Strongly Irresolvable Spaces

Several new characterizations of strongly irresolvable topological spaces are found and precise relationships are noted between strong irresolvability, hereditary irresolvability, and submaximality. It is noted that strong irresolvablity is a faint topological property, while neither hereditary irresolvablity nor submaximality are semitopological.


Introduction
Throughout, (X,τ) denotes a topological space and if A ⊆ X, τ | A is the subspace topology on A. Also, Cl(A) and Int(A) denote the closure and interior, respectively, of A in X.If B ⊆ A, Cl A (B) and Int A (B) are the closure and interior, respectively, of B in the subspace A. A subset D of X is dense if Cl(D) = X and the family of all dense subsets of X is D(X,τ).A subset C of X is codense if X − C = D is dense, or equivalently, Int(C) = ∅.A subset N of X is nowhere dense if Int(Cl(N)) = ∅ and the family of nowhere dense subsets of X is N(X,τ).It is known that a subset N of X is nowhere dense if and only if for each nonempty open set U, there exists a nonempty open subset V ⊆ U with V ∩ N = ∅.Further, N(X,τ) is an ideal.That is, every subset of a nowhere dense set is nowhere dense and every finite union of nowhere dense sets is nowhere dense.This latter property does not hold generally for codense sets.In fact, many spaces exist which are the union of two codense subsets.Equivalently, there exist spaces containing two disjoint dense subsets.Definition 1.1.A topological space is resolvable if it contains two disjoint dense subsets.A space is irresolvable if it is not resolvable.Definition 1.2.A topological space is crowded (or dense in itself) if it has no isolated points.
Since an isolated point must belong to every dense set, the presence of isolated points prevents resolvability.Hewitt [3] observed that all classical topological spaces without isolated points are resolvable.In particular, he proved that all crowded first countable spaces are resolvable.He then constructed crowded spaces which are far from being resolvable.Definition 1.3.A topological space is submaximal if every dense subset is open.Definition 1.4.A space is hereditarily irresolvable if every subspace is irresolvable.
If D is a proper dense subset in a space (X,τ), D is not dense in the space (X,σ) where σ = τ(X − D) = τ [D] is the smallest expansion of τ for which D is closed.By a maximal expansion construction, Hewitt was able to construct submaximal spaces and hereditarily irresolvable T 1 -spaces.Anderson [1], also using expansion topologies, found a construction for submaximal connected Hausdorff spaces.Hewitt did not investigate strongly irresolvable spaces which were later mentioned or used in [2,4].Definition 1.5.A space (X,τ) is strongly irresolvable if each open subspace is irresolvable.
We avoid expansion topologies to present easy examples of submaximal, hereditarily irresolvable, and strongly irresolvable spaces using ultrafilters and Hewitt's decomposition theorem which states that every space X can be expressed as a disjoint union F G of a closed resolvable subspace F with an open hereditarily irresolvable subspace G.It can easily be shown that this decomposition is unique for each space X.
We offer the following obvious but useful fact and then close this section with a proof of the equivalence of various descriptions of strongly irresolvable spaces that have appeared in the literature.
Proposition 1.6.If U is a nonempty open subspace of a resolvable space, then U is a resolvable subspace.
Theorem 1.7.Strong irresolvability of the space (X,τ) is equivalent to each of the following.
(2) Every dense subset has a dense interior.
(4) Every subset is the union of an open set and a nowhere dense set.
Proof.(1)⇔(2).If D is dense and union of an open set with a nowhere dense set.
Proof.If X is submaximal and A is a nonempty resolvable subspace with ).This contradiction shows that X is hereditarily irresolvable.If X is hereditarily irresolvable, all subspaces including the open ones are irresolvable so that X is strongly irresolvable.Certainly strongly irresolvable spaces are irresolvable.
Examples will be used to show that these classes of spaces are nonempty and the implications of this theorem are not reversible.Finite spaces can be used for this purpose but these spaces can only be irresolvable if isolated points are present.In fact, Hewitt showed that first countable crowded spaces are always resolvable.The class of first countable spaces includes all finite spaces and all metric spaces.Thus, our examples of irresolvable spaces without isolated points cannot be first countable.The basic unit of construction in these examples is an infinite set X equipped with a free ultrafilter topology F * = F ∪ {∅}, where F is a free ultrafilter on X.It is easy to see that every infinite set supports a free ultrafilter.For if X is infinite and (P,⊆) is the poset of all filters on X, by Zorn's lemma, there is a maximal filter F in P, called an ultrafilter, containing the free filter of cofinite subsets of X. Clearly F is free and F * is a topology on X. Ultrafilters are known to have the following interesting properties.Proposition 2.2.If F is an ultrafilter on X and B ⊆ X has the property that In fact, the space of this proposition is hyperconnected in the sense that every nonempty open set is dense, for if τ) since submaximal spaces are strongly irresolvable.The irresolvability of this space is strongly dependent on the ultrafilter property of F. If F were replaced by simply a free filter, the space might be resolvable.For example, if X is any infinite set and ρ is the cofinite topology on X, then the infinite subsets are dense and since two disjoint infinite subsets can be found, the space is resolvable.It is also crowded and T 1 with N(X,ρ) is the collection of finite subsets.In particular, if X = N is the set of natural numbers, the finite sets are the nowhere dense sets and E = {2, 4,...} and O = {1, 3,...} are disjoint dense subsets.
We now construct a hereditarily irresolvable space which fails to be submaximal.
Example 2.5.Let Y and Z be disjoint infinite sets.Let F be a free ultrafilter on Y and let G be a free ultrafilter on Z. Equip Y and Z with topologies σ and ρ, respectively, with We claim that (X,τ) is a crowded T 1 -space which is hereditarily irresolvable but not submaximal.First, we see that τ is a topology.
In this last example, had Z been equipped with a free filter topology for which Z is resolvable, the construction of X as above results in a strongly irresolvable space which is not hereditarily irresolvable.This is our next example.
Example 2.6.Let Y and Z be disjoint infinite sets.Let F be a free ultrafilter on Y .Equip Y with the ultrafilter topology σ = F ∪ {∅} and let ρ be the cofinite topology on Z.Let X = Y Z have topology τ = {A B | A ∈ F and B ∈ ρ} ∪ {∅}.We claim that (X,τ) is a crowded T 1 -space which is strongly irresolvable but not hereditarily irresolvable.As in the previous example, τ is a topology and (X,τ) is a crowded T The spaces constructed as disjoint unions of spaces with filter topologies in the two examples just given are connected.By disconnecting the two components in the previous example, we obtain the following example of an irresolvable space which is not strongly irresolvable.
Example 2.7.Let Y and Z be infinite sets, let σ = F ∪ {∅} for some free ultrafilter F on Y , and let ρ be the cofinite topology on Z.Let X = Y Z have topology τ = {A B | A ∈ σ and B ∈ ρ}.As in the previous examples, τ is a topology and (X,τ) is a crowded T 1 -space.The space X is irresolvable since Y is an open irresolvable subspace.But X is not strongly irresolvable since Z is an open resolvable subspace.The space X is disconnected since Z is a nonempty proper clopen subset.

