Re: 1-1/2+1/3-1/4+1/5-1/6+1/7



Han.deBruijn@xxxxxxxxxxxxxx writes:

Let's repeat the definition of S(). I hope this is the latest version:

y is in S(x) iff there is some n e N such that for all m e N ,
with m > n , y is in S_m(x) .
Here, S_m(x) = s o s o ... o s(x), where there are exactly
m compositions. (m e N)

Lemma ( with () -> [] ) :

y is in S[x] iff there is some n e N such that for all m e N ,
with m > n , y is in S_m[x] .
Here, S_m[x] = s o s o ... o s[x], where there are exactly
m compositions. (m e N)

Proof:

Because S[x] = { S(y) , y e x } and S(y) according to the above.

The proof does not look remotely like a proof of this claim to me.

Let's define

y e T(x) <=> there is an n e N such that for all m e N with m > n,
y is in S_m[x].

You want to claim that T(x)=S[x]. Of course, you've noticed that

S[N] = {N} and
T(N) = {},

so this will yield a contradiction. But your "proof" is not a proof
at all, so let's see what we can do.

S[x] = { S(z) | z in x }
= { y | (E z in x)(E n in N)(A m in N)(m > n -> y in S_m(z)) }

I do not see why that should be the same set as

T(x) = { y | (E n in N)(A m in N)(m > n -> (E z in x)(y in S_m(z))) }

So, you have to show that

(E z in x)(E n in N)(A m in N)(m > n -> y in S_m(z))

iff

(E n in N)(A m in N)(m > n -> (E z in x)(y in S_m(z)))

I don't believe it's so.

--
"My personal opinion is, it was a shameful act for someone to disclose
this very important program in a time of war. The fact that we're
discussing this program is helping the enemy."
-- George W. Bush: Questioning Illegal Wiretaps Just Helps The Enemy!
.



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