@
Appl. Gen. Topol. 18, no. 1 (2017), 107-115

doi:10.4995/agt.2017.6376

c© AGT, UPV, 2017

On cardinalities and compact closures

Mike Krebs

Department of Mathematics, California State University - Los Angeles, 5151 State University

Drive, Los Angeles, California 90032 (mkrebs@calstatela.edu)

Communicated by S. Garćıa-Ferreira

Abstract

We show that there exists a Hausdorff topology on the set R of real
numbers such that a subset A of R has compact closure if and only if
A is countable. More generally, given any set X and any infinite set
S, we prove that there exists a Hausdorff topology on X such that a
subset A of X has compact closure if and only if the cardinality of A
is less than or equal to that of S. When we attempt to replace “less
than or equal to” in the preceding statement with “strictly less than,”
the situation is more delicate; we show that the theorem extends to
this case when S has regular cardinality but can fail when it does not.
This counterexample shows that not every bornology is a bornology
of compact closure. These results lie in the intersection of analysis,
general topology, and set theory.

2010 MSC: 54A99.

Keywords: topology; Hausdorff; cardinality; compact closure.

1. Introduction

A bornology on a set X is a covering B of X such that (i) if A,B ∈B, then
A∪B ∈B, and (ii) if B ∈B and A ⊂ B, then B ∈B. Bornologies are objects
of much study in analysis—see, for example, [2]. The prototypical example of
a bornology is the collection of all bounded sets in a metric space. Another
standard example, for a Hausdorff space X, is the collection of all subsets of X
with compact closure. The latter construction is rather general, and one may
well wonder: Given a bornology on X, does there necessarily exist a topology

Received 15 July 2016 – Accepted 18 January 2017

http://dx.doi.org/10.4995/agt.2017.6376


M. Krebs

on X with respect to which B is precisely the bornology of sets with compact
closure? In this paper, we answer that question in the negative by constructing
a set Y for which the bornology of subsets of Y with cardinality strictly less
than that of Y cannot be a bornology of compact closure—see Example 3.10.

This example leads to the following general question. Given two sets X and
S, take the bornology B of subsets of X with cardinality less than or equal to
that of S. Is B a bornology of compact closure? One can also ask this question
for the bornology of subsets of X with cardinality strictly less than that of S.
For the first question, we show that the answer is always yes. For the second
question, our counterexample Y mentioned above shows that the answer is not
always yes; however, we prove that it is whenever S has regular cardinality.

Considerations in general topology may lead one to ask the same questions
without reference to bornologies. We now re-introduce this topic from this new
point of view. When a set X is endowed with the discrete topology, a subset
A of X is compact if and only if A is finite. One may wonder next, does there
necessarily exist a topology on X such that a subset A of X is compact if and
only if A is countable? One quickly realizes that unless X is finite, no such
topology can be Hausdorff. For if so, then let A = {an | n ∈ N} be a countably
infinite subset of X with an 6= am whenever n 6= m. (Here N denotes the set of
natural numbers.) Note that each set Ak := {an | n ≥ k} is countable, hence
compact, hence closed because the topology is Hausdorff. But then {Ak} is a
nested collection of nonempty closed subsets of the compact set A, yet it has
empty intersection, which is a contradiction.

Hausdorff being a typical property to impose a topological space, we there-
fore modify the question slightly: Does there exist a Hausdorff topology on X
in which a set has compact closure if and only if it is countable? In particular,
what about the case X = R, where R is the set of real numbers?

If we assume both the continuum hypothesis (CH) and the Axiom of Choice
(AC), then the answer to this last question is an immediate yes, for the fol-
lowing reason. Recall that CH states that no uncountable set has cardinality
strictly less than that of R. Let Ω be the least uncountable ordinal, that is, an
uncountable well-ordered set such that every subset of the form {y ∈ Ω | y ≤ x}
for x ∈ Ω is countable. It follows from CH that the cardinality of R equals
that of Ω. We may then identify R with Ω and give it the topology induced by
the order on Ω. A straightforward exercise shows that with this topology, the
closure of a set A in R is compact if and only if A is countable.