Local resolvability and decomposition of irresolvability levels
Hewitt's paper [3,Theorem 20] for some j ∈ I, and by Hewitt's result there is a resolvable subspace is an open cover consisting of resolvable subspaces.Moreover, every open cover of X consists of resolvable subspaces since open subspaces of resolvable spaces are resolvable.
It is clear that R(τ) ∈ τ and that R(τ) is the union of all open resolvable subspaces of X.So, X is resolvable if and only if R(τ) = X.That is, locally resolvable spaces are resolvable.In fact, since Hewitt showed that the closure of a resolvable subspace is resolvable, we have that either R(τ) = X or Cl(R(τ)) = X.In this latter case, X is irresolvable.We have the following observation.Proposition 3.3.For a space (X,τ), the following are equivalent.
(2) R(τ ( G is an open dense hereditarily irresolvable subspace.And, if H was an open dense hereditarily subspace of X = F G, then H ∩ F = ∅.For otherwise, this subset would be both resolvable, being an open subspace of F, and irresolvable.Thus, (3)⇒(5).If A is resolvable then Cl(A) is resolvable.If A is not nowhere dense, then Int(Cl(A)) is open, nonempty, and resolvable which implies that Int(Cl(A)) ∩ G = ∅.Thus, Int(F) = ∅ which contradicts F being nowhere dense.
Definition 3.4.A space (X,τ) is homogeneous if for any pair x, y of distinct points, there is a homeomorphism h : X → X with y = h(x).
Topological groups are homogeneous spaces.
Evidently, homogeneity plus irresolvability implies strong irresolvability.But, this does not constitute a decomposition of strong irresolvability since homogeneity is not implied by strong irresolvability.
We introduce three conditions that a space (X,τ) may have (i) C 1 : every proper regular open set is irresolvable, (ii) C 2 : every nowhere dense set is irresolvable, (iii) C 3 : every irresolvable nowhere dense set is closed.We also label the levels of irresolvability as follows: (i) I: The space is irresolvable, (ii) SI: the space is strongly irresolvable, (iii) HI: the space is hereditarily irresolvable, (iv) S: the space is submaximal.If a label of a property is enclosed in brackets, the class of all spaces having the property is intended.For example, SI ⇒ C 1 can also be indicated by [ Example 3.6.Let (N,ρ) be the set of natural numbers with the cofinite topology.Then N is a crowded resolvable T 1 -space and it has property C 1 .For if U is a nonempty regular open set, U = Int(Cl(U)) = Int(X) = X.Thus, N does not have any nonempty proper regular open subset.So, N has C 1 but not I and hence neither SI nor HI.This space also has C 2 , for the nowhere dense sets are the finite subsets.Since infinite sets are dense and if a set F is finite, it is closed in the cofinite topology and has empty interior.Also the subspace (F,ρ | F) is discrete and every point is an isolated point in the subspace.For if x ∈ F, then U = N − (F − {x}) ∈ ρ and U ∩ F = {x} ∈ ρ | F. So, the subspace F is irresolvable.Since all nowhere dense sets are irresolvable and all nowhere dense sets are also closed, C 3 is satisfied as well.Proof.Clearly, HI ⇒ SI ∧C 2 .For the reverse implication, let (X,τ) be a C 2 ∧ SI space and let X = F G be the Hewitt decomposition.Then, G is a nonempty open hereditarily irresolvable subspace and Int(F) = ∅.It follows that F is nowhere dense and resolvable if nonempty.Evidently by C 2 , F = ∅ and X = G is HI.Theorem 3.9.S = HI +C 3 .
Proof.Certainly, S ⇒ C 3 ∧ HI.For the reverse implication let (X,τ) be a C 3 ∧ HI space and let D be dense in X.Then X − D is codense and hence nowhere dense and irresolvable subspace.By C 3 , X − D is closed and D is open.
It might be worth noting that the α-space property, X = X α , has a decomposition C 2 + C 3 .The α-space for the space (X,τ) is X α = (X,τ α ), where τ α = {U − E | U ∈ τ and E ∈ N(X,τ)} is the smallest expansion of τ for which all τ-nowhere dense sets are closed [8].Clearly, X = X α if and only if C 2 ∧ C 3 .It only remains to see that neither C 2 nor C 3 alone implies X = X α .In the example given earlier of a HI space which is not S, clearly C 2 holds but not C 3 for otherwise, so would S.
But, then to be irresolvable, F must be finite and hence closed.Thus, Y has C 3 .

