Divulgaciones Matem´aticas Vol. 23-24, No. 1-2 (2022-2023), pp. 1–11
https://produccioncientificaluz.org/index.php/divulgaciones/
DOI: https://doi.org/10.5281/zenodo.11515886
(CC BY-NC-SA 4.0)
c
Autor(es)
e-ISSN 2731-2437
p-ISSN 1315-2068
Super quasi-topological and paratopological
vector spaces versus topological vector spaces
Super casi-topol´ogicos y paratopol´ogicos espacios vectoriales versus espacios
vectoriales topol´ogicos
Madhu Ram (madhuram0502@gmail.com)
ORCID: https://orcid.org/0000-0001-6583-0978
Department of Mathematics
University of Jammu
Jammu-180006, Jammu & Kashmir, India.
Bijan Davvaz (davvaz@yazd.ac.ir)
Department of Mathematics
Yazd University
Iran.
Abstract
In this paper, we introduce the idea of super quasi-topological vector space which is
an extension of the concept of topological vector space and investigate some of its basic
properties. We extend the existing notion of quasi-topological vector space to all complex
vector spaces and investigate the relationship of super quasi-topological vector spaces with
paratopological and quasi-topological vector spaces.
Palabras y frases clave: Topological vector space, paratopological vector space, quasi-
topological vector space, super quasi-topological vector space, quotient space.
Resumen
En este art´ıculo, presentamos la idea del espacio vectorial supercuasitopol´ogico, que es
una extensi´on del concepto de espacio vectorial topol´ogico, e investigamos algunas de sus
propiedades b´asicas. Extendemos la noci´on existente de espacio vectorial cuasi-topol´ogico
a todos los espacios vectoriales complejos e investigamos la relaci´on de los espacios vecto-
riales s´uper cuasi-topol´ogicos con los espacios vectoriales paratopol´ogicos y cuasi-topol´ogicos.
Key words and phrases: Espacio vectorial topol´ogico, espacio vectorial paratopol´ogico,
espacio vectorial cuasi-topol´ogico, espacio vectorial supercuasi-topol´ogico, espacio cociente.
1 Introduction
Recall that a paratopological group is a group Gwith a topology such that the group operation
of Gis continuous. If in addition, the inversion map in a paratopological group is continuous,
then it is called a topological group.
Recibido 11/03/2022. Revisado 7/04/2022. Aceptado 21/09/2022.
MSC (2010): Primary 57N17; Secondary , 57N99.
Autor de correspondencia: Madhu Ram
2 Madhu Ram - Bijan Davvaz
According to [2], a real vector space Lendowed with a topology τsuch that (L, +, τ) is a
paratopological group, is called:
(1) paratopological vector space if for each neighborhood Uof λx with x∈Land λ∈R+(the
set of non-negative real numbers), there exist a neighborhood Vof xand an > 0 such
that [λ, λ +[.V ⊆U.
(2) quasi-topological vector space if the function Hr:L→Ldefined by Hr(x) = rx with
r∈R+, is continuous.
Hence, all translations and dilations of a paratopological (resp. quasi-topological) vector
space are homeomorphisms. For more details, see [1] and [2]. Paratopological vector spaces were
discussed and many results have been obtained (for example, see [1], [2], [3] and [4]).
Lemma 1.1. (cf. [2]) For a real vector space Lwith a topology τ, the following conditions are
equivalent.
I. (L, τ)is a paratopological vector space.
II. There exists a local basis Bat 0of Lsatisfying the following conditions:
(a) for every U, V ∈ B, there exists W∈ B such that W⊆U∩V;
(b) for each U∈ B, there exists V∈ B such that V+V⊆U;
(c) for each U∈ B and for each x∈U, there exists V∈ B such that x+V⊆U;
(d) for each U∈ B and for each r > 0,rU ∈ B;
(e) each U∈ B is absorbent and quasi-balanced.
Motivated by the papers [2] and [3], the aim of this paper is to introduce and study the
super quasi-topological vector spaces. Relationship of super quasi-topological vector spaces with
paratopological, quasi-topological and topological vector spaces is investigated.
