Re: Family of Sets - finer than in exterior sense

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arkay_ca - 25 Aug 2006 18:27 GMT

> I have couple of questions on this notion.
>
> A family of sets, say AF, is considered partially
> contained in another set X, if A(i) is subset of X.
>
> A family of Sets, AF (i belongs to I)is partially
> contained in another family of Sets BF(j belongs to
> J), if
>
> all A(i) is subset of a X(j).
>
> this is definition of finer than in exterior sense.
>
> Question is : Is this terminology still used ?
>
> Please confirm if my observation is correct:
>
> If AF is finer than BF in exterior sense and vice
> versa, there should be atleast one Ai equal to one
> Bj.
>
> Thanks.

Correction to above def:

finer than BF in exterior sense definition should read:
if AF is partially contained in every set B(j) where j belongs to J.

ie/ Some A(i) is subset of B(j) for all j.

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Paul Sperry - 26 Aug 2006 04:31 GMT

In article
<6107856.1156526876320.JavaMail.jakarta@nitrogen.mathforum.org>,

> > I have couple of questions on this notion.
> >
[quoted text clipped - 10 lines]
> >
> > Question is : Is this terminology still used ?

It is new to me.

> > Please confirm if my observation is correct:
> >
> > If AF is finer than BF in exterior sense and vice
> > versa, there should be atleast one Ai equal to one
> > Bj.

Let A(n) = { 1, 2, ... , 2n - 1} and B(n) = {1, 2, ... , 2n} for
n = 1, 2, ...

A(n) < B(n), B(n) < A(n + 1) but no B(n) is equal to any A(m). I hope
that I am understanding your definitions properly.

> Correction to above def:
>
> finer than BF in exterior sense definition should read:
> if AF is partially contained in every set B(j) where j belongs to J.
>
> ie/ Some A(i) is subset of B(j) for all j.

Signature

Paul Sperry
Columbia, SC (USA)

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Brian M. Scott - 26 Aug 2006 05:56 GMT

On Fri, 25 Aug 2006 13:27:26 EDT, arkay_ca
<rama.kunapuli@hotmail.com> wrote in
<news:6107856.1156526876320.JavaMail.jakarta@nitrogen.mathforum.org>
in alt.math.undergrad:

>> I have couple of questions on this notion.

>> A family of sets, say AF, is considered partially
>> contained in another set X, if A(i) is subset of X.

This isn't clear. Do you mean 'if there is at least one i
such that A(i) is a subset of X'? In other words, if F is a
family of sets, and X is a set, then F is almost contained
in X if at least one member of F is a subset of X?

>> A family of Sets, AF (i belongs to I)is partially
>> contained in another family of Sets BF(j belongs to
>> J), if

>> all A(i) is subset of a X(j).

>> this is definition of finer than in exterior sense.

[moved for greater clarity:]

> Correction to above def:

> finer than BF in exterior sense definition should read:
> if AF is partially contained in every set B(j) where j belongs to J.

> ie/ Some A(i) is subset of B(j) for all j.

>> Question is : Is this terminology still used ?

I've not encountered it.

>> Please confirm if my observation is correct:

>> If AF is finer than BF in exterior sense and vice
>> versa, there should be atleast one Ai equal to one
>> Bj.

If I understand your definition correctly, the hypothesis is
this:

(1) for each B in BF there is an A in AF such that A
is a subset of B, and
(2) for each A in AF there is a B in BF such that B
is a subset of A.

Your conjecture is false if the families are allowed to be
infinite: for each n >= 0 let A(n) = {k in N : k >= 2n}, and
let B(n) = {k in N : k >= 2n + 1}. Clearly no A(n) is equal
to any B(m), but for all n B(n) is a subset of A(n), and
A(n+1) is a subset of B(n).

The conjecture is true if at least one of the families is
finite. Suppose that AF is finite. Then we can recursively
construct a sequence of sets <X(n) : n in N> such that

(1) X(n) is in AF when n is even,
(2) X(n) is in BF when n is odd, and
(3) for every n in N, X(n+1) is a subset of X(n).

Since AF is finite, there are only finitely many different
choices for the sets X(2k), so there must be natural numbers
n and m such that n < m and X(2n) = X(2m). But then X(2m)
is a subset of X(2n+1), which is a subset of X(2n), so X(2n)
= X(2n+1), where of course X(2n) is in AF and X(2n+1) is in
BF.

Brian

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arkay_ca - 28 Aug 2006 19:38 GMT

I have encountered this definition in Berge’s Topological Spaces. He uses two notions: One finer than in exterior sense and other is finer than in interior sense. He uses these to define sub-partitions, sub-base, sub-coverings etc.,

Finer than exterior sense (say FTES) is defined:

AF (FTES) BF means

There is at least one Ai, it is true that, Ai is subset of Bj for each member of Bj

Example: AF={A1,A2,A3}
    BF = { B1, B2,B3,B4}

    A2 is subset of B1
    A2 is subset of B2
    A2 is subset of B3
    A2 is subset of B4

Similarly

BF ( FTES) AF implies

    Say we use B3
    B3 is subset of A1
    B3 is subset of A2
    B3 is subset of A3

There is one B3 subset of A2 and A2 is subset of B3. Hence A2=B3.

But the book doesn’t contain any such assertion. Always wondered…

Brian, your counter example intrigues me. How does one generate such examples?

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Brian M. Scott - 29 Aug 2006 01:34 GMT

On Mon, 28 Aug 2006 14:38:07 EDT, arkay_ca
<rama.kunapuli@hotmail.com> wrote in
<news:9662881.1156790317782.JavaMail.jakarta@nitrogen.mathforum.org>
in alt.math.undergrad:

> I have encountered this definition in Bergeʼs Topological
> Spaces. He uses two notions: One finer than in exterior
> sense and other is finer than in interior sense. He uses
> these to define sub-partitions, sub-base, sub-coverings
> etc.,

> Finer than exterior sense (say FTES) is defined:

> AF (FTES) BF means

> There is at least one Ai, it is true that, Ai is subset of
> Bj for each member of Bj

> Example: AF={A1,A2,A3}
> BF = { B1, B2,B3,B4}

>     A2 is subset of B1
>     A2 is subset of B2
>     A2 is subset of B3
>     A2 is subset of B4

This is different from the definition that I thought you
were giving before. This is:

(1) There is an A in AF such that for each B in BF,
A is a subset of B.

Equivalently:

(1') There is an A in AF that is a subset of the
intersection of the family BF.

What I thought you meant was a weaker statement:

(2) For each B in BF there is an A in AF such that
A is a subset of B.

Here the choice of A may depend on the choice of B; in the
first version there is one A that works for all of the B.

Are you sure that Berge's definition is (1) and not (2)? I
can see (2) being useful in a topological context, but (1)
really doesn't look very useful to me. (And my example and
proof last time were based on (2), not (1).)

[...]

> Brian, your counter example intrigues me. How does one
> generate such examples?

I don't know that I can give you a very useful answer.
Certainly a lot of it is experience -- I took my first
topology course, for instance, over 40 years ago -- but that
isn't the whole story: even amongst experienced
mathematicians you'll find that some are more adept than
others at constructing counterexamples. In this case, to be
honest, the example more or less leaped out at me: I knew
that I wanted an infinite decreasing chain of sets, and it
only remained to find one that was easy to describe.

Brian

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Math Forum / Mathematics / Undergraduate Math / August 2006

Re: Family of Sets - finer than in exterior sense