UDC 513.831

MATHEMATICS

V. GOL’SHTINSKII

ZERO-DIMENSIONAL STATIONARY SUBSETS OF FINITE-DIMENSIONAL METRIC SPACES

(Presented by Academician P. S. Aleksandrov on 1 II 1966)

Definition. A subset \(A\) of a metrizable topological space \((X,T)\) will be called stationary if there exists on the set \(X\) a metric \(\rho\) compatible with the topology \(T\) (i.e., a metric satisfying the condition \(T_\rho=T\), where \(T_\rho\) is the topology induced by the metric \(\rho\)), and moreover the following condition is fulfilled:

\((U)\). For every continuous mapping \(f:X\to X\) for which \(\rho(x,f(x))<1\) for all points \(x\in X\), the relation
\[ A\cap f(X)=\varnothing \]
holds.

Let us note that if the space \((X,\rho)\) is a compact metric space, then the set \(A\subset X\) is stationary in the topological space \((X,T_\rho)\) if and only if there exists a number \(\varepsilon>0\) such that, for every continuous mapping \(f:X\to X\) satisfying the condition \(\rho(x,f(x))<\varepsilon\) for all points \(x\in X\), one has
\[ A\cap f(X)\ne\varnothing . \]

In this note we prove the existence of “small” stationary sets in the sense indicated below.

Theorem. In every finite-dimensional (in the sense of \(\dim\)) metrizable space \((X,T)\) there exists a zero-dimensional closed stationary subset.

Proof. Let \(\dim X=n\), where \(n\) is a natural number (the cases \(n=1\) and \(n=0\) are excluded as trivial). Then, as is known, there exist closed sets \(A_i\subset X\), \(i=\pm1,\pm2,\ldots,\pm n\), such that for any system of closed partitions \(F_i\) between the sets \(A_{-i}\) and \(A_i\), \(i=1,2,\ldots,n\), we have
\[ \bigcap_{i=1}^{n} F_i\ne\varnothing . \]

But, on the other hand, since the space \((X,T)\) is metrizable, \(\dim X=\operatorname{Ind} X\), and therefore there exist such closed partitions \(F_i\) that
\[ \dim \bigcap_{i=1}^{k} F_i=n-k . \]
In particular,
\[ \dim \bigcap_{i=1}^{n} F_i=0 . \]
We shall prove that the set
\[ F=\bigcap_{i=1}^{n} F_i \]
is stationary.

Indeed, let \(\rho_0\) be any metric on the space \(X\), compatible with the topology \(T\), and let \(g_i:X\to I\) be continuous mappings of the space \(X\) into the interval \(I=[-1,1]\) such that \(g_i(x)=\varepsilon\) for all points \(x\in A_{\varepsilon i}\), \(\varepsilon=-1\) or \(1\), and \(g_i(x)=0\) for all points \(x\in F_i\); finally, \(0<|g_i(x)|<1\) for all points \(x\in X\setminus(A_{-i}\cup F_i\cup A_i)\). Define a new metric on the space \(X\) by the equality
\[ \rho(x,y)=\rho_0(x,y)+\max_i |g_i(x)-g_i(y)| . \]

The metric \(\rho\) is also compatible with the topology \(T\).

Let now \(f:X\to X\) be an arbitrary continuous mapping, with \(\rho(x,f(x))<1\) for all \(x\in X\). Then \(g_i f(x)<0<g_i f(y)\) for all points \(x\in A_{-i}\), \(y\in A_i\); hence it follows that the sets \((g_i f)^{-1}(0)\) are partitions between the sets \(A_{-i}\) and \(A_i\), \(i=1,2,\ldots,n\). Therefore there exists a point \(x\in X\) such that \(g_i f(x)=0\) for all indices \(i=1,2,\ldots,n\). But then \(f(x)\in F\), i.e. \(F\cap f(x)\ne\varnothing\). The theorem is proved.

This result can be generalized.

Definition*. A subset \(A\) of a topological space \((X,T)\) will be called stationary if there exists on it a continuous pseudometric \(\rho\) for which condition \((\overline{\mathrm{y}})\) is satisfied.

Then we have

Theorem*. In every perfectly normal space \((X,T)\) for which \(\dim X=\operatorname{Ind}X<\infty\), there exists a closed stationary zero-dimensional subset.

Problem. Does there exist a metrizable space that contains no zero-dimensional closed stationary subset? Is the Hilbert brick \(I^{\aleph_0}\) such a space?

Warsaw University
Warsaw, Polish People’s Republic

Received
17 XII 1965