In the following, all vector spaces are over the field F∈ {R,C}. For any undefined concepts
and terminologies, refer to [8].
2 Relationship among various classes of topological vector
spaces
In this section, we define super quasi-topological vector space and extend the definition of
paratopological and quasi-topological vector space to all complex vector spaces. Then we in-
vestigate the relation between super quasi-topological, quasi-topological, paratopological and
topological vector spaces.
Definition 2.1. Let Lbe a vector space that is equipped with a topology τsuch that (L, +, τ)
is a paratopological group. We say that (L, τ) is
1. paratopological vector space if for each neighborhood Uof rx with x∈Land r∈R+(the
set of non-negative real numbers), there exist a neighborhood Vof xand an > 0 such
that [r, r +[.V ⊆U;
Divulgaciones Matem´aticas Vol. 23-24, No. 1-2 (2022-2023), pp. 1–11
Super quasi-topological and paratopological vector spaces versus topological vector spaces 3
2. quasi-topological vector space if the function ϕr:L→Ldefined by ϕr(x) = rx with r∈R+,
is continuous;
3. super quasi-topological vector space if the function ϕr:L→Ldefined by ϕr(x) = rx with
r∈R, is continuous.
Proposition 2.1. There is a first countable locally connected quasi-topological vector space which
is not a super quasi-topological vector space.
Proof. Suppose that the complex vector space C×Cis endowed with the topology which has a
base of the sets of the form Dr×Dswhere Dr={1
√2(x−y)+ i
√2(x+y): x, y ∈R, x ≥r, i2=−1},
Ds={s+iy :y∈R, i2=−1}and r, s ∈R. Then C×Cis a first countable locally connected
quasi-topological vector space but it is not a super quasi-topological vector space. Furthermore,
C×Cis not a paratopological vector space. Also, it is neither a second countable nor a lindelof
space.
Proposition 2.2. There is a first countable non-connected quasi-topological vector space which
is not a paratopological vector space.
Proof. Endow the complex vector space Cwith the topology generated by the family of sets of
the form Dr={1
√2(x−r) + i
√2(x+r): x∈R, i2=−1}, with r∈R. Then Cis first countable
non-connected quasi-topological vector space. Observe that Cis not a paratopological vector
space.
Proposition 2.3. There is a first countable connected paratopological vector space which is not
a topological vector space.
Proof. Consider the topology on the complex vector space C×Cwhich has a base of the sets
of the form Pr×Qs, where Pr={1
√2(x−y) + i
√2(x+y): x, y ∈R, x > r, i2=−1},
Qs={x+iy :x, y ∈R, y > s, i2=−1}and r, s ∈R. Then C×Cwith this topology is a
first countable connected paratopological vector space which is not a topological vector space.
Moreover, it is second countable as well as lindelof space.
Proposition 2.4. There is a first countable non-connected super quasi-topological vector space
which is not a paratopological vector space.
Proof. Obtain the topology on the complex vector space Cby the family of sets of the form
Qr={1
2(r−√3y) + i
2(√3r+y): y∈R, i2=−1}, with r∈R. Then Cwith this topology
is a first countable super quasi-topological vector space, but it is not a paratopological vector
space.
Proposition 2.5. There is a first countable connected real quasi-topological vector space which
is not a super quasi-topological vector space.
Proof. Consider the topology on the real vector space Rgenerated by the family of sets of the form
[a, +∞), with a∈R. Then Rwith this topology is a first countable connected quasi-topological
vector space which is not a super quasi-topological vector space.
Proposition 2.6. Let (L, τ )be a complex paratopological vector space. Then (L, τθ)is also a
paratopological vector space where τθ={eiθ U:U∈τ, 0≤θ≤2π}.
Divulgaciones Matem´aticas Vol. 23-24, No. 1-2 (2022-2023), pp. 1–11
4 Madhu Ram - Bijan Davvaz
Proof. Let xand ybe any two elements of L, and eiθ Dan open neighborhood of x+y(with
respect to the topology τθ). Then there exist a neighborhood Uof e−iθ xand a neighborhood V
of e−iθy(with respect to the topology τ) such that U+V⊆D. As e−iθx∈Uand e−iθy∈V,
we have x∈eiθUand y∈eiθ V. This gives
x+y∈eiθ(U+V)⊆eiθ D.
