UDC 513.83

MATHEMATICS

E. D. KHALIMSKII

ON TOPOLOGIES OF GENERALIZED INTERVALS

(Presented by Academician P. S. Novikov on 21 IV 1969)

§ 1. By an interval, an ordered set, and a mapping we shall mean, respectively, a generalized continuum interval \((^3)\)*, a linearly ordered set, and a continuous mapping. A \(\Pi\)-topology on an ordered set \(C\) is a topology under which \(C\) is an interval, one of the orders induced by connectedness \((^3)\) in which coincides with the given one.

Lemma 1. Let \((A,B)\) be such a cut in an ordered set \(C\) that \(A\) \((B)\) has no greatest (least) element. In order that the topology on \(C\) be a \(\Pi\)-topology, it is necessary that \(A\) \((B)\) be an open set in this topology.

Theorem 1. If in an ordered set \(C\) there are gaps, then it is impossible to introduce a \(\Pi\)-topology on \(C\).

Theorem 2. If for a point \(a \in \vec C\) every set of the form \(]a_1,a[\) \((]a,a_1[)\), where \(a_1<a\) \((a_1>a)\), is infinite, then \(\{a\}\) is a closed set in \(C\).

Theorem 3. If in an ordered set \(C\) there are neither gaps nor jumps, then every \(\Pi\)-topology** on \(C\) majorizes the interval topology; hence it is a \(T_2\)-topology.

Theorem 4. If there exists a partition of an ordered set \(C\) into three nonempty subsets \(A_1,A_2,A_3\) such that \(a_1<a_2<a_3\) \((a_1,a_2,a_3\) are arbitrary elements of the sets \(A_1,A_2,A_3\), respectively), and \(A_1\) has no greatest element, \(A_3\) has no least element, while the set \(A_2\) consists of a finite and, moreover, even number of elements, then it is impossible to introduce a \(\Pi\)-topology on \(C\).

An element \(a\) of a subset \(U\) of an ordered set \(C\) will be called an end element of the set \(U\) in \(C\) if in \(C\) there exists an interval of one of the forms \([a_1,a[\) or \(]a,a_1]\), whose intersection with \(U\) is the empty set.

Theorem 5. If \(a\) is an end element of a set \(U\) open (closed) in the interval \(\vec C\), then the set \(\{a\}\) is open (closed) in \(\vec C\).

Corollary 1. If \(\{a\}\) is a closed (open) set in \(\vec C\) and there exists a point \(a_1 \in \vec C\) such that \(]a_1,a[=\varnothing\) or \(]a,a_1[=\varnothing\), then every set \(U\) open (closed) in \(\vec C\), containing \(a\), also contains \(a_1\).

Corollary 2. If a point \(a\) is not an end of the interval \(\vec C\) and \(\{a\}\) is open in \(\vec C\), then in \(\vec C\) there exist points \(a_1\) and \(a_2\) such that \(]a_1,a_2[=\{a\}\); the closure of \(\{a\}\) in this case is \([a_1,a_2]\). If \(a\) is the first (last) element of \(\vec C\) and \(\{a\}\) is an open set in \(\vec C\), then in \(\vec C\) there exists a point \(a_1\) such that \(]\leftarrow,a_1[=\{a\}\) \((]a_1,\rightarrow[=\{a\})\); the clos—

* In the definition of an interval in \((^3)\) one should add the item: 3. If the connected component \(V_a\) of the space \(A-\{a\}\) contains the connected component \(V_b\) of the space \(A-\{b\}\) or an end \(a_i\) of the interval \(A\), then the connected component \(V_c\) of the space \(A-\{c\}\), containing exactly \(V_b\) or \(a_i\), is either itself contained in \(V_a\), or contains \(V_a\) \((a \in A,\ b \in A,\ c \in A)\).

** The existence of a \(\Pi\)-topology follows from \((^3)\).

tion $\{a\}$ in this case is equal to $[a,a_1]\setminus([a_1,a])$ (the existence of each of the points $a_1,a_2$ follows from Theorem 2).

Theorem 6. In order that a $\Pi$-topology can be introduced on an ordered set $C$, it is necessary and sufficient that there be no gaps in $C$ and that there exist no partition of the set $C$ satisfying the condition of Theorem 4.

