# Some Characterizations of Open, Closed, and Continuous Mappings

Authors: Navpreet Singh Noorie; Rajni Bala
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

We obtain new characterizations of open maps in terms of closures, of closed maps in terms of interiors, and of continuous maps in terms of interiors. Further open (closed) onto mapsf:X→Y are described in terms of images under f of certain closed (open) sets in X. Continuity of (onto) maps is also characterized in terms of saturated sets.


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## Body

## 1. Introduction

It is well known
that a mapf : X→Y between arbitrary topological spaces is open if and only
if (f(A))O⊃f(AO) for arbitrary subsets
A of X, where AO denotes the interior of A (see [1]). On the other
hand, Schochetman has
shown in [2] that if X and Y are metric spaces, then f is open if and only if f−1(y) ⊂ limsupnf−1(yn),
for every convergent sequence {yn} with limnyn = y in Y. We are motivated by the simple but important observation that not
only is the set limsupnf−1(yn)
closed (special case of [2, Lemma 2.1]) but the set {y∈Y:f−1(y)⊂limsupnf−1(yn)}
is also closed whenever f is an open map. It is then easy to generalize that a
map f : X→Y between arbitrary topological spaces is open if and only if for
any closed set F in X the set f#(F)={y∈Y:f−1(y)⊂F} is closed in Y. This raises the question if
we can describe open maps in terms of inclusion relation involving closures of
certain subsets, perhaps involving f#, rather than in terms of the
well-known inclusion relation involving interiors mentioned in the beginning.Further, it has
been shown in [3, 2.5 Problem 321] that an onto map f : X→Y is closed if and
only if for every open set U in X, the set f#(U)={y∈Y:f−1(y)⊂U} is open in Y. (We see in Theorem 2.10
below that we can drop the condition of onto in this result.) This again
suggests the possibility of describing closed maps in terms of inclusion
relation involving interiors of certain subsets, perhaps involving f#,
rather than in terms of the well-known inclusion relation involving closures,
namely, that a map f : X→Y between arbitrary topological spaces is closed if
and only if Cl(f(A)) ⊂f (Cl(A)) for arbitrary subsets A of X,
where Cl(A) denotes the closure of A (see [1]).In this paper,
we obtain new characterizations of open maps in terms of closures (Theorem2.7
below), closed maps in terms of interiors (Theorem 2.10 below), and continuous
maps in terms of interiors (Theorem 2.13 below). Further open (closed) onto maps
f : X→Y are described in terms of images under f of certain closed (open)
sets in X in Corollary 2.8 (Corollary 2.11). Continuity of (onto) maps is also
characterized in terms of saturated sets in Corollary 2.15 (Theorem 2.14).Notation 1.
For
any setsX and Y, let f : X→Y be any map and E any subset of X. Then
(i)
f#(E)={y∈Y:f−1(y)⊂E};(ii)
E#=f−1(f#(E)).Remark 1.1.
Letf : X→Y be continuous and E be any subset of X, where X and Y are
topological spaces. Then E# is open in X whenever f#(E)
is open in Y. The converse holds, if f is onto and follows from Theorem 2.14
below.