Let rbe any non-negative real number and eiθ Uan open neighborhood of rx (with respect to
the topology τθ). Then there exist a neighborhood Vof e−iθ x(with respect to the topology τ)
and an > 0 such that [r, r +[.V ⊆Uwhich implies that rx ∈[r, r +[.eiθV⊆[r, r +[.eiθU.
Thus (L, τθ) is a paratopological vector space.
Proposition 2.7. Let (L, τ)be a complex quasi-topological vector space. Then (L, τθ)is also
a quasi-topological vector space where τθ={eiθU:U∈τ, 0≤θ≤2π}.
Proof. Follows in a similar way as the proof of Proposition 2.6.
Proposition 2.8. Let (L, τ)be a complex super quasi-topological vector space. Then (L, τθ)is
also a super quasi-topological vector space where τθ={eiθ U:U∈τ, 0≤θ≤2π}.
Proof. Follows in a similar way as the proof of Proposition 2.6.
Definition 2.2. We say that a quasi-topological vector space (L, τ) is strong if it satisfies the
following conditions:
1. there exists a topology =on Lsuch that (L, =) is a topological vector space with = ⊆ τ,
and
2. there exists a local base Bat the zero vector of the quasi-topological vector space (L, τ)
such that V\{0}is open in (L, =) for every V∈ B.
Proposition 2.9. There exists a first countable non-connected strong quasi-topological vector
space which is not second countable.
Proof. Consider the real vector space Rendowed with the topology τwhich has a base of the
sets of the form (a, b) and [c, +∞), where a, b and care real numbers. Then (R, τ) is a first
countable strong quasi-topological vector space. Clearly, it is neither a connected space nor a
second countable space.
Proposition 2.10. There exists a first countable non-connected quasi-topological vector space
which is not strong.
Proof. Consider the complex plane Cendowed with the topology τwhich has a base of the sets
of the form D(z, r) and Dtwhere D(z, r) denotes the open disk with center zand radius r, and
Dt={z∈C:Re(z)≥t, t ∈R}. Then (C, τ) is a quasi-topological vector space which is not
strong.
Proposition 2.11. There exists a regular super quasi-topological vector space which is not strong.
Proof. Let Cand τbe as in Proposition 2.5. Then Cis not a strong quasi-topological vector
space.
Proposition 2.12. There exists a Hausdorff strong quasi-topological vector space which is not a
super quasi-topological vector space.
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Super quasi-topological and paratopological vector spaces versus topological vector spaces 5
Proof. Let Rand τbe as in Proposition 2.11. Then Ris not a super quasi-topological vector
space.
The following result collects the above information and shows that the class of paratopological
vector spaces and the class of quasi-topological vector spaces are sufficiently wide.
Theorem 2.1. The following statements are valid.
1. The class of quasi-topological vector spaces contains the class of super quasi-topological,
strong quasi-topological, paratopological and topological vector spaces.
2. The class of super quasi-topological vector spaces contains the class of topological vector
spaces.
3. The class of super quasi-topological vector spaces is independent of the class of paratopolog-
ical vector spaces.
3 Basic properties of super topological vector spaces
In this section, we investigate some basic properties of super quasi-topological vector spaces. By
definition, every topological vector space is a super quasi-topological vector space, so our results
on a super quasi-topological vector space can be viewed as either improvements or extensions of
results in topological vector spaces. When we say that a topology τis a super quasi-topology on
a vector space L, we mean that (L, τ) is a super quasi-topological vector space.
Theorem 3.1. For a super quasi-topology τon a vector space L,x∈Land a non-zero real r,
the following hold:
1. the function Tx:L→Ldefined by Tx(y) = x+yis a homeomorphism;
2. the function Hr:L→Ldefined by Hr(x) = rx is a homeomorphism.
Consequently for any subset Pof L, we have Cl(x+P) = x+Cl(P);Int(x+P) = x+Int(P);
Cl(rP ) = rCl(P);Int(rP ) = rInt(P)and for any open (closed) subset Qof L,x+Qand rQ
are open (closed).
