• #### 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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