## 2. Results

Definition 2.1.
For any setsX and Y, a subset E of X will be called a saturated subset of X under a map f : X→Y if for some subset B of Y, E = f−1(B).
Equivalently, if E =f−1(f(E)).Remark 2.2.
For any mapf : X→Y, E# is saturated for each subset E
of X.We begin with
the following lemmas. The proof of the Lemma2.3 is straightforward from the
definitions and is omitted.Lemma 2.3.
For any setsX and Y, let f: X→Y be any
map and E be any subset of X. Then
(i)
ifA⊂ B, then f#(A)⊂f#(B);(ii)
f(E#)=f#(E)∩f(X);(iii)
E#={f−1y:y∈Yandf−1y⊂E}⊂E;(iv)
f#(ϕ) = [f(X)]C;(v)
f#(X) = Y,X#=X, and ϕ#=ϕ;(vi)
f(E#)=f#(E)∩f(E);(vii)
f#(EC)=[f(E)]C and so f#(E)=[f(EC)]C and f(E)=[f#(EC)]C;(viii)
f#(E#)=f#(E);(ix)
E is saturated if and
only if E=E#.Remark 2.4.
Iff : X→Y is continuous, then for any saturated set E
in X, E is open in X whenever f#(E) is open in Y. This follows from
Remark 1.1 and (ix) above.The following lemma gives characterization of onto maps.Lemma 2.5.
For any mapf :
X→Y, where X and Y are any sets, the following
conditions are equivalent:
(a)
f is onto;(b)
f#(A)=f(A#)
for each subset A of X;(c)
f#(A)⊂f(A) for each subset A of X;(d)
f#(ϕ)=ϕ.Proof.
(a)⇒(b) follows from the definition of A#.
(b)⇒(c) follows from Lemma 2.3(iii).
(c)⇒(d) is obvious.
(d)⇒(a) follows from Lemma 2.3(iv).Remark 2.6.
From now
onwards, all spaces are assumed to be arbitrary topological spaces.Theorem 2.7.
For any mapf
: X→Y, the following conditions are equivalent.
(a)
f is an open map, that is, for
each subset A of X, f (AO) ⊂[f(A)]O;(b)
for each subsetA of X,Cl(f#(A))⊂f#(Cl(A));(c)
for each closed subsetF of X,f#(F)
is closed in Y.Proof.
(a)⇒(b): By Lemma 2.3(vii), Cl(f#(A))
= Cl((f(AC))C) = [(f(AC))O]C⊂ [f((AC)O)]C,
since (a) holds. Therefore, Cl[f#(A)] ⊂ [f((AC)O)]C=
[f((Cl(A))C)]C= f#(Cl(A)) by Lemma 2.3(vii) again. Hence, Cl[f#(A)] ⊂f#(Cl(A)) and so (b)
holds.
(b)⇒(c): Let F be any closed subset of X. Then Cl(F) = F and so (b) implies that Cl[f#(F)]⊂f#(F). Hence f#(F)
is closed, and (c) holds.
(c)⇒(a): Let U be any open subset of X. Then (c) implies that f#(UC)
is closed in Y. But f#(UC) = [f(U)]C, by Lemma
2.3(vii). Therefore, f(U) is open in Y. Hence f is an open map. This proves
(a).Corollary 2.8.
Letf : X→Y be onto. Then f is an open map if and only
if for each closed subset F of X, f(F#) is closed in Y.Proof.
The proof follows immediately from the equivalence of (a)
and (c) in the above theorem and Lemma2.5(b).In the next corollary, we see that interestingly images of saturated
closed sets underopen surjections are closed sets.Corollary 2.9.
Letf : X→Y be an open onto map. Then for each saturated
closed subset F of X,f(F) is closed in Y. In particular, for any set A, if A# is closed, then f(A#) is closed in Y.Theorem 2.10.
For any mapf : X→Y, the
following conditions are equivalent.
(a)
f is closed, that is, for each
subset A of X, Cl(f(A))⊂f(Cl(A));(b)
for each subsetA of X, f#(AO) ⊂ [f#(A)]O;(c)
for each open subsetU of X, f#(U)
is open in Y.Proof.
(a)⇒(b): By Lemma 2.3(vii) and condition (a), f#(AO)
= [f((AO)C)]C = [f(Cl(AC))]C⊂ [Cl(f(AC)]C = [Cl(f#(A))C]C= [f#(A)]O. Hence, f#(AO) ⊂ [f#(A)]O, and (b)
holds.
(b)⇒(c): If U is any open subset of X, then U = UO and (b) implies that f#(U) ⊂ [f#(U)]O. Therefore, f#(U) is an open set and (c) holds.
