UDC 519.50+519.54

A. G. EL’KIN

\(A\)-SETS IN COMPLETE METRIC SPACES

(Presented by Academician P. S. Aleksandrov, 4 X 1966)

As is known, every \(A\)-set in a complete metric space with a countable base is either at most countable, or contains a topological image of the Cantor perfect set (the Alexandrov–Hausdorff theorem \((^1,^2)\)).

Stone \((^3)\) proved the following theorem: every Borel set in a complete metric space (not necessarily separable) is either \(\sigma\)-discrete, or contains a Cantor perfect set.

We prove the following proposition:

Main theorem. An \(A\)-set in a complete metric space (not necessarily separable) is either \(\sigma\)-discrete, or contains a Cantor perfect set.

Remark 1. From this theorem there follow both the Alexandrov–Hausdorff theorem and Stone’s theorem.

Before proving our theorem, we introduce several necessary definitions and formulate a number of simple auxiliary propositions.

Definition 1. Let \(A\) be some set lying in a metric space \(X\). \(A\) is called relatively discrete if every point \(x \in A\) is an isolated point of the set \(A\); \(A\) is called discrete if \(A\) is relatively discrete and closed in \(X\); \(A\) is called metrically discrete if there exists such an \(\varepsilon < 0\) that for any two points \(x\) and \(y\) of \(A\), \(x \ne y\), one necessarily has \(\rho(x,y) \ge \varepsilon\).

Lemma 1 (see \((^3)\)). The following assertions concerning a subset \(A\) of a metric space \(X\) are equivalent:

1) \(A = \bigcup\limits_{i=1}^{\infty} A_i\), where \(A_i\) is relatively discrete, \(i = 1, 2, \ldots\)

2) \(A = \bigcup\limits_{i=1}^{\infty} B_i\), where \(B_i\) is discrete, \(i = 1, 2, \ldots\)

3) \(A = \bigcup\limits_{i=1}^{\infty} C_i\), where \(C_i\) is metrically discrete, \(i = 1, 2, \ldots\)

Definition 2. A subset \(A\) of a metric space \(X\) is called \(\sigma\)-discrete if \(A\) can be represented in at least one of the forms 1), 2), or 3). The empty set is always \(\sigma\)-discrete.

Remark 2. \(1^\circ\). The property of a set \(A\) of being \(\sigma\)-discrete is topologically invariant.

\(2^\circ\). Every subset of a \(\sigma\)-discrete set is itself \(\sigma\)-discrete.

\(3^\circ\). A \(\sigma\)-discrete space of weight \(\tau\) has cardinality \(\le \tau\). In particular, in a space of countable weight every \(\sigma\)-discrete set is at most countable.

It follows from this that the following properties of a set \(A\) are mutually exclusive: a) \(A\) is \(\sigma\)-discrete; b) \(A\) contains a Cantor perfect set.

\(4^\circ\). If a set \(A\) is not \(\sigma\)-discrete, then it is uncountable.

5°. The sum of a finite or countable number of \(\sigma\)-discrete sets is a \(\sigma\)-discrete set.

Definition 3. A point \(x \in A \subseteq X\) is called a point of local \(\sigma\)-discreteness of the set \(A\) if there exists a neighborhood \(Ox\) of this point \(x\) in \(X\) for which the set \(Ox \cap A\) is necessarily \(\sigma\)-discrete. A point of local \(\sigma\)-discreteness of the set \(A\) will also be called a \(\sigma\)-discrete point of the set \(A\). A point of the set \(A\) that is not a \(\sigma\)-discrete point of this set will be called a non-\(\sigma\)-discrete point of the set \(A\). The set of all points of local \(\sigma\)-discreteness of the set \(A\) will be denoted by \(A_{\sigma}\); the set of non-\(\sigma\)-discrete points of the set \(A\) will be denoted by \(A_{\bar{\sigma}}\). Obviously,
\[ A=A_{\sigma}\cup A_{\bar{\sigma}};\quad A_{\sigma}\cap A_{\bar{\sigma}}=\Lambda. \]

Definition 4. A space \(X\) is called locally \(\sigma\)-discrete if each of its points is a \(\sigma\)-discrete point of this space, i.e. if \(X_{\sigma}=\Lambda\). Obviously, a \(\sigma\)-discrete set is locally \(\sigma\)-discrete.