Finite products
A basic problem that remained unsolved for several decades following Hewitt's discovery of irresolvable spaces was the question of irresolvability of finite product spaces.A property is said to be finitely productive if the product space X × Y has the property whenever both factor spaces X and Y have the property.It was incorrectly stated in [2], that strong irresolvability is finitely productive.This claim is strongly negated by the following simple counterexample.
Example 4.1.Let X be an infinite set, let F be a free ultrafilter on X, and equip X with the topology τ = F ∪ {∅}.Then X is a crowded submaximal T 1 -space.Let X 2 = X × X have the product topology π and let D = {(x, x) | x ∈ X} ⊆ X 2 be the diagonal subset.We will show that D is dense and codense in X 2 .Thus X = D (X − D) is resolvable being a disjoint union of dense sets.To see that D is dense, note that every nonempty open set Malyhin showed in [6] that any infinite sets X and Y have topologies for which these spaces are irresolvable T 1 -spaces and yet the product space X × Y is maximally resolvable.Maximal resolvability requires that the number of pairwise disjoint dense subsets that exist equals the least number of elements in a nonempty open set, which is the dispersion character.Malyhin also showed that if a free ultrafilter exists with the property that countably infinite intersections of its members are still members, then T 1 -spaces exist whose product is irresolvable.Malyhin [7] has shown that it is consistent with ZFC (Zermelo-Frankel + choice) set theory that all finite products of infinite crowded spaces are resolvable.