Corollary 3.1. Every super quasi-topological vector space is a homogeneous space.
A subset Aof a super quasi-topological vector space Lis called semi-balanced if for each
x∈A,λx ∈Awhenever −1≤λ≤1. It is semi-absorbent if for each x∈L, there is a real r > 0
such that λx ∈Afor each real λsatisfying −r < λ < r. Moreover, Ais called bounded if for
every neighborhood Uof 0, there is a real t > 0 such that A⊆sU for all reals ssatisfying |s| ≥ t.
As a consequence of Theorem 3.1, it can be shown in a similar way to that of topological
vector spaces, the following result:
Theorem 3.2. Suppose that (L, τ )is a super quasi-topological vector space, x∈L,06=r∈R
and A, B are subsets of L. The following assertions are valid:
1. Ais open if and only if x+Aand rA are open;
2. Ais closed if and only if x+Aand rA are closed;
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6 Madhu Ram - Bijan Davvaz
3. Ais compact if and only if x+Aand rA are compact;
4. if Ais convex, then so are Cl(A)and Int(A);
5. if Ais semi-balanced, then so is Cl(A);
6. if Aand Bare compact, then A+Bis compact;
7. if Aand Bare connected, then A+Bis connected;
8. if Aand Bare bounded, then so are Cl(A)and A∪B;
9. any finite subset of Lis bounded.
Theorem 3.3. Let τbe a super quasi-topology on a vector space L. There exists a local base B
at the origin satisfying the following conditions:
1. for every U, V ∈ B, there is W∈ B such that W⊆U∩V;
2. for each U∈ B, there is V∈ B such that V+V⊆U;
3. for each U∈ B, there is a symmetric V∈ B such that V+V⊆U;
4. for each U∈ B and for each x∈U, there is V∈ B such that x+V⊆U;
5. for each U∈ B and r∈R, there is V∈ B such that rV ⊆Uand V r ⊆U.
Conversely, let Lbe a vector space and let Bbe a family of subsets of Lsatisfying (1)-(5) and
that each member of Bcontains the origin. Then there is a super quasi-topology on Lwith Bas
a base of neighborhoods of the origin.
Proof. From Definition 2.1, and Theorem 3.1, it is easy to check that conditions (1)-(5) hold.
To prove the converse part, let Bbe a family of subsets of Lsatisfying the conditions (1)-(5)
and that each member of Bcontains 0. Let =={W⊆L: for every x∈W, there exists U∈ B
such that x+U⊆W}.
Claim 1. =is a topology on L.
Clearly, L∈ = and ∅ ∈ =. It is also easy to see that =is closed under unions. To show that
=is closed under finite intersections, let P, Q ∈ = and let x∈P∩Q. Then there exist U, V ∈ B
such that x+U⊆Pand x+V⊆Q. From condition (1), it follows that there exists O∈ B such
that O⊆U∩V. Then x+O⊆P∩Q. Hence P∩Q∈ =, and =is a topology on L.
Claim 2. If W∈ B and x∈L, then x+W∈ =.
Let y∈x+Wbe an arbitrary element. Then −x+y∈W. From condition (4), it follows
that there exists U∈ B such that −x+y+U⊆W. This means that y+U⊆x+W. Hence
x+W∈ =.
Claim 3. The family TB={x+U:x∈L, U ∈ B} is a base for the topology =on L.
Obviously, it follows from Claim 2.
Claim 4. The vector addition mapping in Lis continuous with respect to the topology =.
Let x, y be arbitrary elements of Land let Wbe an element of =such that x+y∈W. Then
there exists U∈ B such that x+y+U⊆W. For U, there is V∈ B such that V+V⊆Uby
condition (2). Then x+Vand y+Bbe two elements of TBcontaining xand y, respectively
such that
(x+V)+(y+V)⊆x+y+V+V⊆x+y+U⊆W.
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Super quasi-topological and paratopological vector spaces versus topological vector spaces 7
This ends claim 4.
Claim 5. The function Hr:L→Ldefined by Hr(x) = rx is continuous with r∈R.
