MATHEMATICS

G. I. KAC

ON THE FUNCTIONAL CLOSEDNESS OF COMPLETELY REGULAR SPACES

(Presented by Academician P. S. Aleksandrov, 11 II 1958)

A completely regular space (G) is called functionally closed (a (Q)-space) if for every completely regular space (S) containing (G) as a dense subspace, there exists a continuous function defined on (G) (possibly unbounded) which cannot be continuously extended to (S) ((^1)).

A completely regular space (G) is called a complete topological space if it is complete with respect to some uniform structure of its own.

As usual, by (\beta G) we shall denote the maximal bicompact extension of (G) ((^2)). The following definition of a complete topological space is equivalent to the one given above.

A completely regular space (G) is called a complete topological space if, for every point (x_0 \in \beta G - G), there is a continuous generalized metric (\rho(x,x')) on (G) which cannot be extended to a continuous generalized metric on (x_0 \cup G^*).

A. A. Kubenskii ((^4)) studied the connection between functionally closed and complete topological spaces. He proved that every normal complete topological space of attainable cardinality is functionally closed **. In the present note this theorem is generalized to the case of completely regular spaces. More precisely, it is proved:

Theorem 1. A complete topological space of attainable cardinality is functionally closed.

The validity of Theorem 1 follows from the result of A. A. Kubenskii cited above and from the following theorem.

Theorem 2. If discrete spaces of cardinality not exceeding (m) are functionally closed, then any complete topological spaces of cardinality (m) are functionally closed ***.

Lemma 1. In order that a completely regular space (G) be functionally closed, it is necessary and sufficient that for every point (x_0 \in \beta G - G) there exist a continuous function defined on (G) which cannot be continuously extended to (x_0 \cup G) ((x_0 \cup G) is considered in the relative topology as a subspace of (\beta G)).

* The equivalence to the usual definition follows easily from Theorem 1 ((^3)). A generalized metric is a function (\rho(x,x')) satisfying all the conditions valid for an ordinary metric, except for the condition (\rho(x,x') = 0 \to x = x'). A generalized metric (\rho(x,x')) is continuous if the set of all points such that (\rho(x,x') < \varepsilon) is open ((\varepsilon > 0,\ x') fixed).

** That is, of cardinality less than the first inaccessible. A cardinal (p > c) is inaccessible if it cannot be represented in the form

[
p = \sum_{\alpha \in A} 2^{m_\alpha},
]

where all (m_\alpha) and the set (A) have cardinalities less than (p).

*** The formulations of Theorems 1 and 2 were communicated to the author by Yu. M. Smirnov.

The necessity of the assertion is obvious. Let us prove sufficiency. Suppose that (G) is not a functionally closed space. Then there exists a completely regular space (S) containing (G) as a dense subspace, and moreover every continuous function (f(x)) given on (G) can be extended to a continuous function (F(y)) defined on (S). By (K) denote the bicompact extension of the space (S) and, consequently, of the space (G). According to a well-known theorem ((^{2})), there exists a continuous mapping (y=\psi(x)) of the bicompactum (\beta G) onto (K), leaving invariant the points of (G) ((\psi(x)=x) for (x\in G)). By (x_0) denote the preimage of some point of (S-G). Obviously, (x_0\in \beta G-G). The function (F(\psi(x))) is continuous on (x_0\cup G) and is an extension of the function (f(x)). Thus, any function continuous on (G) can be extended continuously to (x_0\cup G). This proves the sufficiency of the assertion of the lemma.

Lemma 2. Let the function (f(x)) be continuous and defined on a completely regular space (G). Put

[
f_N(x)=
\begin{cases}
f(x) & \text{for points } x\in G \text{ at which } |f(x)|\leq N;\
N & \text{” ” } x\in G,\ \text{” ” } f(x)>N;\
-N & \text{” ” } x\in G,\ \text{” ” } f(x)<-N.
\end{cases}
]

The function (f_N(x)) is bounded and continuous on (G), and therefore can be extended continuously to (\beta G) ((^{2})). Denote the extended function by (F_N(x)). The function (f(x)) can be extended continuously to (x_0\cup G) ((x_0\in\beta G)) if and only if there exists a (finite) limit
[
\lim_{N\to\infty} F_N(x_0).
]
(\lim_{N\to\infty} F_N(x_0)) is the value of the extended function at the point (x_0).

