INTERNATIONAL JOURNAL OF LATEST TECHNOLOGY IN ENGINEERING,
MANAGEMENT & APPLIED SCIENCE (IJLTEMAS)
ISSN 2278-2540 | DOI: 10.51583/IJLTEMAS | Volume XIII, Issue X, October 2024
www.ijltemas.in Page 232
II. Development of Lattice of Topologies
Let C = {τ
α
: α ∆} be a chain of reductions of a topology τ on a set X. Then C, with the relation of set inclusion is a poset. We
also see that C is totally ordered. Ifτ
α
and τ
α
are two topologies in C such that, say, τ
α
is weaker than τ
α
, then the g.l.b. of the
sub-family T = {τ
α
,τ
α
} of C, that is, inf(T), is τ
α
. Also sup(T) = τ
α
. Hence C is a lattice of topologies by set inclusion.
Let R be another relation on the chain C, where τ
α
Rτ
r
if τ
α
≤ τ
r
. That is, the relation R (≤) on C, now, is that of comparison of
topologies. With this relation on C, we see again that C is a lattice of topologies. What we have just established is the following.
Corollary 1.12 Every chain C of reductions of a topology on a set X is a lattice in at least two ways.
Observations
Every set on which a partial order is defined is not a lattice; that is, not every poset that is a lattice. In particular, every family of
topologies is not a lattice. For example, if the topologies in a family F are not comparable, then the family F would not be a lattice
in either of the ways; but F would still be a poset in the two ways.
If a family of subsets of a set X is pairwise comparable by set inclusion (i.e. totally ordered by set inclusion), then it generates a
topology (on X) which has a chain of reductions. This indeed is a theorem which marks the end and climax of this section.
Theorem 1.1 Any (set inclusion) pairwise comparable family F of subsets of a set X generates a reducible topology τ on X. And
the chain C of reductions of τ can be constructed in such a way that card(F) = card(C).
Proof: Let F = {A
α
: A
α
X}
α
∆
be a family of (set inclusion) pairwise comparable subsets of X. Let A
α1
and A
α2
be two elements
of F such that, say, A
α1
A
α2
. Let γ
1
= A
α1
-induced topology on X and γ
2
= A
α2
-induced topology on X. If γ
1
and γ
2
are not
comparable, let τ
1
= γ
1
and τ
2
= γ
1
▽ γ
2
, the join of γ
1
and γ
2
(defined as the weakest topology, on X, finer than both γ
1
and γ
2
). Then
τ
1
and τ
2
are two comparable topologies on X. Precisely, τ
1
is strictly weaker than τ
2
.
Since F is pairwise comparable, the sets in F can be arranged such that
A
α
A
r
···.
It follows from the construction above that these sets in F have, corresponding to them, a family C = {τ
α
}
α
∆
of topologies on X,
which is pairwise comparable in that
τ
α
≤ τ
r
≤ ···.
It is easy to see that C is equivalent to F; that is, card(C) = card(F).
■
It is easier to see the existence of the chain C, constructed in the proof of the theorem if we remember that the construction can
actually be done through inducement by the discrete topologies of A
α1
and A
α2
; or, by what is similar, first getting a topology on
A
α2
and then using this to induce a topology on A
α1
; and then finally using these two topologies to construct subset-induced
topologies on X.
III. Subset-induced Topologies
Proposition 2.1 If X E, then any topology, say τ
X
, on X induces a topology, say τ
XE
, on E, given by τ
X
(E) = τ
X
{E}.
Proof: It is easy to see that τ
X
(E), since τ
X
. Also E τ
X
(E), by definition. Let {G
i
: i = 1, . . ., n} be a sub-collection of
τ
X
(E). We show that the intersection
belongs to τ
X
(E). Clearly
τ
X
(E) if any of the
comes from τ
X
. If all the
are each equal to E, then E =
is an element of τ
X
(E). Hence in any case τ
X
(E) is closed under finite intersections. Let
α
α be any family of sets in τ
X
(E). If one of these sets equals E, then their union would equal E, which belongs to τ
X
(E).
If none of these sets equals E, then each of them belongs to τ
X
and hence their union belongs to τ
X
which is itself a subfamily of
τ
X
(E). These imply that τ
X
(E) is also closed under arbitrary unions, and is therefore a topology on E.
■
Definition 2.1 The topology τ
X
(E), on E, is called an X-topology on E; or a topology induced on E by the topology τ
X
on X.
Observe that one subset can induce several topologies on its superset.
Proposition 2.2 Let (E,τ) be a topological space, and let X τ be a τ-open subset of E. Let τ
X
= {G τ : G X}. Then τ
X
is a
topology on X.
Proof:
1. τ
X
, since τ and X.
2. X τ
X
, since X τ and X X.