Let Wbe an element of =containing rx with x∈L. Then there exists U∈ B such that
rx +U⊆W. By condition (5), there is V∈ B such that rV ⊆U. Then r(x+V) = rx +rV ⊆
rx +U⊆W. This shows that Hris continuous.
Theorem 3.4. Let (L, τ )be a super quasi-topological vector space. If Vis the neighborhood filter
of the origin, then for each x∈L,F(x) = {x+V:V∈ V} is the neighborhood filter of the point
x. Consequently, a topology of a super quasi-topological vector space is completely determined by
the neighborhood filter of the origin.
Theorem 3.5. Let (L, τ)be a super quasi-topological vector space. If Nis the neighborhood
filter of the origin, then for every A⊆L,Cl(A) = T{A+U:U∈ N}.
Proof. Suppose that x∈U+Afor each U∈ N, and let Wbe a neighborhood of x. By Theorem
3.4, there is a symmetric V∈ N such that x+V⊆W. By assumption, there is some a∈Asuch
that x∈a+V. Since Vis symmetric, a∈A∩(x+V). Thus, x∈Cl(A).
Conversely, if x∈Cl(A), then every neighborhood U+x,U∈ N, contains a point of A, so
for some u∈U,x+u∈A. Without loss of generality, we assume that Uis symmetric. Then
x∈A+U. It ends the proof.
Theorem 3.6. Let (L, τ)be a super quasi-topological vector space and Nthe neighborhood filter
of zero in L.
1. The open symmetric neighborhoods of the origin form a fundamental system of neighbor-
hoods of the origin.
2. The closed symmetric neighborhoods of the origin form a fundamental system of neighbor-
hoods of the origin.
Proof. (1) Simple.
(2) If Vis a neighborhood of zero, then there is U∈ N such that U+U⊆V. By Theorem 3.6,
Cl(U)⊆U+U. Thus, Vcontains a closed neighborhood of zero. If Pis a closed neighborhood
of zero, P∩(−P) is a closed symmetric neighborhood of zero contained in Vby Theorem 3.1.
Example 3.1. Consider the real vector space C={x+iy :x, y ∈R, i2=−1}where the addition
and multiplication operation of Care the usual addition and multiplication of complex numbers.
Endow Cwith the topology which has a base of the sets of the form Dr={r+ix:y∈R, i2=−1},
with r∈R(the set of real numbers). Then Cwith this topology is a super quasi-topological vector
space which is neither a paratopological vector space nor a topological vector space.
Theorem 3.7. Let (L, τ)be a super quasi-topological vector space. Then the following conditions
are equivalent:
1. {0}is closed;
2. {0}is the intersection of neighborhoods of the origin;
3. Lis Hausdorff.
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8 Madhu Ram - Bijan Davvaz
Proof. By Theorem 3.6, (1) and (2) are equivalent. (3) ⇒(2) is obvious. Let x, y be two
elements of Lsuch that x6=y. Then x−y6= 0. By part (2), there is a neighborhood Vof
0 such that x−y /∈U. By Theorem 3.4, there is a symmetric neighborhood Vof 0 such that
V+V⊆U. Then it is easy to check that x+Vand y+Vare disjoint neighborhoods of xand
y, respectively. It ends the proof.
Example 3.2. Consider the vector space Cas in Example 3.8. For each z0∈C, with y0=
Im(z0), denote by Ly0={x+iy0:x∈R, i2=−1}, the horizontal line passing through y0, and
B(z0), the open ball with center z0and radius . Let
Uy0, z0, =Ly0∩B(z0) (3.1)
Obtain the topology on Cgenerated by the family of sets of the form (3.1). Then Cis a
Hausdorff super quasi-topological vector space which is not a paratopological vector space.
Example 3.3. Let Lbe the vector space of all continuous functions on (0,1). For ϕ∈Land
> 0, let U(ϕ, ) = {h∈L:|h(x)−ϕ(x)|< , for all x ∈(0,1)}. Obtain the topology on
Lthat these sets U(ϕ, )generate. Then Lwith this topology is a super quasi-topological vector
space, but not a topological vector space.