Nearly open sets
The collection of all almost open subsets is denoted AO(X,τ) and SO(X,τ) denotes the family of semiopen subsets of (X,τ).
It is easy to show that AO(X,τ) and SO(X,τ) are closed under arbitrary union but not generally under arbitrary intersection.It is also known that τ α = AO(X,τ) ∩ SO(X,τ).Further, A ∈ AO(X,τ) if and only if A = U ∩ D for some U ∈ τ and some dense subset D. Note that every dense set D is almost open.For any set

The collection of faintly open sets is FO(X,τ).
The nonempty faintly open sets are the noncodense sets.Evidently, complements of nonempty faintly open sets cannot be dense.In particular, a set is dense if and only if it intersects nonemptily every nonempty faintly open set.

Proposition 5.3. A space (X,τ) is irresolvable if and only if every dense set is faintly open. That is, D(τ) ⊆ FO(X,τ).
Proof.A space is irresolvable if and only if no dense set is codense.
Theorem 5.6.The following are equivalent for a space (X,τ).
In [1] a method is found for constructing connected Hausdorff crowded submaximal spaces.If (X,τ) is such a space, then τ = τ α = AO(X,τ) = τ A and AO(X,τ) SO(X,τ).This last inequality is forced by the fact that the space is Hausdorff and connected and hence not ED.

Functions and irresolvability
Definition 6.1.A bijection f : X → Y is a faint (semi)homeomorphism if both f and f −1 preserve faintly (semi)open sets.A property P transmitted by faint (semi)homeomorphism is a faint (semi)topological property.
It is clear that every semihomeomorphism is a faint homeomorphism so that faint topological properties are semitopological.The following example shows that not every faint homeomorphism is a semihomeomorphism.Example 6.2.Let (R,σ) be the Sorgenfrey line.That is, the set {[a, b) | a < b} is a base for σ.Also, let (R,τ) be the usual space of reals.Then, the identity function f : (R,τ) → (R,σ) is a faint homeomorphism but not a semihomeomorphism.For [0,1] = Cl τ ((0,1) is nowhere dense.So, f and, by symmetry of argument, f −1 preserve nowhere dense sets.
It is a corollary that X and X α share the same nowhere dense sets since the identity function f : X → X α is a faint homeomorphism.
Proof.If i : X → X α and j : Y → Y α are identity maps, then is a nonempty subset of f (A).The argument for inverse images is the same.Example 6.8.Let (R,τ) be the usual space of reals and let f : (R,τ) → (R,τ) be defined by It is known [9] that semitopological properties are precisely those properties shared by both X and X α .Apparently, both submaximality and hereditary irresolvability are not semitopological but strong irresolvability is semitopological.In fact, more can be said for strong irresolvability.Theorem 6.9.Strong irresolvability is a faint topological property.
Proof.Let f : X → Y be a faint homeomorphism, let X = ∅ be strongly irresolvable, and let E be dense in Y .Then which is a contradiction.So, Int(E) is dense and Y is strongly irresolvable.Remark 6.10.It may be noted that submaximality is preserved by open surjections, and hence expansions of submaximal topologies are submaximal.
It was shown in [5] that a space (X,τ) is submaximal if and only if τ = AO(X,τ).We extend this result slightly.Proposition 6.11.A space (X,τ) is submaximal if and only if τ = τ A .Proposition 6.12.A space (X,τ) is strongly irresolvable if and only if τ α = τ A .
Proof.X is SI if and only if X α is submaximal.

