TWO COUNTABLE , BICONNECTED , NOT WIDELY CONNECTED HAUSDORFF SPACES

We construct two countable, biconnected spaces, not widely connected, not having a dispersion point, and not being strongly connected. The first is Hausdorff and the second is Urysohn and almost regular.


Introduction.
The first example of a biconnected space with a dispersion point was constructed by B. Knaster and K. Kuratowski in [23], and the first example of a biconnected space without a dispersion point by E. W. Miller in [26].Two stronger examples of biconnected spaces without a dispersion point were constructed by M. E. Rudin in [30,31].The example in [30] has the property that the complement of every connected subset containing more than one point is at most countable and the example in [31] has the property of being widely connected.All spaces in [26,30,31], are subsets of the plane.The first two are constructed under the Continuum Hypothesis and the third one under Martin's Axiom.In [7], G. Gruenhage constructed a countable connected Hausdorff space under Martin's Axiom and a perfectly normal connected space under the Continuum Hypothesis in which the complement of every connected subspace containing more than one point is finite.In [36], we constructed a countable widely connected Hausdorff space and a countable widely connected and biconnected Hausdorff space.Now, we construct two countable spaces which are biconnected without being widely connected and without a dispersion point.The first is Hausdorff, and the second is Urysohn almost regular.In addition, as it is the case with widely connected spaces and spaces with a dispersion point, both have the property of not being strongly connected [13].The construction is based on a modification of [16] or [20].It can be also based on [37].From the construction, it follows that there exist 2 c non-homeomorphic such spaces.
A space X is called (1) Urysohn if every pair of distinct points of X have disjoint closed neighborhoods.
(2) Almost regular if X contains a dense subset at every point of which the space is regular.
A connected space X is called (1) Biconnected (K.Kuratowski [24]) if it admits no decomposition into two connected disjoint proper subsets containing more than one point.
A point x of a connected space X is called (1) A cut point if X\{x} is disconnected.
(2) A dispersion point if X\{x} is totally disconnected.A connected space (X, τ) is called (1) Maximal connected if, for every strictly finer topology σ , the space (X, σ ) is not connected.

Results
The space T .For the construction of the countably, biconnected and not widely connected Hausdorff space T , we first construct an appropriate countable Hausdorff totally disconnected space X containing a specific point p and a closed discrete subspace N which cannot be separated by disjoint open sets.Then keeping fixed the subspace N and condensing the point p (instead of condensing pairs of points as in [16,20], or [37]), we construct the space T .
On the set where N is the space of natural numbers, we define the following topology: every point a ki is isolated.For the points of N a basis of open neighborhoods in X is defined as follows: let ᏼ be a free ultrafilter on N and let ᏼ k be the copy of ᏼ in {a ki : i = 1, 2,...}.
If U ∈ ᏼ, we denote the copy of U in {a ki : i = 1, 2,...} by U k .Then, for every k ∈ N, a basis of open neighborhoods is the collection of sets For the point p, a basis of open neighborhoods is the collection of sets Obviously, the space X is Hausdorff and totally disconnected but not regular since the point p and the closed subset N cannot be separated by disjoint open sets.
We observe that every basic open neighborhood of p is defined by some U ∈ ᏼ, and every U ∈ ᏼ defines a basic open neighborhood U(p).Obviously, U(p)\U(p) = U .
Let X 1 (n), n = 1, 2,... be disjoint copies of X and let N 1 (n) and p 1 (n) be the copies of N and p, respectively, in X 1 (n).The copies of U(k) and U(p) in X 1 (n) are denoted by U(k 1 (n)) and U(p 1 (n)), respectively.Since the set P 1 = {p 1 (n) : n = 1, 2,...} and the dense subset D = X\N ∪ {p} of isolated points of X are countable, there exists one-to-one function f 1 of P 1 onto D. We attach the spaces X 1 (n), n = 1, 2,... to the space X identifying simultaneously each point p 1 (n) with the point On the set we define the following topology: every point of T 1 \X is isolated.For every k ∈ N, a basis of open neighborhoods is the collection of sets (2.5) For every isolated point x of X, a basis of open neighborhoods is the collection of sets where For the point p, a basis of open neighborhoods is the collection of sets It can be easily proved that the space T 1 is Hausdorff, totally disconnected, and contains the space X as a closed nowhere dense subset.We observe that every basic open neighborhood in T 1 , of every x ∈ X is defined by some U ∈ ᏼ, and every Furthermore, for every pair of points x, y of D and every basic open neighborhoods which implies that every continuous real-valued function of T 1 is constant on D and, hence, on X since D is dense in X.