Although this logic no longer holds when we do not assume CH, it suggests
an approach. Begin by taking a well-ordering of R. Recursively define a topol-
ogy on R by constructing a neighborhood basis at each point, assuming one
has been constructed at each previous point. We carefully select this neighbor-
hood basis so that the sets with compact closure are precisely the countable
sets. Indeed, we may generalize this reasoning considerably. The details are
carried out in Section 3, where we prove our two main theorems. The first
(Theorem 3.1) states that given any set X and any infinite set S, there exists

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On cardinalities and compact closures

a Hausdorff topology on X such that the sets with compact closure are pre-
cisely those whose cardinality is less than or equal to that of S. We obtain the
first sentence in the abstract by taking X = R and S = N. In Section 2, we
discuss some set-theoretic preliminaries, including the definitions of “regular”
and “singular” cardinals. The second main theorem (Theorem 3.9) states that
when the cardinality of S is regular, the phrase “less than or equal to” in The-
orem 3.1 can be replaced by the phrase “strictly less than.” We conclude with
an example to show that the regularity condition in Theorem 3.9 cannot be
eliminated.

2. Background from set theory

Throughout this paper, we work within the Zermelo-Fraenkel axiom system
(ZF).

Recall that a linear ordering on a set X is said to be a well-ordering if every
nonempty subset of X has a smallest element.

Theorem 2.1 (Well-Ordering Principle). Every set admits a well-ordering.

It is well-known that the well-ordering principle is equivalent to the Axiom
of Choice (AC). The first step in our proofs will be to well-order the set X, so
the proofs depend on the well-ordering principle, and hence AC, right from the
git-go.

It is also well-known that a countable union of countable sets is countable.
More generally, a union over a set no bigger than X of sets no bigger than X
is no bigger than X. More precisely, we have the following theorem, where the
notation |B| ≤ |C| means that the cardinality of B is less than or equal to that
of C. (Likewise, we will later use the notation |B| < |C| to indicate that the
cardinality of B is strictly less than that of C.)

Theorem 2.2. Let X be an infinite set, and let I be a set with |I| ≤ |X|. For
each i ∈ I, let Ai be a set with |Ai| ≤ |X|. Then |

⋃
i∈I Ai| ≤ |X|.

Theorem 2.2 is proved in [1]. The proof depends on AC.
Our third and final use of AC comes as we define the terms regular and

singular for cardinalities. Roughly speaking, we say that the cardinality of a
set is regular if the set cannot be written as a smaller union of smaller sets,
and that it is singular otherwise. We now make this concept more precise.

Definition 2.3. Let X be a set. We say that X has singular cardinality if
there exists a set I with |I| < |X| such that for each i ∈ I there exists a set Ai
with |Ai| < |X|, and that X =

⋃
i∈I Ai. We say that X has regular cardinality

if X does not have singular cardinality.

We say that this definition relies on AC because although it is not the stan-
dard definition, it is equivalent to the standard definition under the assumption
of AC. We refer to [1] for details.

There is no purpose to Definition 2.3 unless both regular and singular car-
dinals exist. As the name suggests, regular cardinals are not hard to find. For

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M. Krebs

instance, ℵ0 := |N| is regular, because N does not equal a finite union of finite
sets. Producing a singular cardinal requires a deliberate construction, such as
the following.

Example 2.4. Let X1 = N. For each n ∈ N, define Xn+1 to be the power set
of Xn, i.e., the set of all subsets of Xn. By Cantor’s theorem, |Xn| < |Xn+1|.
Hence Y :=

⋃
n∈N Xn has singular cardinality.

3. Main theorems

Throughout this section, fix a nonempty set X and an infinite set S. If A
is a subset of a topological space, then we denote its closure by A. Our first
objective in this section is to prove the following theorem.

Theorem 3.1. There exists a Hausdorff topology on X so that if A ⊂ X, then
A is compact if and only if |A| ≤ |S|.

Choose a well-ordering ≤ on X. Assume that with respect to this ordering,
X has a maximal element M. (If not, then create a new ordering by reversing all
inequalities involving the minimal element.) To prove Theorem 3.1, we begin
by defining a topology. The definition is recursive and depends on knowing
that what has been defined so far already forms a topology, a fact that in turn
requires proof. So we must simultaneously make a recursive definition and an
inductive proof. For y ∈ X, we define the closed ray (−∞,y] := {x ∈ X | x ≤
y} and the open ray (−∞,y) := {x ∈ X | x < y}. Observe that X = (−∞,M].

