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## Alternate proofs in Tutorial 1?

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• In Tutorial 1 (dynkin systems), Ex. 7, we show in step 1 that, given D is a pi-system and a dynkin system, if A,B in D, then A U B in D. My proof of this
Message 1 of 5 , Sep 2, 2001
In Tutorial 1 (dynkin systems), Ex. 7, we show in step 1 that, given
D is a pi-system and a dynkin system, if A,B in D, then A U B in D.
My proof of this differs from yours, and I'm not certain it's
correct. I proceeded as follows (don't know how to really do this in
ASCII, but below, T means intersection):

(A T B) in D, because D is a pi-system
B = B U (A T B) in D
B U (A T B) = (B U A) T (B U B) = (A U B) T B in D
thus (A U B) in D, because B in D and D is a pi-system.

This seems a lot simpler than the given solution, but I'm
inexperienced in set theory so don't know if I'm doing something
silly.

For step 2, I proceeded inductively. My suspicion is that you can't,
but I can't explain why - again, as a relative newbie, I'm wondering
if I'm falling into some hidden trap. Here's the reasoning:

union{n=1->k-1}Bn = union{n=1->k-1}An (induction hypothesis)
union{n=1->k}Bn = Bk U (union{n=1->k-1}Bn)
= Bk U (union{n=1->k-1}An)
= union{n=1->k}An U (union{n=1->k-1}An)
= (Ak U union{n=1->k-1}An) U (union{n=1->k-1}An)

But by the associative law, this is just Ak U union{n=1->k-1}An =
union{n=1->k}An. So now all that's left is to establish the base
condition, which is just

union{n=1->1}Bn = union{n=1->1}An

or A1 = A1, which is certainly true. Thus, by induction it's shown
that union{n=1->inf}Bn = union{n=1->inf}An. Is this legal?

Thanks,
Allan
• Hi Allan, ... in ... Yes ... True ... All this is true, since all these sets are equal to B which is in D ... D being a pi-system tells you that if A and B
Message 2 of 5 , Sep 3, 2001
Hi Allan,

> My proof of this differs from yours, and I'm not certain it's
> correct. I proceeded as follows (don't know how to really do this
in
> ASCII, but below, T means intersection):
>
> (A T B) in D, because D is a pi-system

Yes

> B = B U (A T B) in D

True

> B U (A T B) = (B U A) T (B U B) = (A U B) T B in D

All this is true, since all these sets are equal to B which is in D

> thus (A U B) in D, because B in D and D is a pi-system.

D being a pi-system tells you that if A' and B' are in D, then A'TB'
is also in D. It does not tell you that if A'TB' is in D and B' is in
D, then A' is in D... Hence I do not agree with your last line.

>
> if I'm falling into some hidden trap. Here's the reasoning:
>
> union{n=1->k-1}Bn = union{n=1->k-1}An (induction hypothesis)
> union{n=1->k}Bn = Bk U (union{n=1->k-1}Bn)
> = Bk U (union{n=1->k-1}An)
> = union{n=1->k}An U (union{n=1->k-1}An)
> = (Ak U union{n=1->k-1}An) U (union{n=1->k-1}An)
>
> But by the associative law, this is just Ak U union{n=1->k-1}An =
> union{n=1->k}An. So now all that's left is to establish the base
> condition, which is just
>
> union{n=1->1}Bn = union{n=1->1}An
>
> or A1 = A1, which is certainly true. Thus, by induction it's shown
> that union{n=1->inf}Bn = union{n=1->inf}An. Is this legal?

This is a correct induction proof which establishes the equality
between the finite union \/_1^nAk and \/_1^n Bk. You need one little
more step to argue that we have equality between the countable unions
\/Ak and \/Bk

Regards. Noel.
• ... Ah - gotcha. Kind of smuggled in my own conclusion, didn t I. ;) Thanks. ... What is that step? I follow your proof, I m just trying to make the leap
Message 3 of 5 , Sep 3, 2001
> D being a pi-system tells you that if A' and B' are in D,
> then A'TB' is also in D. It does not tell you that if A'TB'
> is in D and B' is in D, then A' is in D... Hence I do not
> agree with your last line.

Ah - gotcha. Kind of smuggled in my own conclusion, didn't I. ;)
Thanks.

> ...
> This is a correct induction proof which establishes the equality
> between the finite union \/k_1^n Ak and \/k_1^n Bk. You need one
> little more step to argue that we have equality between the
> countable unions \/Ak and \/Bk.

What is that step? I follow your proof, I'm just trying to make the
leap from the inductive-style proofs I'm used to over to the sort you
use. If the "little more step" is to then do what you've already
done in the solution, then I suppose my induction was a waste of time.

Can I simply take a limit? If I've established the result

(\/k_1^n Ak) \ (\/k_1^n Bk) = 0

by induction, then can I argue for:

lim(n->infinity) [(\/k_1^n Ak) \ (\/k_1^n Bk)] = 0?

This seems to work, since it just becomes lim(n->infinity) [0] which
is clearly = 0.

Cheers,
Allan
• Er, on self-correction...I guess ( /k_1^n Ak) ( /k_1^n Bk) = 0 only covers ( /k_1^n Bk) = ( /k_1^n Ak) so at least two limits are required, the other being
Message 4 of 5 , Sep 3, 2001
Er, on self-correction...I guess (\/k_1^n Ak) \ (\/k_1^n Bk) = 0 only
covers

(\/k_1^n Bk) >= (\/k_1^n Ak)

so at least two limits are required, the other being

lim (n->inf) [(\/k_1^n Bk) \ (\/k_1^n Ak)] = 0

to cover (\/k_1^n Bk) <= (\/k_1^n Ak).

Showing that those two differences both = 0 (the empty set, not zero)
in the limit n->infinity seems to me to establish the equality of the
two countable unions...am I wrong? Again, I'm not certain that
taking such a limit is in fact valid; I'm just working off of the
intuitive equality of

(empty set) = lim (n->inf) (empty set)

> Can I simply take a limit? If I've established the result
>
> (\/k_1^n Ak) \ (\/k_1^n Bk) = 0
>
> by induction, then can I argue for:
>
> lim(n->infinity) [(\/k_1^n Ak) \ (\/k_1^n Bk)] = 0?
>
> This seems to work, since it just becomes lim(n->infinity) [0]
which
> is clearly = 0.
>
> Cheers,
> Allan
• ... Let x in /Ak. There exists k such that x in Ak. In particular, x in /_1^kAj= /_1^kBj. So x lies in Bj for some j. Hence x lies in /Bk. This shows the
Message 5 of 5 , Sep 4, 2001
> > You need one
> > little more step to argue that we have equality between the
> > countable unions \/Ak and \/Bk.
>
> What is that step?

Let x in \/Ak. There exists k such that x in Ak. In particular, x in
\/_1^kAj=\/_1^kBj. So x lies in Bj for some j. Hence x lies in \/Bk.
This shows the inclusion \/Ak < \/Bk . The reverse inclusion is
handled likewise.

> Can I simply take a limit?

The notion of limits for sets is not a very useful one. It is common
to handle 'liminf' and 'limsup' for sets, and even limits (when the
sequence is monotone with respect to imclusion, in which case a lim
is just a union or an intersection). In general any arguement
involvng limits of set is bound to be wrong and/or cumbersome and
unnecessary. Unions and intersections are best handled directly using
their simple definition:

x in /\Ai <=> x in Ai for all i
x in \/Ai <=> x in Ai for some i

Regards. Noel.
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