Lemma 2 (see \((^3)\)). A locally \(\sigma\)-discrete paracompact space is necessarily \(\sigma\)-discrete.

Remark 3. 1°. For every set \(A\) in a metric space \(X\), the set \(A_{\sigma}\) is always \(\sigma\)-discrete.

2°. If \(A\) is not \(\sigma\)-discrete, then \(A_{\bar{\sigma}}\) is also not \(\sigma\)-discrete. If, however, \(A\) is \(\sigma\)-discrete, then, obviously, \(A_{\bar{\sigma}}=\Lambda\). Thus either \(A_{\bar{\sigma}}=\Lambda\), or \(A_{\bar{\sigma}}\) is not \(\sigma\)-discrete (and hence is uncountable).

Lemma 3. Let \(G\) be an open set of the space \(X\), and let \(A \subseteq X\). If the set \(M=G\cap A_{\bar{\sigma}}\) is nonempty, then \(M\) is not \(\sigma\)-discrete, and moreover \(M_{\sigma}=\Lambda\).

Proof of Lemma 3. Let \(M\ne\Lambda\). Suppose the contrary, namely that \(M\) is \(\sigma\)-discrete. Let \(x\in M\). \(M=G\cap A_{\bar{\sigma}}\). Then \(G\) is a neighborhood of the point \(x\) in \(X\); the set
\[ G\cap A=(G\cap A_{\sigma})\cup M \]
is \(\sigma\)-discrete, which contradicts the fact that \(x\in A_{\bar{\sigma}}\). Thus the set \(M\) is not \(\sigma\)-discrete.

We now prove that \(M_{\sigma}=\Lambda\). Suppose the contrary, and let \(M_{\sigma}\ne\Lambda\). Take an arbitrary point \(y\in M_{\sigma}\). By the definition of the set \(M_{\sigma}\), there exists a neighborhood \(Oy\) of the point \(y\) in \(X\) for which the set \(Oy\cap M\) is \(\sigma\)-discrete. Since \(y\in G\), \(O_{1y}=G\cap Oy\) is a neighborhood of the point \(y\) in \(X\). Then
\[ O_{1y}\cap A=(O_{1y}\cap A_{\sigma})\cup(G\cap Oy\cap A_{\bar{\sigma}}) =(O_{1y}\cap A_{\sigma})\cup(M\cap Oy). \]

It follows from this that \(O_{1y}\cap A\) is necessarily \(\sigma\)-discrete, which contradicts the fact that \(y\in A_{\bar{\sigma}}\). Consequently, \(M_{\sigma}=\Lambda\). The lemma is proved.

Proof of the main theorem. In view of Remark 2, 3°, the formulation of the theorem is correct. The proof of this theorem repeats the arguments of Alexandrov—Hausdorff in the proof of the theorem on the cardinality of \(A\)-sets lying in complete spaces with a countable base (see \((^1,^2)\)). The difference is that, in constructing spherical neighborhoods, we shall use not condensation points but non-\(\sigma\)-discrete points. Thus, let \(A\) be an \(A\)-set lying in a complete metric space \(X\), i.e.
\[ A=\bigcup_{(i,k,l,\ldots)}(F_i\cap F_{ik}\cap F_{ikl}\cap\cdots), \]
where the summation extends over all possible sequences \((i,k,l,\ldots)\) of natural numbers, and all the sets \(F_{ikl\ldots s}\) are closed in \(X\). Obviously, one may assume that for each numerical sequence \((i,k,l,\ldots)\) the inclusions
\[ F_i\supseteq F_{ik}\supseteq F_{ikl}\supseteq\cdots \]
hold.

