the null set.
If x in A, then x in N, x = x + 1.
Thus 0 = 1 which is contradiction.
Consequently x not in A.
For all x in A, x not in A
<->
for all x, x not in A
<->
not some x, x in A
<->
A = nulset
> The answer makes sense but not sure how to work through the steps
another
> example in the book shows
>
> t1 := { y E N | y = y +1}
>
> so y must be an element of N which satisfies y = y +1 which there are
none
>
> t2 := { y | y E t1 }
>
As A = { x | x in A }, t1 = t2.
> y = null ..
>
That is nonsense.
> so not sure how to work through the equation ... dealing with the null
set
> ... any advice
The empty set, nulset = { x | x /= x }.
Riddle of the day. Can somebody when he just not himself,
apply for member****p in the empty set?
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