Lemma/Definition 3.2. For any given x ∈ X, define Nx, Bx, Tx, and Wx
according to (1)–(6) below with y = x, assuming that (1)–(6) are true for all
y < x.

(1) We define Ny to be the collection of all sets of the form (−∞,y] \ K
such that K is Ty-closed in (−∞,y) and such that if C is a Ty-closed
subset of K with |C| ≤ |S|, then C is Ty-compact. Here Ty is defined
as in (3).

(2) We define By :=
⋃
z<y Nz.

(3) We have that By is a basis for a topology Ty on (−∞,y).
(4) We have that Ny ∪By is a basis for a topology Wy on (−∞,y].
(5) For all z ≤ y, if K is a Ty-closed subset of (−∞,y), then K ∩(−∞,z)

is a Tz-closed subset of (−∞,z).
(6) For all z ≤ y, we have that (−∞,z] is a Wy-closed subspace of (−∞,y].

Proof. Note that in (1), (5) and (6) above, as well as in the proof below,
we always take Wp as the topology on any closed ray (−∞,p] and Tp as the
topology on any open ray (−∞,p). This should be assumed when not explicitly
stated.

Let x ∈ X, and assume that (1)–(6) have been established for all y < x.
Items (1) and (2) are definitions and so do not require proof. Hence it suffices
to show that (3), (4), (5),and (6) hold when y = x.

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On cardinalities and compact closures

Proof of (3): We must show that Bx is a basis for a topology on (−∞,x).
From (1), we have that (−∞,y] ∈ Ny for all y < x (take K = ∅), and so Bx
covers (−∞,x). Next let U,V ∈ Bx, and let p ∈ U ∩ V . We will show that
p ∈ E ⊂ U∩V for some E ∈Bx. By definition of Bx, we have that U ∈Nq and
V ∈Nz for some q < x and z < x. So by (4), we have that K1 := (−∞,q] \U
is closed in (−∞,q] and K2 := (−∞,z] \ V is closed in (−∞,z]. By (6), we
have that L1 := K1 ∩ (−∞,p] and L2 := K2 ∩ (−∞,p] are closed in (−∞,p].
Then L1 ∪ L2 is closed in (−∞,p], so E1 := (−∞,p] \ (L1 ∪ L2) is open in
(−∞,p]. So by (4), E1 is a union of members of Np ∪Bp. Because p ∈ E1 and
members of Bp are subsets of (−∞,p), we must have p ∈ E ∈Np for some E.
Then p ∈ E ⊂ E1 ⊂ U ∩V , and E ∈Bx.

Proof of (4): Next, we prove that Nx∪Bx is a basis for a topology on (−∞,x].
As in our proof of (3), we have that Nx∪Bx covers (−∞,x]. Let U,V ∈Nx∪Bx,
and let p ∈ U ∩V . We will show that p ∈ M ⊂ U ∩V for some M ∈Nx ∪Bx.
Case 1: U,V ∈ Nx. Then U = (−∞,x] \K1 and V = (−∞,x] \K2 for some
K1,K2 of the form specified by (1). We will show that K1 ∪K2 also has this
form. Let C be a Tx-closed subset of K1∪K2 with |C| ≤ |S|. Then C∩K1 is a
Tx-closed subset of K1 with |C∩K1| ≤ |S|, so C∩K1 is Tx-compact. Similarly,
C∩K2 is Tx-compact. Hence we may take M = U ∩V = (−∞,x]\(K1 ∪K2).
Case 2: U ∈Bx,V ∈Nx. Then for some y < x we have that U = (−∞,y]\K1
for some K1 closed in (−∞,y) such that every Ty-closed subset C of K1 with
|C| ≤ |S| is Ty-compact, and V = (−∞,x] \K2 for some K2 closed in (−∞,x)
with the same property, appropriately modified. Let p ∈ U ∩ V . We’ll show
that there exists E ∈ Bx such that p ∈ E ∈ U ∩ V . Let L1 = K1 ∩ (−∞,p]
and L2 = K2 ∩ (−∞,p]. Our argument from the proof of (3) will go through
provided that (−∞,p] is closed in (−∞,x) and in (−∞,y). But this follows
from (6), because if p < z < x, then (−∞,z] \ (−∞,p] is open in (−∞,z],
hence a union of members of Nz ∪Bz, hence open in (−∞,x). Similarly for
(−∞,y). Case 3: U ∈Nx,V ∈Bx. Similar to Case 2. Case 4: U,V ∈Bx. The
proof is identical to that of (3).