Call for Papers
Space dynamics is a very general title that can accommodate a long list of activities.This kind of research started with the study of the motion of the stars and the planets back to the origin of astronomy, and nowadays it has a large list of topics.It is possible to make a division in two main categories: astronomy and astrodynamics.By astronomy, we can relate topics that deal with the motion of the planets, natural satellites, comets, and so forth.Many important topics of research nowadays are related to those subjects.By astrodynamics, we mean topics related to spaceflight dynamics.
It means topics where a satellite, a rocket, or any kind of man-made object is travelling in space governed by the gravitational forces of celestial bodies and/or forces generated by propulsion systems that are available in those objects.Many topics are related to orbit determination, propagation, and orbital maneuvers related to those spacecrafts.Several other topics that are related to this subject are numerical methods, nonlinear dynamics, chaos, and control.
The main objective of this Special Issue is to publish topics that are under study in one of those lines.The idea is to get the most recent researches and published them in a very short time, so we can give a step in order to help scientists and engineers that work in this field to be aware of actual research.All the published papers have to be peer reviewed, but in a fast and accurate way so that the topics are not outdated by the large speed that the information flows nowadays.
Before submission authors should carefully read over the journal's Author Guidelines, which are located at http://www .hindawi.com/journals/mpe/guidelines.html.Prospective authors should submit an electronic copy of their complete manuscript through the journal Manuscript Tracking System at http://mts.hindawi.com/according to the following timetable: SI] ⊆ [C 1 ].The equation SI = C 1 + I denotes a decomposition of strong irresolvability into the join of two strictly weaker conditions, that is, [SI] = [C 1 ] ∩ [I] and [SI] [C 1 ] and [SI] [I].An earlier example showed that I SI.The following example shows that C 1 SI.It also shows that C 2 HI, and the same example shows that C 3 S.

Proposition 6 . 3 .Proposition 6 . 5 .
On the other hand f is a faint homeomorphism since each nonempty open interval (a,b) contains a nonempty right open interval [c,d) and vice versa.That is, each topology τ and σ is a π-base for the other.A π-base could be called a faint base.If f : X → Y is a bijection, f is a faint homeomorphism if and only if both f and f −1 preserve dense sets.Corollary 6.4.A composition of two faint homeomorphisms is a faint homeomorphism.Every faint homeomorphism directly and inversely preserves nowhere dense sets.Proof.If f : X → Y is a faint homeomorphism and E ⊆ X is nowhere dense, let V be any nonempty open subset of Y .Then, Int( f −1 (V )) = U = ∅ and so there exists a nonempty open subset U

Proposition 6 . 7 .
If f : X → Y is a faint homeomorphism, then direct and inverse images under f of nonempty almost open sets contain nonempty almost open sets.Proof.Suppose that A ⊆ X is almost open and nonempty.Then, A = U ∩ D for some nonempty open set U ⊆ X and for some dense subset D states that a space is resolvable if and only if every nonempty open subset contains a subset which is resolvable as a subspace.He proved this by transfinite induction.One immediate consequence is that a space is resolvable if and only if each nonempty basic open subset contains a resolvable subspace.Corollary 3.1.A space (X,τ) is resolvable if and only if there is an open cover of X consisting of resolvable subspaces.Proof.If X = i∈I U i with each U i an open resolvable subspace of X, and if a faint homeomorphism which is not a semihomeomorphism since images of open sets may be neither semiopen nor almost open.In particular, f (