• #### Cocompact

From William Elliot@21:1/5 to All on Sun Sep 6 05:50:26 2015
Let S_c be the cocompact space of S.  S_c is S with the topology
{ empty set, S\K | K compact closed within S }.

If K is compact within S_c, is K closed within S_c?

If K is compact within S_c, is K closed within S?

If not, would either hold, if S is assumed to be
locally compact or locally compact Hausdorff?

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• From David Cullen@21:1/5 to William Elliot on Fri Apr 29 06:17:51 2016
On Sunday, September 6, 2015 at 5:50:30 AM UTC-6, William Elliot wrote:

Let S_c be the cocompact space of S.  S_c is S with the topology
{ empty set, S\K | K compact closed within S }.

If K is compact within S_c, is K closed within S_c?

If K is compact within S_c, is K closed within S?

If not, would either hold, if S is assumed to be
locally compact or locally compact Hausdorff?

Hi William,

Counterexample 1:
Consider the discrete space S = {0,1} with the Sierpinski topology {
{}, {0}, {0,1} }.
Since S is discrete, all subsets are compact, so the closed compact
sets are { {}, {1}, {0,1} }.
Moreover, S is trivially a locally compact space.
The S_c-open sets are the members of the original topology, and S = S_c. However, the set {0} is S_c-compact but is neither S-closed nor
S_c-closed.

Counterexample 2:
Consider the space of real numbers R with the usual topology.
R is a locally compact Hausdorff space. For a set X contained in R,
write X' to mean R\X.
Any open set in R can be written as a the union over a countable set of
open intervals I.
Using DeMorgan's law:
Any closed set can be written as an intersection over a countable set
of complements of open intervals I.
By Heine-Borel, any compact set can be written as the intersection of a
closed symmetric interval B
with an intersection over a countable set of complements of open
intervals I.
Thus, again by DeMorgan's law:
an R_c-open set is one that is the union of the complement of a closed symmetric interval B
with the union over a countable set of open intervals I: B' \/
\/(i=1 to infininty)(I_i).
Consider the interval [0, infinity), and let C be an R_c-open cover of
this interval.
C can be written as {(B_j)' \/ \/(i= 1 to infinity)(I_ij)} for j in
some index set J.
Let y be some element of J, and suppose B_y is the interval [-r, r] for
some real r > 0.
D := { I_ky | k is a natural number} union with { (B_j)' \/ \/(i= 1 to infinity)(I_ij) | j in J\{y} }is an R-open cover of [0, r].
[0,r] is an R-compact set.Thus, D admits a finite subcover:
E = {I_ky | k is in some finite subset of the natural numbers} union
with
{ (B_j)' \/ \/(i = i to infinity)(I_ij) | j is in some finite subset K
of J\{y} }.
Thus { (B_y)' \/ \/(i= 1 to infinity)(I_iy) } union { (B_j)' \/ \/(i =
i to infinity)(I_ij) | j in K }
is a finite R_c-open subcover of C, hence [0, infinity) is R_c-compact.
[0, infinity) is not R_c-closed, or else (-infinity, 0) would be
R_c-open, and so [0, infinity) would need to be R-compact,
which it is not.  Therefore not every S_c-compact set in an S-(locally
compact) S-Hausdorff space is S_c-closed.

We do however have the following
Theorem:
Every S_c-compact set in an S-(locally compact) S-Hausdorff space is
S-closed.
Proof:
Recall that one of the equivalent definitions of local compactness for Hausdorff spaces is that every point has a local
base consisting of compact neighborhoods.  From this, the Hausdorff
property implies that any two points are
separable by disjoint compact neighborhoods. Also note that compact
subsets of Hausdorff spaces are closed.
Let K be an S_c-compact set, and let x be in S\K.
For each k in K there exist disjoint, S-compact neighborhoods X_k of x
and Y_k of k.
{ (X_k)' | k in K } is then an S_c-open cover of K,
so admits a finite refinement { (X_k_1)', ..., (X_k_m)' } for some
natural m.
Now, /\(i = 1 to m)(S-interior(X_k_i)) is an S-open set containing x
which is disjoint from K,
so S\K is S-open, and thus K is S-closed.

Please check my work and let me know what you think!

David

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• From William Elliot@21:1/5 to David Cullen on Sat Apr 30 07:47:57 2016
On Fri, 29 Apr 2016, David Cullen wrote:

On Sunday, September 6, 2015 at 5:50:30 AM UTC-6, William Elliot wrote:

Let S_c be the cocompact space of S.  S_c is S with the topology
{ empty set, S\K | K compact closed within S }.

If K is compact within S_c, is K closed within S_c?

