- 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,

> My proof of this differs from yours, and I'm not certain it's

in

> correct. I proceeded as follows (don't know how to really do this

> ASCII, but below, T means intersection):

Yes

>

> (A T B) in D, because D is a pi-system

> 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.

>

This is a correct induction proof which establishes the equality

> 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?

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. > D being a pi-system tells you that if A' and B' are in D,

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

> 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.

Thanks.

> ...

What is that step? I follow your proof, I'm just trying to make the

> 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.

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

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

which

>

> (\/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]

> is clearly = 0.

>

> Cheers,

> Allan > > You need one

Let x in \/Ak. There exists k such that x in Ak. In particular, x in

> > little more step to argue that we have equality between the

> > countable unions \/Ak and \/Bk.

>

> What is that step?

\/_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.