Proof of (5): If z = x then (5) is tautological, so assume that z < x. Let
K be a Tx-closed subset of (−∞,x). Let L = K ∩ (−∞,z), and suppose that
p ∈ (−∞,z) \L. We will show that there exists E ∈ Tz such that p ∈ E and
E∩L = ∅. We know that there exists Q ∈Bx such that p ∈ Q and Q∩K = ∅.
Then Q ∈ Ny for some y < x, so Q is open in (−∞,y]. So by (6), we know
that Q∩ (−∞,z] is open in (−∞,z]. Hence, by (4), we have that Q∩ (−∞,z]
is a union of members of Nz ∪Bz. If U ∈ Nz, then U ∩ (−∞,z) is open in
(−∞,z), by (1). If U ∈Bz, then U ∩ (−∞,z) = U is open in (−∞,z), by (3).
So E := (Q∩(−∞,z])∩(−∞,z) = Q∩(−∞,z) is a union of Tz-open sets and
is therefore open in (−∞,z). Moreover, p ∈ E and E ∩L = ∅.

Proof of (6): If z = x then (6) is tautological, so assume that z < x.
First we show that (−∞,z] is closed in (−∞,x]. The proof of (4), Case 2,
shows that (−∞,z] is Tx-closed. Let C be a Tx-closed subset of (−∞,z] with
|C| ≤ |S|. We will show that C is Tx-compact. From this it will follow that
(−∞,x] \ (−∞,z] ∈ Nx and therefore that (−∞,z] is closed in (−∞,x]. To

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M. Krebs

show that C is Tx-compact, it suffices to show that C ∪{z} is Tx-compact,
because C is Tx-closed. Let {Uα} be an open cover of C ∪{z} by members of
Bx. Choose a set Uα0 from this cover with z ∈ Uα0 . Then Uα0 ∈Ny for some y
with z ≤ y < x. So Uα0 = (−∞,y]\K for some K closed in (−∞,y) such that
if D is a Ty-closed subset of K with |D| ≤ |S|, then D is Ty-compact. By (5),
we know that C ∩K is Ty-closed. Also, |C ∩K| ≤ |C| ≤ |S|. Hence C ∩K is
Ty-compact. It follows from (5) that Uα ∩ (−∞,y) is Ty-open for all α. Hence
finitely many sets Uα1 ∩ (−∞,y), . . . ,Uαn ∩ (−∞,y) cover C ∩ K. Therefore
Uα0,Uα1, . . . ,Uαn is our desired finite subcover of C ∪{z}.

Next, we show that if K is a Wz-closed subset of (−∞,z], then K is Wx-
closed. By (4), we have that (−∞,z]\K is a union of members of Nz∪Bz ⊂Bx,
so (−∞,z] \K is open in (−∞,x]. Therefore (−∞,x] \K = ((−∞,z] \K) ∪
((−∞,x] \ (−∞,z]) is Wx-open.

Finally, we show that if K is a Wx-closed subset of (−∞,x], then K∩(−∞,z]
is Wz-closed. Let p ∈ (−∞,z] \ K. Then p ∈ U ⊂ (−∞,x] \ K for some
U ∈Nx∪Bx. It suffices to show that U∩(−∞,z] ∈Wz. Case 1: U ∈Bx. Then
U ∈Ny for some y < x. So U is open in (−∞,y]. Therefore U ∩(−∞,z] ∈Wz
by (6), inductively. Case 2: U ∈ Nx. Then U = (−∞,x] \ K1 for some K1
closed in (−∞,x), by (1). So U equals a union of sets Wβ ∈ Bx. Arguing as
in Case 1, we see that for all β, we have that Wβ ∩ (−∞,z] ∈ Wz. Therefore
U ∩ (−∞,z] =

⋃
(Wβ ∩ (−∞,z]) ∈Wz. �

Regrettably, the statement and proof of Lemma/Definition 3.2 contain many
unavoidable technicalities. The idea, though, is fairly straightforward. Recall
that if A is a Hausdorff topological space and p is a point not in A, then we
define the co-compact topology on A∪{p} by choosing the neighborhood basis
at p to consist of all complements of compact subsets of A. The definition
of Ny is somewhat similar to this; we take the neighborhood basis at y to be
the collection of all complements of closed sets, every “small” closed subset of
which is compact. By “small” here we mean, “of cardinality no greater than
that of S.”