If we fix the first index, we obtain the sets

\[ A_i=\bigcup_{(k,l,\ldots)}(F_i\cap F_{ik}\cap F_{ikl}\cap\ldots),\qquad A_i\subseteq F_i,\qquad i=1,2,\ldots,\qquad A=\bigcup_{i=1}^{\infty}A_i . \]

When two indices are fixed, we obtain the sets

\[ A_{ik}=\bigcup_{(l,\ldots)}(F_i\cap F_{ik}\cap F_{ikl}\cap\ldots),\qquad A_{ik}\subseteq F_{ik},\qquad i=1,2,\ldots;\ k=1,2,\ldots, \]

\[ A_i=\bigcup_{k=1}^{\infty} A_{ik} \]

and so on.

Let \(A\) not be \(\sigma\)-discrete. We must prove that then it contains a Cantor perfect set. By virtue of remark 3, \(2^0\), the set \(A_{\bar\sigma}\) is not \(\sigma\)-discrete. Therefore (see remark 2, \(4^0\)) in this set one can choose two distinct points \(a_1\) and \(a_2\). Surround them by disjoint closed balls \(V_1\) and \(V_2\). The corresponding (i.e. having the same center and radius) open balls will be denoted by \(U_1\) and \(U_2\). Let \((p,q,r,\ldots)\) be a binary numerical sequence composed of the digits 1 or 2.

Consider the sets \(A_i\), \(A=\bigcup_{i=1}^{\infty}A_i\). Since \(A\) is not \(\sigma\)-discrete, there exist two natural numbers \(i_1\) and \(i_2\) (not necessarily distinct) such that \(A_{i_p}\) is not \(\sigma\)-discrete. At the same time, we may assume that \(i_p\) has been chosen in such a way that
\[ A_{i_p\bar\sigma}\cap U_p\ne \Lambda . \]
We prove this. Suppose the contrary; let, for example, for \(p=1\) and for every \(i=1,2,\ldots\) one have
\[ A_{i\bar\sigma}\cap U_1=\Lambda . \]
Then
\[ A\cap U_1=\bigcup_{i=1}^{\infty}(A_i\cap U_1)=\bigcup_{i=1}^{\infty}(A_{i\sigma}\cap U_1), \]
and, consequently, \(A\cap U_1\) is \(\sigma\)-discrete (see remarks 3, \(1^0\); 2, \(2^0\); 2, \(5^0\)). But this means that the point \(a_1\in A\cap U_1\) is a \(\sigma\)-discrete point of the set \(A\) (by definition 3), which contradicts the fact that \(a_1\in A_{\bar\sigma}\). Thus, we may assume that \(i_p\), moreover, has been chosen in such a way that \(A_{i_p\bar\sigma}\cap U_p\ne \Lambda\).

But if \(A_{i_p\bar\sigma}\cap U_p\ne \Lambda\), then, by lemma 3, this set is not \(\sigma\)-discrete. Therefore in this set one can choose two distinct points \(a_{p1}\) and \(a_{p2}\). Surround them by disjoint closed balls \(V_{p1}\) and \(V_{p2}\), \(V_{pq}\subseteq U_p\), \(q\) equal to 1 or 2. The corresponding open balls will be denoted by \(U_{p1}\) and \(U_{p2}\).