We shall prove sufficiency. For each $a\in C$ there are only two possibilities: I. in $]\leftarrow,a[$ there is no greatest element, or in $]a,\to[$ there is no least element. II. In $]\leftarrow,a[$ there is a greatest element and in $]a,\to[$ there is a least element.

In case I, $a$ is a limit element; in case II, $a$ is an isolated element of the set $C$. We divide the set of all isolated elements into equivalence classes, assigning to one equivalence class all and only those elements between any two of which there is not a single limit element. Choose one representative $a$ from each equivalence class. There are only the following possibilities for $a$: 1) there exists a limit element $a'$ such that the number of elements between $a$ and $a'$ is finite; 2) there exists a sequence
\[ \cdots<a_{-n}<\cdots<a_{-2}<a_{-1}=a=a_1<a_2<\cdots<a_n<\cdots \]
such that
\[ ]a_i,a_j[=\bigcup_{i<k<j} a_k \]
\[ (n=1,2,\ldots,<\omega); \]
3) there exists a sequence
\[ a_{-m}<a_{-m+1}<\cdots\cdots<a_{-1}=a=a_1<a_2<\cdots<a_n<\cdots \]
such that
\[ ]\leftarrow,a[=\bigcup_{i=1}^{m} a_{-i} \]
and
\[ ]a_i,a_j[=\bigcup_{i<k<j} a_k \]
\[ (n=1,2,\ldots,<\omega); \]
4) there exists a sequence
\[ \cdots<a_{-n}<\cdots\cdots<a_{-2}<a_{-1}=a=a_1<a_2<\cdots<a_m \]
such that
\[ [a,\to[=\bigcup_{i=1}^{m} a_i \]
and
\[ ]a_i,a_j[=\bigcup_{i<k<j} a_k \]
\[ (n=1,2,\ldots,<\omega); \]
5) there exists a sequence
\[ a_{-m}<\cdots<a_{-1}=a=a_1<a_2<\cdots<a_n \]
such that
\[ C=\bigcup_{-m<i<n} a_i . \]

We introduce a topology in $C$. We declare the following sets open in $C$: $]\leftarrow,a[$ and $]a,\to[$ in cases I and 1), if the number of elements between $a$ and $a'$ is odd; $]\leftarrow,a]$ and $[a,\to[$ in case 1), if the number of elements between $a$ and $a'$ is even; in cases 2)—5), either sets of the form $]\leftarrow,a_{\pm(2k-1)}]$ and $]a_{\pm(2k-1)},\to[$, or sets of the form $]\leftarrow,a_{\pm 2k}[$ and $]a_{\pm 2k},\to[$, indifferently which $(k=1,2,\ldots)$.

The open sets described are taken as a subbase of a topology, which we shall call interval-alternating (i.a.). From the definition of the i.a. topology it follows that for each point $a$ of $C$ the sets $]\leftarrow,a[$ and $]a,\to[$ are either both open or both closed in $C$. Hence, if $C_1\subset C$ and in $C$ there exist points $a_1<a_2<a_3$ such that $a_1$ and $a_3\in C_1$, $a_2\notin C_1$, then the nonempty sets $]\leftarrow,a_2[\cap C_1$ and $]a_2,\to[\cap C_1$ are either both open or both closed in $C_1$, which indicates that $C_1$ is disconnected. We shall show that a subspace $C_2$ of the space $C$ is connected if, for any points $a_1$ and $a_2$ of $C_2$ $(a_1\ne a_2)$, every point of $C$ lying between them belongs to $C_2$. Let $U$ be a nontrivial open-and-closed subset of $C_2$. From the definition of the i.a. topology it follows that every closed nontrivial subset $U$ in $C_2$ has at least one end element $a_0$ in $C_2$; moreover, if $a_0$ is an end element of a subset closed (open) in $C_2$, then $\{a_0\}$ is closed (open) in $C_2$. Hence, by supposition, there exists an end element $a_0$ of the set $U$ in $C_2$, and $\{a_0\}$ is an open-and-closed set in $C_2$. Consequently, there exist sets $U_1$, closed, and $U_2$, open in $C$, such that $U_i\cap C_2=\{a_0\}$ $(i=1,2)$. It is obvious that $a_0$ is an end element of $U_i$ in $C$ $(i=1,2)$. Hence $\{a_0\}$ is an open-and-closed set in $C$, which contradicts the definition of the i.a. topology. Consequently, $C_2$ is connected, and from Proposition 1 in (3) we conclude that the i.a. topology is a $\Pi$-topology.