Example 3.3. Let X = N ∪{M}, where M /∈ N. We order X by taking the
usual order on N and declaring that n < M for all n ∈ N. Let S = N. For
any n ∈ N, the topology Wn on (−∞,n] defined by Lemma/Definition 3.2 is
the discrete topology. Then the topology TM on N = (−∞,M) is also discrete.
The compact subsets of (−∞,M) are precisely the finite subsets. It follows
that the topology WM on X = (−∞,M] coincides with the order topology.
Example 3.4. Let Ω be the least uncountable ordinal, and let X = Ω∪{M},
where M /∈ Ω. We order X by taking the usual order on Ω and declaring
that x < M for all x ∈ Ω. Let S = N. We claim that the topology WM on
X = (−∞,M] cannot coincide with the order topology on X in this case. For
suppose otherwise. Then K := (−∞,M) is a closed subspace of (−∞,M),
and moreover every closed subset C of K with |C| ≤ |S| is compact. So
{M} = (−∞,M] \ K is open in X. But in the order topology on X, the set
{M} is not open.

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On cardinalities and compact closures

From this point on, we endow X = (−∞,M] with the topology WM as in
Lemma/Definition 3.2. The following four lemmas establish that WM possesses
the property required by Theorem 3.1.

Lemma 3.5. X is Hausdorff.

Proof. Let a,b be distinct points in X. Without loss of generality, assume that
a < b. We have that (−∞,a] ∈Na and (−∞,b] ∈Nb, so (−∞,a] and (−∞,b]
are open in X. By (6) in Lemma/Definition 3.2, we have that (−∞,a] is closed
in X. So (−∞,a] and (−∞,b] \ (−∞,a] are disjoint open neighborhoods of
a,b, respectively. �

Lemma 3.6. If A is a closed subset of X with |A| ≤ |S|, then A is compact.

Proof. It suffices to show that A ∪{M} is compact. Let {Uα} be a cover of
A∪{M} by members of NM∪BM . Then M ∈ Uα0 for some member Uα0 of the
cover with Uα0 ∈NM . Then L = X \Uα0 is closed in (−∞,M) and every TM -
closed subset C of L with |C| ≤ |S| is TM -compact. From Lemma/Definition
3.2, we see that A ∩ (−∞,M) is TM -closed. Hence L ∩ A ∩ (−∞,M) is TM -
closed as well. Also, |L∩A∩ (−∞,M)| ≤ |S|. Therefore L∩A∩ (−∞,M) is
TM -compact. For all α, we have that Uα ∩ (−∞,M) ∈ TM . So finitely many
sets Uα1, . . . ,Uαn cover L∩A∩ (−∞,M). Then Uα0,Uα1 . . . ,Uαn give us the
desired subcover. �

Lemma 3.7. If A ⊂ X and |A| ≤ |S|, then |A| ≤ |S|.

Proof. Suppose that |A| > |S|. Let m be the smallest element of X such that
|(−∞,m]∩A| > |S|. Let ` be the least upper bound of A∩(−∞,m). (The fact
that X is well-ordered guarantees that m and ` exist.) Let Q = {a ∈ A | a <
m}. Let C =

⋃
a∈Q((−∞,a] ∩ A). By Theorem 2.2, |C| ≤ |S|. Observe that

(−∞,m] \ (−∞,`] is an open set disjoint from A; therefore it is disjoint from
A. So (−∞,m]∩A contains C and at most two other points, namely ` and m.
Becasue S is infinite, therefore |(−∞,m] ∩A| ≤ |S|, a contradiction. �

Lemma 3.8. If A ⊂ X and |A| > |S|, then A is not compact.