We construct by induction the spaces T 2 ,T 3 ,...,T m , where and where X m−1 (n), n = 1, 2,... are disjoint copies of the initial space X, and , where f m−1 is one-to-one function of the set It can be easily proved that the space T m is Hausdorff, totally disconnected, and contains the space T m−1 as a closed nowhere dense subset.We observe that every basic open neighborhood in T m , of every x ∈ T m−1 is defined by some U ∈ ᏼ, and every U p m (j) \ p m (j) . (2.9) A basis of open neighborhoods of t in T is the collection of sets (2.10) If t ∈ T \N, then either t ∈ X\N or t is an isolated point of T l , l = 1, 2,..., where l is the minimal integer for which t ∈ T l .
In the first case, we first consider the basic open neighborhood A basis of open neighborhoods of t in T is the collection of sets (2.12) In the second case, we first consider the basic open neighborhood O 1 U (t) of t in T l and then its corresponding basic open neighborhood in T l+m , U p l+m (j) \ p l+m (j) . (2.13) A basis of open neighborhoods of t in T is the collection of sets From the definition of topology on T , it follows that, for every t ∈ T and for every U ∈ ᏼ, the set O U (t) is open-and-closed in T \N and that O U (t)\O U (t) = U .

Proposition 1. The space T is countable biconnected Hausdorff not widely connected and without a dispersion point.
Proof.That T is countable Hausdorff is obvious.To prove that T is connected, we consider two arbitrary points x, y of T and let m be the minimal integer for which both x, y ∈ T m .But then every continuous real-valued function of T m+1 is constant on T m and, hence, for every continuous real-valued function g of T , g(x) = g(y), which implies that T is connected.
Suppose now that T is not biconnected and let A, B be two connected, proper disjoint That T is not widely connected is obvious observing that, for every U ∈ ᏼ and every t ∈ T , the subset O U (t) is connected.That T has no dispersion point is obvious by its construction.
Corollary 1.The space T is not strongly connected.
Proof.let τ denote the topology on T and let τ max denote a maximal connected topology finer that τ.By [13, Cor.14A], it follows that the space (T , τ max ) has infinitely many cut points.Hence, if t is such a point, then there exist two disjoint subsets K and L such that K and L are open-and-closed in T \{t}, contain more than one point, and K ∪ L = T \{t}.Since the sets K ∪ {t},L ∪ {t}, are connected in (T , τ max ), they are also connected in (T , τ).But by the proof of Proposition 1, it follows that, for every pair of connected subsets of (T , τ), which contain more than one point, their intersections include a member of ᏼ.Therefore, the set (K ∪{t})∩(L∪{t}) = {t} must be a member of ᏼ, which is impossible.
Corollary 2. There exists 2 c mutually non-homeomorphic countable biconnected Hausdorff spaces not widely connected and without a dispersion point.
Proof.Because [19,Thm. 10], there exists 2 c different types of free ultrafilters on the discrete subspace N of the initial space X.
The space S. For the construction of the countable biconnected Urysohn almost regular space S, we first construct an appropriate countable Urysohn almost regular non-regular space and then, using the method of F. B. Jones [17], we construct a space Y having the additional property of containing a point ∞ at which the space Y is regular.The condensation process of this regular point is the same as in the construction of the space T .
We consider the initial space X and, for every n ∈ N, we consider a sequence b ni i∈N converging to n and consisting of isolated points not belonging to X.We set B = {b ni : n, i = 1, 2,...} and we consider the space C = X ∪ B. Let C 1 , C 2 be disjoint copies of C and let p 1 , p 2 and N 1 , N 2 be the copies of p and N in C 1 , C 2 , respectively.We attach the space C 1 to C 2 identifying the point p 1 with p 2 .We set q = p 1 = p 2 and we consider the space Z = (C 1 \{p 1 }) ∪ {q} ∪ (C 2 \{p 2 }) which is obviously Hausdorff but not regular since the point q and the closed subset N 1 ∪ N 2 = K cannot be separated by disjoint open sets.
Let Z n , n = 1, 2,... be disjoint copies of Z and let ∞ n=1 Z n be their disjoint union (topological sum).We add one more point r and, on the set L = {r } ∪ ∞ n=1 Z n , we define a basis of open neighborhoods of r as follows: we consider the copies B 1 , B 2 of B in C 1 , C 2 , respectively.We set B 1 ∪ B 2 = R and let R n , n = 1, 2,... be the copy of R in Z n .Let be a free ultrafilter on the closed discrete subspace Q = {q n : n = 1, 2,...}, where q n is the copy of q in Z n .Then, for every U ∈ , a basis of open neighborhoods of r is the collection of sets U(r ) = {r }∪{∪R i : q i ∈ U}.
