An example of a space which has a closed, H-bounded subset which is not contained in any H-set is presented in this section.
A good way to visualize this space is to consider the space in
Example 1.9,
but with the topology of the point at infinity changed to represent a free
ultrafilter on 
instead of a cofinite filter. Our space then is 
copies of this space (each with its point at infinity corresponding to a
different free ultrafilter) with their sets identified.
To simplify the discussion, each subset of the form
will be called a plank and referred to
as the 
plank. u is the point at infinity for the plank. 
is the set of all points at infinity. Within a plank, a set of the form
will be called a column and be referred
to as the 
column of the plank. Thus a basic open neighborhood of
n consists of n along with all but finitely many points of the column
of each plank. In the context of the space X, the symbol will always refer to the copy of which is not part of any plank or .
Proof: It is easy to see that is closed, for every point in 
has a basic open neighborhood that misses .
We claim that is H-bounded and hence countably
H-bounded. To see this begin by
considering an arbitrary
open cover of by basic open sets in X (i.e., by sets of the
form
where 
). Within the space 
,
which implies that 
which is clopen. Then 
is an open cover
of the compact space . There exists 
such that
Thus 
covers all but finitely many
points of . But 
so 
covers all but finitely many points of .
Returning again to the space X, notice that
covers all but finitely many points of . Thus every open cover of X
has a finite subcollection whose closures cover .
Proof. Suppose B is an infinite H-set. Consider covers consisting of
basic open sets. First consider the intersection of B with . No
point of 
is in the closure of a basic neighborhood of any point
of 
. Also if C is a basic open neighborhood of
then u is the only element of in cl(C) (because
the neighborhoods of u ``live" only on the plank). Thus is
finite.
Next consider the intersection of B with a single plank, say the
plank. The closures of basic open neighborhoods of points in other planks are
just the points themselves. The closures of basic neighborhoods of points at
infinity of other planks do not pick up any points in any other plank (only
points in ). Thus only basic neighborhoods of points in
and the point u have closures which contribute to covering B.
If B meets an infinite
number of columns, then there is a set such that
misses an infinite number of the
columns that B meets. Since B is an H-set, only a finite number of
neighborhoods of points in can help cover what the neighborhood of u
did not cover. But each basic neighborhood of a point in covers only
part of a single column in a given plank. Thus B can meet only finitely many
columns of any given plank.
The set B can meet only a finite number of planks. To see this, suppose B meets an
infinite number of planks. It was just shown that B meets each plank in a
finite number of columns and that one can always choose neighborhoods of the
points in to miss these columns. Thus all but finitely many of
these points must be picked up by neighborhoods of points in
. So not only does B meet each plank in a finite number of columns,
it meets each plank in the same finite number of columns. Suppose some column,
say the column, is met by B in an infinite number of planks, say 
for some infinite cardinal 
.
may be chosen so that 
misses at least one point in B in the
column. Defining 
arbitrarily on the remaining planks gives an open
neighborhood of n that misses an infinite number of points of B, only
finitely many of which can be picked up in some other way by the closures of a
finite number of basic open sets. Therefore B can meet each column on only a
finite number of planks, and since B meets only finitely many columns, it
follows that B meets only finitely many planks.
Finally if B meets in an infinite set, then since
is finite, choose basic open neighborhoods of the points in
whose closures miss an infinite number of points in
. The closures of basic neighborhoods of points in the planks
miss and the closures of basic neighborhoods of points in
pick up only one point in . Since B is an H-set, it is possible to
use only neighborhoods of a finite number of points of 
which then
leaves infinitely many points of B uncovered. Therefore B must meet
in a finite set.
Proof: The proof of this proposition is similar to the
proof of Proposition 2.3, but with
the necessary modifications to work with countable covers. To simplify
notation, basic open neighborhoods of points in will be denoted as
where 
and 
.
Sometimes 
will be used to indicate any basic open neighborhood of u if
the set A is not important. Also of interest will be neighborhoods of
points in which consist of the point n and all of the column.
They will be denoted as 
.
Let B be an infinite countably H-set. Suppose is infinite. Let b be a
countably infinite subset of . Cover B by
where 
is any set in u. This is a countable open cover of B. Since the 
are clopen and if 
then 
and if 
then 
it follows that there is no
finite subcollection whose closures cover. Thus B can contain only finitely many elements
of .
Next consider B's intersection with the plank. Suppose that B meets an infinite
number of columns. Choose such that 
misses infinitely many of the columns
that B touches. Arbitrarily pick 
for 
(which is
finite). Cover B with 
. This is
a countable cover, but infinitely many are needed in any subcollection whose
closures cover. Thus B meets only finitely many columns of the plank.
Now suppose that B meets infinitely many planks. If B is covered with
, then it is evident that
in order to have a finite subcollection with closures dense, at most the same
finitely many columns can meet B in each plank. Next suppose that B hits
the column on infinitely many planks. Let b be an countably infinite
subset of such that if 
then B hits the column of the
plank. Let 
if B misses the column of the plank,
and let 
otherwise where x is an element of B in
the column of the plank. Let 
,
is countable and infinitely many of the singletons are required in
order for the closures of any subcollection to cover. Thus B meets only
finitely many planks.
Finally suppose is infinite. Choose 
for 
such that for each u and 
is a countable open cover of B with no finite dense
subcollection. Therefore B meets at most a finite subset of .
Proof: Every infinite H-set or countably H-set meets at most finitely many points of
.