Proof. Temporarily assume that A is compact. Observe that |A| > |S|. Let m
be the smallest element of X such that |(−∞,m]∩A| > |S|. By Lemma/Definition
3.2, we have that (−∞,m] is a closed subspace of X = (−∞,M], so (−∞,m]∩A
is Wm-compact. Let K = (−∞,m) ∩ A. Note that |K| > |S|, by definition
of m. Let C be any Tm-closed subset of K such that |C| ≤ |S|. We will
show that C is Tm-compact. By definition of Wm, this will show that K is a
Wm-closed subset of the Wm-compact set (−∞,m] ∩A and therefore that K
is Wm-compact.

Let ` be the least upper bound of C in X. We must have that ` < m, for
otherwise, by Theorem 2.2, we would have that

|K| =

∣∣∣∣∣⋃
c∈C

((−∞,c] ∩A)

∣∣∣∣∣ ≤ |S|.

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M. Krebs

So C ⊂ (−∞,`]. It follows then from Lemma/Definition 3.2 that because C is
Tm-closed, therefore C is W`-closed, and therefore C is Wm-closed. But C is a
subset of the Wm-compact set A, so C is Wm-compact, therefore Tm-compact.

Note that {(−∞,k] : k ∈ K} is a Wm-open cover of K. Hence

K ⊂ (−∞,k1] ∪·· ·∪ (−∞,kn]

for some k1, . . . ,kn. So

K ⊂
n⋃
j=1

((−∞,kj] ∩A).

But each kj < m, so |(−∞,kj] ∩ A| ≤ |S|, by definition of m. But then
|K| ≤ |S|, because S is infinite. This contradicts the fact that |K| > |S|. �

Theorem 3.1 follows at once from Lemmas 3.5, 3.6, 3.7, and 3.8.

Theorem 3.9. Let X be a set, and let S be an infinite set with regular cardi-
nality. Then there exists a Hausdorff topology on X so that if A ⊂ X, then A
is compact if and only if |A| < |S|.

Proof. The proof is identical to that of Theorem 3.1 with two small modifica-
tions. One must replace every instance of “≤ |S|” with “< |S|.” Also, one must
use Definition 2.3 in place of Theorem 2.2 whenever the latter is invoked. �

The following example illustrates how Theorem 3.9 can fail when S has
singular cardinality.

Example 3.10. Consider the sets defined in Example 2.4. We will show that
there does not exist a topology on Y such that A ⊂ Y has compact closure if
and only if |A| < |Y |.

Suppose otherwise. The fact that Y does not have strictly smaller cardinality
than itself implies that Y = Y is not compact. Let {Uα} be an open cover of Y
with no finite subcover. We know that |Xn| < |Y | for all n, so Xn is compact.
Cover Xn with finitely many sets Uαn,1, . . . ,Uαn,jn from the collection {Uα},
and let Vn =

⋃jn
`=1 Uαn,` . By our assumption on {Uα}, there exists a point

bn ∈ Y such that bn /∈ Wn, where Wn =
⋃n
k=1 Vk. Note that the sets Wn form

an increasing chain of open sets. Also note that Xn ⊂ Xn ⊂ Vn ⊂ Wn.
Let S = {bn | n ∈ N}. Then |S| < |Y |. We will show that S is not

compact, thereby producing the desired contradiction. For each ` ∈ N, let
S` = {bn | n ≥ `}. Then {S`} is a collection of closed subsets of S with the
finite intersection property. It suffices to show that

⋂∞
`=1 S` = ∅. Let y ∈ Y .

Then y ∈ Xm for some m. Hence y is in the open set Wm. Observe that
Wm ∩ Sm = ∅, because of how we chose the points bn. Therefore y /∈ Sm.
Therefore {S`} has empty intersection, and so S cannot be compact.

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On cardinalities and compact closures

Acknowledgements. The author would like to acknowledge Gerald Beer for
discussions of bornologies that prompted the questions answered in this paper,
and would like to thank the anonymous referee for several useful suggestions.

References

[1] A. Dasgupta, Set theory, Birkhäuser/Springer, New York, 2014.

[2] H. Hogbe-Nlend, Bornologies and functional analysis, North-Holland Publishing Co.,
Amsterdam-New York-Oxford, 1977.

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