(c)⇒(a): Let F be any closed subset of X. Then
(c) implies f#(FC) is open in Y. Therefore, part (vii)
of Lemma 2.3 implies that [f(F)]C=f#(FC) is open in Y. Thus, f(F) is closed in Y,
and hence f is a closed map.Corollary 2.11.
Letf : X→Y be onto. Then f is a closed map if and only
if for each open subset U of X, f(U#) is open in Y.Proof.
The proof follows immediately from the equivalence of (a)
and (c) in the above theorem, sincef(U#)=f#(U) if f
is onto by Lemma 2.5(b).In the next corollary we see that interestingly images of saturated open
sets underclosed surjections are open sets.Corollary 2.12.
Letf : X→Y be a closed onto map. Then for each
saturated open set U in X, f(U) is open in Y. In particular, for any set A, if
A# is open, then f(A#) is open in Y.The remaining results of this section give characterizations of
continuous maps.The following Theorem2.13 is an analog, in terms of interiors, of the
result that a map f:X→Y is continuous if and only if for each subset
A of X, f(Cl(A)) ⊂ Cl(f(A)).Theorem 2.13.
A mapf : X→Y is continuous if and only if for each subset
A of X, [f#(A)]O⊂f#(AO).Proof.
We know that f is
continuous if and only if for each subsetA of X, f[Cl(A)] ⊂ Cl(f(A)). But by part (vii) of Lemma 2.3, f
[Cl(A)] ⊂ Cl(f(A)) if and only if [f#((Cl(A))C)]C⊂ Cl[(f#(AC))C]
= [(f#(AC))O]C. Therefore, f
[Cl(A)] ⊂ Cl(f(A)) if and only if [f#(AC)]O⊂f#((AC)O).
Since this equivalence holds for arbitrary subsets A of X, we have f is
continuous if and only if [f#(A)]O⊂f#(AO) for each subset
A of X.The following theorem characterizes continuity for onto maps.Theorem 2.14.
Letf : X→Y be any onto map. Then the following
conditions are equivalent.
(a)
f is continuous;(b)
[f(A#)]O⊂f [(AO)#] for all
subsets A of X;(c)
A# is open in X
whenever f(A#) is open in Y;(d)
for any saturated setE in X, E is
open in X whenever f(E) is open in Y;(e)
for any saturated setE in X, E is
closed in X whenever f(E) is closed in Y.Proof.
Since (a)⇔ (b) by Theorem 2.13 above and Lemma 2.5(b), we
prove (a) ⇒ (c) ⇒ (d) ⇒ (e) ⇒ (a).
(a)⇒(c). Since f#(A) = f(A#),
by Lemma 2.5(b), (c) follows from (a) by Remark 1.1.
(c)⇒(d). Since E is saturated if and only if E = E#, (d) follows from (c) for arbitrary f.
(d)⇒(e). Let f(E) be closed in Y, where E is
saturated in X. Then [f(E)]C is open in Y and [f(E)]C=
f#(EC), by Lemma 2.3(vii). Then f is onto implies [f(E)]C= f[(EC)#] by Lemma 2.5(b). Also E is
saturated implies EC is also saturated and so (EC)# = EC, by Lemma 2.3(ix). Therefore, [f(E)]C = f(EC)
and so f(E) closed in Y implies that f(EC) is open in Y and
therefore, by (d), EC is open and so E is closed in X, proving
thereby that (e) holds.
(e)⇒(a). Let S be any closed subset of Y and let E
= f-1(S). Then f(E) = S, since f is onto, and so f(E) is closed for
a saturated subset E of X. Therefore, by condition (e), E = f-1(S)
is closed in X. Hence, f is continuous.Corollary 2.15.
A mapf : X→Y is continuous if and only if for any
saturated set E in X, E is open (closed) in X, whenever f(E) is open (closed)
in f(X).Proof.
The proof follows from the equivalence of (a) and (d) ((a)
and (e)) above since the mapf : X→f(X) is onto.



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