The closure of the union equals the union of the closures
Edward Green
stated as a theorem to be proved shortly after defining the closure of
a set, and I've been beating my head against the silly thing. Can
anybody give me a hint in the right direction? Just how should I
approach such a problem?
Daniel Giaimo
This is quite clearly false as stated. For an easy counterexample, take
any topological space in which points are closed. Then every set is the
union of its singleton subsets all of which are closed.
--
Dan G
Gerry
Suppose x is in the closure of the union. Then there's a sequence
of points in the union converging to x. That sequence has
a subsequence consisting entirely of points in one of the original
sets.
--
GM
christian.bau
How do you approach it: There is a family F of sets, and you are to
prove that the closure C of the union of all sets in F is equal to the
union U of the closures of all the sets in F. You need to prove C = U.
So assume that x is in C and prove that it is in U. Then assume that x
is in U and prove that it is in C. With both directions proved you
proved C = U.
BTW. I think it depends on whether F is finite or not. Consider sets
of real numbers. Let Sn = { 1 / n } for all integers n >= 1. Is 0 in
C? Is 0 in U?
Rob Johnson
If you are talking about a finite union, then it is true. In this
case, Gerry Myerson gives an excellent hint.
If you are talking about an infinite union, then it is false. In
this case, consider the union of all single rational points in the
reals. The closure of the union is the reals, while the union of
the closures is the rationals.
Rob Johnson <r...@trash.whim.org>
take out the trash before replying
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Arturo Magidin
This is quite clearly false. Suppose your sets are all finite, and
your sequence is never constant. How are you going to get a sequence
of points in a single set that converges to the point?
For an explicit example, take the union of {x} over all x in (0,1).
The closure of the union is [0,1]. An easy sequence of points in the
union converging to 0 is 1/n. Where is your subsequence that consists
entirely of points in one of the sets {x} and which converges to 0?
--
Arturo Magidin
William Elliot
> The closure of the union equals the union of the closures: this is
> stated as a theorem to be proved shortly after defining the closure of
> a set, and I've been beating my head against the silly thing.
That is false except for finite unions.
Doesn't the author make that clear?
For finite unions, just prove for all A,B,
cl A\/B = cl A \/ cl B. (1)
That is the dual result of the easy theorem, for all A,B,
int A/\B = int A /\ int B. (2)
and DeMorgan's rules of complementation for topological sets.
Those are usual rules for sets and the theorem,
cl S\A = S - int A
where S is the space and int A is A^o, the interior of A.
(1) is the topological dual statement of (2) and can
be immediately derived from (2) by taking complements.
> Can anybody give me a hint in the right direction?
> Just how should I approach such a problem?
First prove int A/\B = int A /\ int B.
Gerry
I was reading the question as, closure of A-union-B equals
(closure of A) union (closure of B). OP said it was stated
as a theorem, so it seems reasonable to assume that it was
only ever meant to apply to two (or, at any rate, finitely
many) sets.
--
GM
Edward Green
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).
Anyway, I'd like to ask a follow up question. I'm thinking there might
be a role for De Morgan's theorems somewhere, what with all the
complements flying around with this business of closed and open sets
(well, William Eliot said as much, I see). Just how hard is it to
prove De Morgan's laws in the case of infinite or even uncountable
index sets? Or perhaps that is not needed, but I'd like to know just
the same.
Gerry Myerson
<89a1177b-c1f3-46b8...@w30g2000yqw.googlegroups.com>,
Edward Green <spamsp...@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.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
William Elliot
> Edward Green <spamsp...@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.
>
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.
----
Daniel Giaimo
> 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
William Elliot
>
> 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.