Distrubutive Property of Limit Points

[Narcoleptic Insomniac]

Let A and B be two subsets and let A' denote the set of
all limit points of the set A. I'm wondering if the
following two properties hold:

1) (A \/ B)' = A' \/ B' and

2) (A /\ B)' = A' /\ B'.

The definition of a "limit point" (or accumulation point)
that I'm given is that a point x in A is said to be a
limit point of A if every neighborhood of x contains at
least one point in A other than x itself.

I'm pretty sure that both properties hold, since I think
that I've done inclusion and reverse inclusion properly
for both (1) and (2).

Instead of posting my proofs for both of these, I'd be
grateful if somebody would just give me a yes or no. Then
I can go back and check my work if equality should fail
to hold.

Thanks in Advance,
Kyle

P.S. No, this is not homework; although it is related to
one of my courses.

[Dave L. Renfro]

Narcoleptic Insomniac wrote (in part):

> Let A and B be two subsets and let A' denote the set of
> all limit points of the set A. I'm wondering if the
> following two properties hold:
>
> 1) (A \/ B)' = A' \/ B' and
>
> 2) (A /\ B)' = A' /\ B'.

In the case of (2), inclusion holds only one way.
You can use the rational and irrational numbers
(in R, with the usual topology) to find out which
direction doesn't hold in (2). You might also find
it instructive to go through your proof of (2) and
see what goes wrong in the specific case of the
rational and irrational numbers.

Dave L. Renfro

[W. Dale Hall]

Here's a simpler example:

let A,B be the open intervals:

A = (-1, 0), B = (0, 1)

Dale.

[Narcoleptic Insomniac]

I asked my professor right after I posted this and the
counter-example with the rationals and irrationals was
precisely the argument he gave (assuming the usual topology).

If Q is the set of rationals and R - Q is the set of
irrationals then (1) holds because Q \/ (R - Q) = R.
(1) also holds for the example Dale since A' = [-1, 0],
B' = [0, 1], and (A \/ B)' = [-1, 1].

However, my professor said that's it might be possible
for the equality of (1) to fail, although he wasn't sure
off hand. Either way, like Dave said, it would be a
good idea for me to go back and see why (2) fails using
these examples. Thanks agains guys ^_^

Regards,
Kyle

[Dave L. Renfro]

W. Dale Hall wrote:

>> Here's a simpler example:
>>
>> let A,B be the open intervals:
>>
>> A = (-1, 0), B = (0, 1)

Yep, that's simpler than my example of using
the rationals and irrationals!

Narcoleptic Insomniac wrote:

> I asked my professor right after I posted this and the
> counter-example with the rationals and irrationals was
> precisely the argument he gave (assuming the usual topology).
>
> If Q is the set of rationals and R - Q is the set of
> irrationals then (1) holds because Q \/ (R - Q) = R.
> (1) also holds for the example Dale since A' = [-1, 0],
> B' = [0, 1], and (A \/ B)' = [-1, 1].
>
> However, my professor said that's it might be possible
> for the equality of (1) to fail, although he wasn't sure
> off hand. Either way, like Dave said, it would be a
> good idea for me to go back and see why (2) fails using
> these examples. Thanks agains guys ^_^

In general, the following two results hold in any
topological space:

Let I be a nonempty index set.

1. union(i in I) of (A_i)' subset [ union(i in I) of A_i]'

2. inter(i in I) of (A_i)' superset [ inter(i in I) of A_i]'

Equality holds in #1 when I is finite, but #2 can fail
even when card(I) = 2.

Something you might find interesting to play around with
is to find subsets E of [0,1] verifying the following:

3. There exists a subset E of [0,1] such that
E, E', and E'' are pairwise different.

4. There exists a subset E of [0,1] such that
E, E', E'', and E''' are pairwise different.

5. For each positive integer n, there exists a subset E
of [0,1] such that E, E', E'', ..., <the n'th derived
set of E> are pairwise different.

6. There exists a subset E of [0,1] such that all finite
iterations of the derived set operation applied to E
are pairwise different.

For what it's worth, it was Cantor's realization that
certain countable sets can have these properties, while
he was proving various theorems about the uniqueness of
the coefficients of trigonometric expansions of functions,
that led Cantor to become involved with the ideas that
brought about the creation of set theory. I don't have
time to look up some references right now (let me know
if you'd like me to, sometime in the next few days) other
than to mention Alexander S. Kechris' 1997 (unpublished?)
manuscript "Set theory and uniqueness for trigonometric
series", which you can find at his web page

http://www.math.caltech.edu/people/kechris.html

Dave L. Renfro

[Narcoleptic Insomniac]

On Oct 10, 2005 7:00 PM CT, Dave L. Renfro wrote:

[...]

What is meant by "pairwise different"; the differences
between the various sets comes in pairs?

Regards,
Kyle

[Dave L. Renfro]

Narcoleptic Insomniac wrote:

> What is meant by "pairwise different";
> the differences between the various
> sets comes in pairs?

By "pairwise different", I mean that no
two of the sets are equal to each other.
Thus, to say that E, E', and E'' are pairwise
different means that E is different from E',
E' is different from E'', and E is different
from E''. You'll often see this described as
"the sets are all different", but to me the
phrase "pairwise different" or "pairwise distinct"
is more precise and less ambiguous.

Dave L. Renfro