Theorem 3.8. If Mis a subspace of a super quasi-topological vector space L, then Cl(M)is
a vector subspace of Lover the field of reals. Furthermore, if Lis a dense vector subspace of a
super quasi-topological vector space Eand if Mis a vector subspace of L, then the closure of M
in Eis a vector subspace of Eover the field of reals.
Proof. Follows from Theorem 3.1.
Theorem 3.9. Let (L, τ )be a super quasi-topological vector space. If Cis the connected com-
ponent of the origin and ra non-zero real, then
1. x+Cand rC are connected for each x∈L;
2. Cis a vector subspace of Lover the field of reals.
Proof. Straightforward.
A topological space Xis totally disconnected if for each x∈X, the singleton {x}is connected
component of X. By Theorem 3.6, a super quasi-topological vector space is totally disconnected
if and only if {0}is the connected component of 0.
Theorem 3.10. Let ϕbe a linear map from a super quasi-topological vector space Lto a super
quasi-topological vector space E, and let Vbe the neighborhood filter of the origin in L.
1. ϕis continuous if and only if it is continuous at 0.
2. ϕis open if and only if for every V∈ V,ϕ(V)is a neighborhood of 0in E.
Proof. Follows from Theorem 3.1.
Theorem 3.11. If a vector subspace Mof a super quasi-topological vector space Lhas an interior
point, then Mis open.
Proof. Let xbe an element of Mand Va neighborhood of 0 in Lsuch that x+V⊆M. Then
for any s∈M, we have
s+V= (s−x)+(x+V)⊆M.
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Super quasi-topological and paratopological vector spaces versus topological vector spaces 9
4 Quotients of super quasi-topological vector spaces
A super quasi-topology on vector space Lclearly induces a topology on any vector subspace of
Lmaking it a super quasi-topological vector space, and unless the contrary is mentioned, we
shall assume that a vector subspace of a super quasi-topological vector space is furnished with
its induced topology.
Let Mbe a vector subspace of a super quasi-topological vector space L. Then there is the
canonical map πof Lonto L/M, which induces a topology on L/M , called the quotient topology.
Given a vector subspace Mof a super quasi-topological vector space Land x∈L, denote by
π(x) or ˜x, the coset of Mthat contains x.
Theorem 4.1. If Mis a vector subspace of a super quasi-topological vector space L, then the
quotient map πfrom Lonto L/M is linear, continuous and open.
Proof. The continuity and linearity of πare obvious. Let Vbe an open subset of L. Since the
map x7→ a+xfrom Lto L, with a∈Lis a homeomorphism, π−1(π(V)) = V+M, an open
subset of L, so π(V) is open in L/M.
Theorem 4.2. If Mis a vector subspace of a super quasi-topological vector space L, then L/M
is a super quasi-topological vector space.
Proof. Let π(x) and π(y) be two elements of L/M, and let Ube an open neighborhood of π(x+y).
Then π−1(U) is an open neighborhood of x+yin L, so there exist open neighborhoods V1and V2
of xand y, respectively in Lsuch that V1+V2⊆π−1(U). Then π(V1) + π(V2)⊆U. By Theorem
4.1, π(V1) and π(V2) are open sets in L/M and hence the addition map (π(x), π(y)) 7→ π(x+y)
from L/M ×L/M to L/M is continuous.
Let rbe any real number. We have to show that the map π(x)7→ π(rx) from L/M to L/M is
continuous. As Lis a super quasi-topological vector space, so for any neighborhood Uof π(rx),
there exists an open neighborhood Vof xin Lsuch that rV ⊆π−1(U). Then rπ(V)⊆U. It
ends the proof.
Theorem 4.3. If Vis the neighborhood filter of 0in a super quasi-topological vector space L, and
if Mis a vector subspace of L, then π(V)is the neighborhood filter of ˜
0for the quotient topology
of L/M.
Proof. By Theorem 4.1, π(V) is a neighborhood of ˜
0 in L/M for each V∈ V. Conversely, if U
is a neighborhood of ˜
0 in L/M, then π−1(U) is a neighborhood of 0 in L; so there is V∈ V such
that V⊆π−1(U). Thus, π(V)⊆U.