Now consider the sets \(A_{i_p k}\). We have
\[ A_{i_p}=\bigcup_{k=1}^{\infty} A_{i_p k}. \]
Since \(A_{i_p}\) is not \(\sigma\)-discrete, there exist two natural numbers \(k_1\) and \(k_2\) (not necessarily distinct) for which the set \(A_{i_p k_q}\) is not \(\sigma\)-discrete. At the same time, we may assume that \(k_q\) has been chosen in such a way that
\[ A_{i_p k_q\bar\sigma}\cap U_{pq}\ne \Lambda . \]
We prove this. Suppose the contrary; let, for example, for \(q=1\) and for every \(k=1,2,\ldots\) one have
\[ A_{i_p k\bar\sigma}\cap U_{p1}=\Lambda . \]
Then
\[ A_{i_p}\cap U_{p1}=\bigcup_{k=1}^{\infty}(A_{i_p k}\cap U_{p1}) =\bigcup_{k=1}^{\infty}(A_{i_p k\sigma}\cap U_{p1}), \]
and, consequently, the set \(A_{i_p}\cap U_{p1}\) is necessarily \(\sigma\)-discrete, whence it follows that \(a_{p1}\in A_{i_p\sigma}\), which cannot be, since \(a_{p1}\in A_{i_p\bar\sigma}\). Thus, \(k_q\) can be chosen in such a way that
\[ A_{i_p k_q\bar\sigma}\cap U_{pq}\ne \Lambda, \]
and \(A_{i_p k_q}\) is not \(\sigma\)-discrete. But if

\[ A_{i_p k_q\bar\sigma}\cap U_{pq}\ne \Lambda, \]

then by lemma 3 this set is not \(\sigma\)-discrete. Therefore in this set one can choose two distinct points \(a_{pq1}\) and \(a_{pq2}\), and so on. Thus,

to each binary sequence \((p, q, r, \ldots)\) there corresponds a sequence of natural numbers \((i_p, k_q, l_r, \ldots)\) such that all the sets
\(A_{i_p} \cap U_p,\ A_{i_p k_q} \cap U_{pq},\ A_{i_p k_q l_r} \cap U_{pqr}, \ldots\)
are nonempty (indeed, they are even uncountable). But then the sets

\[ F_{i_p} \cap V_p,\quad F_{i_p k_q} \cap V_{pq},\quad F_{i_p k_q l_r} \cap V_{pqr}, \ldots \]

will also be nonempty.

If the radii of the balls \(V_p, V_{pq}, V_{pqr}, \ldots\) are chosen respectively less than \(1,\ 1/2,\ 1/3,\ldots\), then the latter sets form a sequence of nonempty decreasing closed sets with diameters tending to zero, and, consequently, their intersection in the complete metric space is nonempty and consists of a single point \(x\), which belongs both to the set \(A\) and to the Cantor perfect set
\[ D=\bigcup_{(p,q,r,\ldots)} (V_p \cap V_{pq} \cap V_{pqr} \cap \ldots). \]
Since this holds for every binary sequence \((p, q, r, \ldots)\), every point \(x \in D\) is contained also in the set \(A\). Thus, \(D \subseteq A\), as was required to prove.

Additional properties of the set \(A_{\sigma}\).

\(1^\circ.\) \(A_{\sigma}\) is closed in \(A\).

\(2^\circ.\) \(A_{\sigma}\) is dense in itself.

\(3^\circ.\) From \(1^\circ\) and \(2^\circ\) it follows that \(A_{\sigma}\) is perfect in \(A\).

\(4^\circ.\) \(A_{\sigma\sigma}=\Lambda\).

\(5^\circ.\) Every point \(x \in A_{\sigma}\) is a condensation point of this set (and hence of the set \(A\)).

The following simple result holds.

Theorem. A scattered set is \(\sigma\)-discrete.

Remark 4. A set is called scattered if it contains no (nonempty) dense-in-itself subset.

This work was written under the supervision of V. I. Ponomarev, to whom I express my sincere gratitude.

Moscow State University
named after M. V. Lomonosov

Received
9 VII 1966

REFERENCES

1. P. S. Aleksandrov, C. R., 162, 323 (1916).
2. F. Hausdorff, Set Theory, 1937. A. H. Stone, Am. Math. J., 85, No. 4, 655 (1963).