Compactness and families of closed subsets having nonempty intersections
sto
I'm trying to prove that K is compact if every family of sets {F_i} such
that
1) each F_i is a closed subset of K
2) the intersection of any finite subfamily of {F_i} is nonempty
has also the property that the intersection the entire family {F_i} is
nonempty.
I've already tried first assuming that K is not compact and then,
starting with an open cover that has no finite subcover, constructing a
family of sets that satisfies these properties, yet has an empty
intersection. But the complement of an open cover does not yield closed
*subsets* of K. I've also tried starting with an open cover of K and
trying to construct a finite subcover by using a family of sets
satifying all the properties, but this was a total dead end.
THanks
-sto
William Elliot
> Let M be a metric space and K subset M.
> I'm trying to prove that K is compact if every family of sets {F_i} such
> that
> 1) each F_i is a closed subset of K
> 2) the intersection of any finite subfamily of {F_i} is nonempty
>
> has also the property that the intersection the entire family {F_i} is
> nonempty.
>
That proposition, a topological dual definition of compact,
is true not just for metric spaces, but all topological spaces.
> I've already tried first assuming that K is not compact and then, starting
> with an open cover that has no finite subcover, constructing a family of sets
> that satisfies these properties, yet has an empty intersection. But the
> complement of an open cover does not yield closed *subsets* of K. I've also
> tried starting with an open cover of K and trying to construct a finite
> subcover by using a family of sets satifying all the properties, but this was
> a total dead end.
>
Start by assuming the intersection is empty.
David Hartley
<s...@address.invalid> writes
>But the complement of an open cover does not yield closed *subsets* of
>K.
--
David Hartley
sto
{F_i} is a family of closed subsets of M each of which is contained in
K. If I take the intersection of {F_i} with K, what I get is a family
of subsets of K that are closed in K, but not necessarily closed in M.
That's where the whole proof breaks down.
sto
> On Fri, 17 Jun 2011, sto wrote:
>
>> Let M be a metric space and K subset M.
>
>> I'm trying to prove that K is compact if every family of sets {F_i}
>> such that
>> 1) each F_i is a closed subset of K
in K
>> 2) the intersection of any finite subfamily of {F_i} is nonempty
>>
>> has also the property that the intersection the entire family {F_i} is
>> nonempty.
>>
> That proposition, a topological dual definition of compact,
> is true not just for metric spaces, but all topological spaces.
>
>> I've already tried first assuming that K is not compact and then,
>> starting with an open cover that has no finite subcover, constructing
>> a family of sets that satisfies these properties, yet has an empty
>> intersection. But the complement of an open cover does not yield
>> closed *subsets* of K. I've also tried starting with an open cover of
>> K and trying to construct a finite subcover by using a family of sets
>> satifying all the properties, but this was a total dead end.
>>
> Start by assuming the intersection is empty.
Assume that every family {F_i} of closed subsets of M which has the
finite intersection property has a nonempty intersection. Then if {H_i}
is a family of closed subsets of M which has an empty intersection,
there exists a finite subfamily {H_1, ... , H_n} having an empty
intersection.
Now if {G_i} is any open cover of M then, by de Morgan, its complement
{G'_i} is a family of closed subsets of M whose intersection is empty.
It follows by assumption that there exists a finite subfamily {G'_1, ...
, G'_n} having an empty intersection, in which case {G_1, ... , G_n} is
a finite subcover of M.
but I don't see how this would work in the case that K subset M because
in that case the complement of an open cover of K would not be a family
of closed sets in M that are contained in K.
Butch Malahide
> On 6/18/11 4:25 AM, David Hartley wrote:> In message <hN6dnbSieIicgWHQnZ2dnUVZ_o2dn...@earthlink.com>, sto
> > <s...@address.invalid> writes
> >> But the complement of an open cover does not yield closed *subsets* of K.
> > But if you then take the intersections with K ...
>
> The statement of the proposition, as I understand it, stipulates that
> {F_i} is a family of closed subsets of M each of which is contained in
> K.
Your understanding is wrong. For the proposition to be correct,
"closed subset of K" should be understood as "subset of K which is
closed in the relative topology of K."
A family of sets is said to have the finite intersection property if
every finite subfamily has a nonempty intersection. Suppose M is a
compact metric space; then every family of closed subsets of M which
has the finite intersection property, has nonempty intersection. In
particular, if K is any subset of M, then every family of closed
subsets of M (each of which happens to be contained in K) which has
the finite intersection property has nonempty intersection. Thus the
proposition, as you understand it, would imply that every subset of a
compact metric space is compact. But this is blatantly false, e.g.,
consider M = [0,1] and K = [0,1).
William Elliot
> On 6/18/11 2:01 AM, William Elliot wrote:
>> On Fri, 17 Jun 2011, sto wrote:
>>
>>> Let M be a metric space and K subset M.
>>
>>> I'm trying to prove that K is compact if every family of sets {F_i}
>>> such that
>>> 1) each F_i is a closed subset of K
> I take this to mean that each F_i is a closed set in M that is contained
> in K
That also means that each F_i is a closed set within the subspace K.
>>> 2) the intersection of any finite subfamily of {F_i} is nonempty
>>>
>>> has also the property that the intersection the entire family {F_i} is
>>> nonempty.
>> Start by assuming the intersection is empty.
> I can see how this works just fine in the particular case that K=M:
>
> Assume that every family {F_i} of closed subsets of M which has the finite
> intersection property has a nonempty intersection. Then if {H_i} is a family
> of closed subsets of M which has an empty intersection, there exists a finite
> subfamily {H_1, ... , H_n} having an empty intersection.
>
> Now if {G_i} is any open cover of M then, by de Morgan, its complement {G'_i}
> is a family of closed subsets of M whose intersection is empty. It follows by
> assumption that there exists a finite subfamily {G'_1, ... , G'_n} having an
> empty intersection, in which case {G_1, ... , G_n} is a finite subcover of M.
>
> but I don't see how this would work in the case that K subset M because in
> that case the complement of an open cover of K would not be a family of
> closed sets in M that are contained in K.
>
The complest of an open cover is the set of open sets not in the cover.
The cmplements of an open cover is the set of the complements the sets
in the cover.
A subset K of a space S, is compact iff the subspace S|K is compact.
Assume /\_j Fj = nulset.
Then \/_j M\Fj = M
\/_j K\Fj = K
for all j, K\Fj open within K, ie the subspace M|K.
some F1,.. F_k with K\F1 \/..\/ K\F_k = K
F1 /\../\ F_k = nulset, a contradiction because
the F's have fip, finite intersection property.
Alternatively, staying within M:
{ M\Fj | j in I (the index set) } covers M
{ M\Fj | j in I } covers K
some M\F1,.. M\F_k cover K
K subset M\F1 \/..\/ M\F_k
K = K\F1 \/..\/ K\F_k
nulset = F1 /\../\ F_k and the F's fip out.