Theorem 4.4. Let Mbe a vector subspace of a super quasi-topological vector space L.
1. L/M is Hausdorff if and only if Mis closed.
2. L/M is discrete if and only if Mis open.
Proof. Straightforward.
Theorem 4.5. If Mand Nare vector subspaces of a super quasi-topological vector space Lsuch
that N⊆M, then the quotient topology of M/N is identical with the subspace topology of M/N.
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10 Madhu Ram - Bijan Davvaz
Proof. Since Mis a vector subspace of L, it is a super quasi-topological vector space with the
topology induced by the topology of L. Let ϕand πbe the canonical mappings from Mto M/N
and from Lto L/N, respectively. Let Ube open for the quotient topology of M/N. Then ϕ−1(U)
is open in M, so ϕ−1(U) = M∩Vwhere Vis an open subset of L.
Claim: U= (M/N)∩π(V).
Let η∈(M/N)∩π(V). Then η=x+Nfor some x∈Mand η=v+Nfor some v∈V.
This implies that v−x∈N, so v∈x+N⊆M+N=M. Therefore, v∈M∩V=ϕ−1(U), so
η=v+N∈U. Clearly, U⊆(M/N)∩π(V) and the claim follows.
Now let Abe open in M/N for the topology on M/N induced by the quotient topology of
L/N. Then A= (M/N)∩Bfor some open subset Bof L/N . Obviously, ϕ−1(A) = M∩π−1(B)
is an open subset of M. This means that Ais open for the quotient topology of M/N.
Corollary 4.1. If Mand Nare vector subspaces of a super quasi-topological vector space L, then
the quotient topology on (M+N)/N is identical with the topology on it induced by the quotient
topology of L/N .
Theorem 4.6. Let fbe a linear map from a super quasi-topological vector space Lto a super
quasi-topological vector space E, and let Mbe a vector subspace of Lthat is contained in the
kernel of f. The linear map gfrom L/M to Esatisfying g◦π=fis continuous (open) if and
only if fis continuous (open).
Proof. The necessity part follows from Theorem 4.1. Conversely, assume fis continuous. Let U
be a neighborhood of 0 in E. Then g−1(U) = π◦f−1(U), so gis continuous at 0. By Theorem
3.14, gis continuous.
Theorem 4.7. If Mis a vector subspace of a super quasi-topological vector space L, and if M
and L/M are both Hausdorff, then Lis Hausdorff.
Proof. Let xbe an element of Lsuch that x6= 0 and let x∈Ufor each U∈ V, the neighborhood
filter of 0 in L. Since Mis Hausdorff, x /∈M. Then x+Mand Mare two distinct elements of
L/M. As L/M is Hausdorff, there are disjoint open sets Aand Bfor the quotient topology of
L/M containing x+Mand M, respectively. By Theorem 3.14, π−1(A) is a neighborhood of x
and π−1(B) is a neighborhood of 0 in L. By assumption, x∈π−1(B), so x∈π−1(A)∩π−1(B),
a contradiction. By Theorem 3.9, Lis Hausdorff.
Theorem 4.8. If Mis the connected component of zero in a super quasi-topological vector space
L, and Ma vector subspace, then L/M is totally disconnected.
Proof. Let Kbe a closed subset of L/M such that π−1(K) is disconnected. We will show that K
is disconnected. Let Aand Bbe non-empty subsets of π−1(K) such that A∪B=π−1(K) and
A∩B=∅. As for each x∈A,x+Mis connected subset of π−1(K) and hence A=A+M=
π−1(π(A)).
Similarly, B=π−1(π(B)).
Since π(A)∩π(B) = π(A∩B) = ∅and (L/M )\π(A) = π(L\A) which is open, so π(A) is
closed subset of L/M . Similarly, π(B) is closed in L/M . As
π(A)∪π(B) = π(A∪B) = π(π−1(K)) = K,
so Kis disconnected. Now,
if Cis the connected component of zero in L/M, and if there is a point π(x) of L/M such
that π(x)∈Cand x /∈M, then π−1(C) would be disconnected, which is a contradiction. It ends
the proof.
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