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From: William Elliot on 13 Jul 2010 01:25
On Tue, 13 Jul 2010, Gerry Myerson wrote:
> Edward Green <spamspamspam3(a)netzero.com> wrote:
>
>> Just how hard is it to prove De Morgan's laws in the case of infinite
>> or even uncountable index sets?
>
> Not very.
> E.g., if x is in the complement of the intersection of a family of sets,
> then there's at least one set it's not in, so it's in the union of the
> complements.
>
DeMorgan's rules of complementation are a direct result of
the logical rules of negation. Notice in this proof where
logical duality is used.

x in S - /\C iff x not in /\C iff not for all U in C, x in U
iff some U in C with x not in U iff some U in C with x in S\U
iff x in \/{ S\U | U in C }.

Topological duality S - int A = cl S\A is a similar
tour de force again using logical rules of negation.

x in S - int A iff x not in int A
iff not some open U nhood x with U subset A
iff for all open U nhood x, U not subset A
iff for all open U nhood x, not for all u in U, u in A
iff for all open U nhood x, some u in U with u not in A
iff for all open U nhood x, some u in U with u in S\A
iff for all open U nhood x, U /\ S\A not empty
iff x in cl S\A

for all A, S - int A = cl S\A (1)

The dual statement
for all A, S - cl S = int S\A

is immediate from (1) by setting A to S\A
and taking complements of both sides.

It's easy to directly prove
for all A,B, int A/\B = int A /\ int B. (2)

It's not easy to directly prove the dual statement
for all A,B, cl A\/B = cl A \/ cl B.

The easy way is to use the topological duality of interior
and closure as shown above. So as before set A to S\A, B
to S\B and take complements of both sides of (2). Notice
the multiple use of the complementation rules.

----
From: Daniel Giaimo on 13 Jul 2010 15:42
On 7/12/2010 6:42 PM, Edward Green wrote:
> Many thanks to all who replied. If I still can't quite manage the
> proof, no doubt I have only my own debility to blame.
>
> On a note of clarification, yes, it was the finite (actually pairwise)
> case considered. Also, the problem appeared after we had barely
> finished defining topologies, interiors and closures, so I don't think
> any more advanced concepts, like metrics, are allowed (did I see some
> metric creeping into some of the responses? I'm not sure).

Where are you still having trouble? Both inclusions follow almost
immediately from the definition of closure. I'll provide the full proof
below some spoiler space in case you still want to keep thinking about it.

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All right. We wish to prove that cl(X U Y) = cl(X) U cl(Y). First note
that X U Y is clearly contained in cl(X) U cl(Y). Since the latter is a
closed set, this means that cl(X U Y) is contained in cl(X) U cl(Y).
Now note that X is contained in cl(X U Y). Since the latter is a closed
set, cl(X) is contained in cl(X U Y). Similarly, cl(Y) is contained in
cl(X U Y). Therefore cl(X) U cl(Y) is contained in cl(X U Y).
Therefore they are equal. QED

--
Dan G
From: William Elliot on 15 Jul 2010 01:43
On Tue, 13 Jul 2010, Daniel Giaimo wrote:
>
> We wish to prove that cl(X U Y) = cl(X) U cl(Y). First note that X U Y
> is clearly contained in cl(X) U cl(Y). Since the latter is a closed
> set, this means that cl(X U Y) is contained in cl(X) U cl(Y). Now note
> that X is contained in cl(X U Y). Since the latter is a closed set,
> cl(X) is contained in cl(X U Y). Similarly, cl(Y) is contained in cl(X
> U Y). Therefore cl(X) U cl(Y) is contained in cl(X U Y). Therefore they
> are equal. QED

Nice.
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