If K is compact within S_c, is K closed within S?

If not, would either hold, if S is assumed to be
locally compact or locally compact Hausdorff?

Counterexample 1:
Consider the discrete space S = {0,1} with the Sierpinski topology {
{}, {0}, {0,1} }.

The two point Sierpinski space Sr, is not discrete
because {1} is not open.

Since S is discrete, all subsets are compact, so the closed compact
sets are { {}, {1}, {0,1} }.

All subsets of a finite space are compact.

Moreover, S is trivially a locally compact space.

All finite spaces are locally compact.

The S_c-open sets are the members of the original topology, and S = S_c. However, the set {0} is S_c-compact but is neither S-closed nor
S_c-closed.

Yes, Sr is a counter example for all the conjectures except for locally
compact Hausdroff spaces.  Discrete S is no counter example.

Counterexample 2:
Consider the space of real numbers R with the usual topology.
R is a locally compact Hausdorff space. For a set X contained in R,
write X' to mean R\X.
Any open set in R can be written as a the union over a countable set of
open intervals I.
Using DeMorgan's law:
Any closed set can be written as an intersection over a countable set
of complements of open intervals I.
By Heine-Borel, any compact set can be written as the intersection of a closed symmetric interval B
with an intersection over a countable set of complements of open
intervals I.
Thus, again by DeMorgan's law:
an R_c-open set is one that is the union of the complement of a closed symmetric interval B
with the union over a countable set of open intervals I: B' \/
\/(i=1 to infininty)(I_i).
Consider the interval [0, infinity), and let C be an R_c-open cover of
this interval.
C can be written as {(B_j)' \/ \/(i= 1 to infinity)(I_ij)} for j in
some index set J.
Let y be some element of J, and suppose B_y is the interval [-r, r] for
some real r > 0.
D := { I_ky | k is a natural number} union with { (B_j)' \/ \/(i= 1 to infinity)(I_ij) | j in J\{y} }is an R-open cover of [0, r].
[0,r] is an R-compact set.Thus, D admits a finite subcover:
E = {I_ky | k is in some finite subset of the natural numbers} union
with
{ (B_j)' \/ \/(i = i to infinity)(I_ij) | j is in some finite subset K
of J\{y} }.
Thus { (B_y)' \/ \/(i= 1 to infinity)(I_iy) } union { (B_j)' \/ \/(i =
i to infinity)(I_ij) | j in K }
is a finite R_c-open subcover of C, hence [0, infinity) is R_c-compact.
[0, infinity) is not R_c-closed, or else (-infinity, 0) would be
R_c-open, and so [0, infinity) would need to be R-compact,
which it is not.  Therefore not every S_c-compact set in an S-(locally compact) S-Hausdorff space is S_c-closed.

Assume C is a R_c.open cover of A = [0,oo)
Pick some U in C.  There's some R.compact K with U = R\K.
C is a R.open cover of R.compact [0,oo) /\ K.  (R_c is coarser than R.)
Thus some finite subcover Cf of [0,oo) /\ K.
Whence Cf \/ {U} is a finite R_c.subcover of C.

Ok, that's a locally compact Hausdorff counter example
for the first conjecture.

We do however have the following
Theorem:
Every S_c-compact set in an S-(locally compact) S-Hausdorff space is S-closed.
Proof:
Recall that one of the equivalent definitions of local compactness for Hausdorff spaces is that every point has a local
base consisting of compact neighborhoods.  From this, the Hausdorff
property implies that any two points are
separable by disjoint compact neighborhoods. Also note that compact
subsets of Hausdorff spaces are closed.
Let K be an S_c-compact set, and let x be in S\K.
For each k in K there exist disjoint, S-compact neighborhoods X_k of x
and Y_k of k.
{ (X_k)' | k in K } is then an S_c-open cover of K,
so admits a finite refinement { (X_k_1)', ..., (X_k_m)' } for some
natural m.

Now, /\(i = 1 to m)(S-interior(X_k_i)) is an S-open set containing x
which is disjoint from K,
so S\K is S-open, and thus K is S-closed.

Assume K is S_c compact. x not in K.
For all k in K, there's some S.open U_k,V_k
with compact closures, that completely separate x,k.
C = { S - cl U_k | k in K } is S_c cover of K.
Thus some finite subcover Cf = { S - cl U1,.. S - cl Uj } covers K.
U = \/Cf is S.open;  x in S.open V = /\{ V1,.. Vj } where each Vk is
paired
with the Uk as described in the second line.
Thus S.open U,V separate x and K.

In conclusion, for all x not in K, x not in cl_S K, ie
cl K subset K, whence K is S.closed

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