Corollary 1. Every i.a. topology is uniquely determined by the set of all one-sided open sets in this topology.

Corollary 2. For every Π-topology on an ordered set \(C\) there exists a unique i.c. topology majorized by it.*

Corollary 3. The interval \([a_1,\ a_2]\) with the i.c. topology is a quasicompact space \((^{1})\).

This follows from Alexandroff’s theorem in \((^{2})\).

Corollary 4. If an ordered set \(C\) can be endowed with a quasicompact Π-topology, then such a topology on \(C\), generally speaking, is not unique, and any two such topologies on \(C\) are incomparable \((^{1})\).

Theorem 7. In order that the topology of the interval \(\vec C\) be an i.c. topology, it is necessary and sufficient that the connected components of every set \(U\) open in \(\vec C\) be sets open in \(\vec C\).

Theorem 8. In order that the topology of the interval \(\vec C\) be an i.c. topology, it is necessary and sufficient that, for every \(a \in \vec C\) such that \(\{a\}\) is a closed set in \(\vec C\), and for every set \(U\) open in \(\vec C\) and containing \(a\), one of the following conditions hold: 1) if \(a\) is not an end of the interval \(\vec C\), then in \(\vec C\) there exists an interval \([a_1,\ a_2]\) \((a_1<a<a_2)\) such that \([a_1,\ a_2]\subset U\); 2) if \(a\) is the first (last) element of \(\vec C\), then in \(\vec C\) there exists an interval \(]\leftarrow,\ a_1]\) (\([a_2,\ \to[\)), where \(a_2<a<a_1\), such that \(]\leftarrow,\ a_1]\subset U\) (\([a_2,\to[\subset U\)).

Theorem 9. In order that a topology majorizing some i.c. topology in an ordered set \(C\) be a Π-topology, it is necessary and sufficient that the sets \(U\) open in this topology satisfy the following two conditions: 1. In \(U\) there is no end element \(a_0\) such that \(\{a_0\}\) is a closed set in \(C\). 2. The nonempty complement** of the nonempty intersection of the set \(U\) with an arbitrary interval to the whole interval is not a set open in this interval.

Theorem 10. If in an ordered set \(C\) there is at least one limit point \(a_0\) and there exists at least one Π-topology on \(C\), then on \(C\) there exist infinitely many Π-topologies that are not i.c. topologies.

Theorem 11. In order that a connected topological space, distinct from the antidiscrete two-point set, be an interval, it is necessary and sufficient that there exist a condensation of it onto some interval endowed with an i.c. topology.

We shall call a topological space \(X\) orderable if on it there exists a transitive relation \(<\), called an order, satisfying the following conditions: 1. If \(x,y\in X\), then one and only one of the relations \(x<y\), \(x=y\), \(y<x\) holds. 2. Each connected component \(A\) of the space \(X\) is an interval \(\bar A\), and the order induced by the relation \(<\) on \(\bar A\) coincides with one of the orders induced by the connectedness of the interval \(\bar A\). 3. If \(x,y\) belong to distinct connected components \(A(x)\) and \(A(y)\) of the space \(X\) and \(x<y\), then \(x'<y'\) for all \(x'\in A(x)\) and \(y'\in A(y)\).

Theorem 12. If the space \(X\) is orderable by a relation \(<\) in such a way that, under this ordering, in the set of points of the space \(X\) there are neither gaps nor jumps, then for \(X\) there exists an interval \(\bar A\), endowed with an interval topology, such that \(X\) condenses onto \(\bar A\). If the topological space \(X\) condenses onto some interval \(\bar A\), then \(X\) is an orderable space.

Theorem 13. In order that every Π-topology in an ordered set \(C\) be an i.c. topology, it is necessary and sufficient that every cut in \(C\) be a jump (except for the antidiscrete two-point set).

§ 2. We introduce the notion of the complete connected oriented union of arbitrary non-one-point intervals distinct from the antidiscrete

* Except for the case when \(\vec C\) is the antidiscrete two-point set \((^{2})\).