It can be easily verified that the space L is Urysohn but not normal since the closed subsets Q and ∞ n=1 K n , (K n is the copy of K in Z n ) cannot be separated by disjoint open sets.Also, the subsets ∞ n=1 K n , and the point r cannot be separated by disjoint open sets, while Q and r can be separated by disjoint open sets but not by disjoint open-and-closed sets.However, L is not regular at r .Since the closed subsets Q and {r } of L cannot be separated by a continuous real-valued function, it follows that if we consider L n , n = 1, 2,... disjoint copies of L, we can apply the construction in [17] and obtain a space Y with the following properties (1) It is countable Urysohn containing a dense subset of isolated points.
(2) It contains a point ∞ at which Y is regular.
(3) The point ∞ and each copy Q n , n = 1, 2,... of the subset Q, in L n cannot be separated by disjoint open-and-closed subsets, that is they cannot be separated by a continuous real-valued function of Y .Proposition 2. There exists 2 c mutually non-homeomorphic countable biconnected Urysohn almost regular spaces, not widely connected, not having a dispersion point, and not being strongly connected.
Proof.We imitate the condensation process that we used in the construction of the space T using the space Y in place of the space X and the point ∞ and the set Q 1 in place of p and N, respectively.Let S m , m = 1, 2,... and S = ∞ m=1 S m be the corresponding spaces to T m and T , respectively.It can be easily proved that S is Urysohn.Since the different copies of the regular point ∞ are attached in each step to the isolated points of S m , m = 1, 2,..., it follows that these points remain regular in the final space S. Obviously, the set of all these points is dense in S and, hence, S is almost regular.
All the other properties of S are proved as in Proposition 1 and Corollaries 1 and 2.
Remark.In [37], E. K. van Dowen constructed a regular space with a dispersion point on which every continuous real-valued function is constant.We can modify his method and construct a countable biconnected Hausdorff space not widely connected, not having a dispersion point, and not being strongly connected.For this, we consider again the initial space X and let X i , i = 1, 2,... be disjoint copies of X.We denote by x i the copy of x ∈ X in X i , and by N i the copy of N. We attach the spaces X i , i = 2, 3,... to X 1 identifying each copy N i with N 1 , that is by putting each n i to n 1 .We denote this point by n.In the space Z = N ∪ ∞ i=1 (X i \N i ), the subset P = {p i : i = 1, 2,...} and the subset D consisting of all isolated points of the copies X i are countable and, therefore, there exists a one-to-one function g of P onto D. On the quotient space T X = N ∪{(p i ,g(p i )) : i = 1, 2,...}, we define a second topology τ in a similar manner as in the construction of the space T .Obviously, the topology τ is weaker than the quotient topology of T X .It can be proved, as in Proposition 1 and Corollaries 1 and 2, that (T X ,τ) is the required space.
In a similar manner, we can construct a Urysohn almost regular space having all the above properties.For this, it suffices to consider space Y as the initial space.
Furthermore, for every pair x, y of isolated points of T m−1 and every basic open neighborhood O m U (x), O m V (y), U, V ∈ ᏼ of x, y respectively, in T m , it holds that O m U (x) ∩ O m V (y) = ∅, which implies that every continuous real-valued function of T m is constant on the set of isolated points of T m−1 and, hence, it is constant on T m−1 since this set is dense in T m−1 .Finally, we consider the set T = ∞ m−1 T m on which we define the following topology : If t ∈ N, we first consider the basic open neighborhood O 1 U (t) of t in T 1 and then its corresponding basic open neighborhood in T m , subsets containing more than one point and A ∪ B = T .By the construction of the space T , it follows that T \N is totally disconnected.Hence, there exists b ∈ B\N.Let O U(b) be the basic open neighborhood of b defined by some U ∈ ᏼ.Suppose thatO U (b) ∩ B ∩ N = W = ∅.If W ∈ ᏼ,then N\W ∈ ᏼ and, hence, for the set O N\W (b), it holds that O N\W (b)∩N = N\W .Therefore, O U (b) ∩ O N\W (b)∩B ∩N = ∅, which implies that the set O U (b) ∩ O N\W (b) ∩ B is open-and-closed in B. Consequently, B ⊆ O U (b) for every U ∈ ᏼ and, hence, B is a singleton, which is a contradiction.Hence, W ∈ ᏼ.But then if we consider a point a ∈ A\N and the basic open neighborhood O U (a) of a, it follows, in a similar manner, that the relation O U (a)∩A∩N = V = ∅ implies that V ∈ ᏼ, which is impossible because B ∩ A = ∅.Therefore, either O U (a) ∩ A ∩ N = ∅ or O U (b) ∩ B ∩ N = ∅.Since O U (a)\O U (a) ⊆ N and O U (b)\O U (b) ⊆ N, it follows that either O U (a)∩N is open-and-closed in A or O U (b)∩N is open-and-closed in B. Hence, either the subset A is a singleton or not connected, or the subset B is a singleton or not connected.