** In item 3b) in Theorem 2 in \((^{3})\) one should also take only the nonempty complement.

semicolons. Take a family of segments $\{\vec A_\lambda\}$, where $\lambda$ ranges over the ordered set $\Lambda$. To each point $a \in \vec A_\lambda$ assign the pair $(\lambda,a)$. Order the set of all pairs by the lexicographic rule. Identify each pair of points $(\lambda_1,a_1)$ and $(\lambda_2,a_2)$ such that $] (\lambda_1,a_1);(\lambda_2,a_2) [ = \varnothing$, in the following cases: 1) each of the sets $\{a_1\}$, $\{a_2\}$ is closed respectively in $\vec A_{\lambda_1}$, $\vec A_{\lambda_2}$; 2) each of the sets $\{a_1\}$, $\{a_2\}$ is open respectively in $\vec A_{\lambda_1}$, $\vec A_{\lambda_2}$. Denote the resulting set by $C'$.

Consider the set of all cuts $(A,B)$ in the set $C'$. If in $A$ ($B$) there is no greatest (least) element, while in $B$ ($A$) there is a least (greatest) element $(\lambda,a)$ such that $\{a\}$ is an open subset of $\vec A_\lambda$, or the cut $(A,B)$ is a gap, then add to $C'$ an element $\alpha$, which we place immediately after every element of $A$ and before every element of $B$. Denote the resulting ordered set by $C$. Obviously, for each ordered set $\vec A_\lambda$ there exists a natural similarity correspondence $f_\lambda:A_\lambda \to C_\lambda \subset C$. For each $\lambda \in \Lambda$, declare open in $C'$ the image under $f_\lambda$ of every set open in $\vec A_\lambda$ that contains no endpoint $a_\lambda$ of the segment $\vec A_\lambda$ such that $\{a_\lambda\}$ is a closed set in $\vec A_\lambda$ and $f_\lambda(a_\lambda)$ is neither the first nor the last element of the set $C$; also declare open all sets open in the following o.t. topology in $C$, in which the open one-point sets in $C$ are precisely those one-point sets each of which is the image under some $f_\lambda$ of a set open in $\vec A_\lambda$, and all sets of the form $U_1 \cup U_2$, where $U_1$ is the image under some $f_\lambda$ of a set open in $\vec A_\lambda$ containing such an endpoint $a_\lambda$ of the segment $\vec A_\lambda$ that $\{a_\lambda\}$ is a closed set in $\vec A_\lambda$, and $U_2$ is a set open in the above-described o.t. topology in $C$ such that $a_\lambda$ is an endpoint of $C' - U_2$ in $C$ and $C_\lambda \subset C - U_2$. We take all the listed open sets as a prebase for the topology in $C$.* From Theorem 9 and Corollaries 1 and 2 to Theorem 6 it follows that a P-topology has been obtained. The segment $\vec C$ constructed in this way will be called the complete connected oriented union of the segments $\vec A_\lambda$, and we shall write
\[ \vec C=\overline{\bigcup_{\lambda \in \Lambda}}\vec A_\lambda . \]
The topology constructed is the weakest P-topology in $C$ for which every $f_\lambda$ ($\lambda \in \Lambda$) is a homeomorphism.

Moscow State Pedagogical Institute
named after V. I. Lenin

Received
14 IV 1969

REFERENCES CITED

$^1$ N. Bourbaki, General Topology, Moscow, 1968.
$^2$ John L. Kelley, General Topology, Moscow, 1968.
$^3$ E. D. Khalimskii, Dokl. Akad. Nauk SSSR 185, No. 2 (1969).**

* The topology in $\overline{\cup \vec A_\lambda}$ in (3) should be introduced in the same way as here.

** The following corrections must be made in paper (3).
1. Two mappings $f:\dot e_n \times \vec A \to X$ and $\bar f:\dot e_n \times \overleftarrow A \to X$ from the class of mappings under consideration should be regarded as mutually opposite if $f(x)=\bar f(x)$ $(x \in \dot e_n \times \vec A;\ \overleftarrow A$ differs from $\vec A$ only by orientation). 2. In Proposition 3, instead of the words “neither one nor $\{e\}$ equivalence classes,” one should read, respectively, “neither one, $e$.” In the footnote to Proposition 3, instead of $\{f\}$ and $\{f_i\}\ne\{e\}$, one should read, respectively, $f$ and $f_i\ne e$.