Suppose that the function (f(x)) can be extended continuously to (x_0\cup G), and let (\Phi(x)) be the extended function. Choose an arbitrary (N>|\Phi(x_0)|) and take so small a neighborhood (U(x_0)) of the point (x_0) that at its points the inequality (N>|\Phi(x)|) holds. Thus, (\Phi(x)=F_N(x)) for (x\in U(x_0)\cap G), and therefore (\Phi(x_0)=F_N(x_0)). Since (N) is an arbitrary sufficiently large number, we have
[
\Phi(x_0)=\lim_{N\to\infty} F_N(x_0).
]

Now suppose that there exists a finite limit (\lim_{N\to\infty} F_N(x_0)). We show that the function

[
\Phi(x)=
\begin{cases}
\displaystyle \lim_{N\to\infty} F_N(x_0) & \text{when } x=x_0;\
f(x) & \text{when } x\in G,
\end{cases}
]

defined on (x_0\cup G), is a continuous extension of the function (f(x)). It suffices to prove the continuity of the function (\Phi(x)) at the point (x_0). Choose arbitrary (\varepsilon>0) and then (N) so large that the inequalities

[
N-|\Phi(x_0)|>2\varepsilon,\qquad |\Phi(x_0)-F_N(x_0)|<\varepsilon.
\tag{1}
]

hold. Next choose so small a neighborhood (U(x_0)) of the point (x_0) that for (x\in U(x_0)\cap G) the inequality

[
|F_N(x_0)-F_N(x)|=|F_N(x_0)-f_N(x)|<\varepsilon.
\tag{2}
]

holds. From inequalities (1), (2) it follows that
[
|f_N(x)|<|F_N(x_0)|+\varepsilon<|\Phi(x_0)|+2\varepsilon<N,
]
and therefore (f_N(x)=f(x)) for all (x\in U(x_0)\cap G). But then, from inequalities (1), (2), we have
[
|\Phi(x_0)-f(x)|=|\Phi(x_0)-\Phi(x)|<2\varepsilon
]
for all (x\in U(x_0)\cap G). The assertion follows.

From the lemma just proved, the validity of the following lemma follows easily.

Lemma 3. Let (G) and (H) be completely regular spaces; let (R(x)) be a continuous mapping of (\beta G) onto (\beta H), with (R(G)=H). Let (x_0\in\beta G-G), (y_0=R(x_0)\in\beta H-H), and let (\varphi(y)) be a continuous function on (H), not extendable

cannot be extended continuously to (x_0 \cup G).

Lemma 4. Let (y=r(x)) be a continuous mapping of a completely regular space (G) onto a completely regular space (H). The mapping (r(x)) can be extended to a continuous mapping of the bicompactification (\beta G) onto (\beta H).

Proof of Theorem 2. Let (G) be a complete topological space of cardinality (m); let (x_0) be an arbitrary point of (\beta G-G). Below we shall prove the existence of a function continuous on (G) which cannot be extended continuously to (x_0\cup G). By Lemma 1 it follows from this that (G) is functionally closed.

In accordance with the definition of a complete topological space given above, there exists a continuous generalized metric (\rho(x_1,x_2)) on (G) which cannot be extended to (x_0\cup G).

Divide all points of the space (G) into disjoint classes, assuming that two points (x_1) and (x_2) of (G) belong to one and the same class under the condition (\rho(x_1,x_2)=0) (and only in this case). It is not difficult to see that the values of the function (\rho(x_1,x_2)) depend only on the classes (y_1) and (y_2) to which the points (x_1) and (x_2) belong. Therefore the equality (\rho^(y_1,y_2)=\rho(x_1,x_2)), where (x_1) and (x_2) are points of the classes (y_1) and (y_2), respectively, defines a function (\rho^(y_1,y_2)) on the set of classes (H).

It is easy to verify that the function (\rho^(y_1,y_2)) defines a metric on the set (H). Further, (H) is regarded as a metric space with metric (\rho^(y_1,y_2)).

Put (y=r(x)), where (y) is the class to which the point (x\in G) belongs; (y=r(x)) is a single-valued mapping of (G) onto (H). From the continuity of the generalized metric (\rho(x,x')) follows the continuity of the mapping (y=r(x)). Thus (y=r(x)) is a continuous mapping of the completely regular space (G) onto the metric (and, consequently, completely regular) space (H). In accordance with Lemma 4, extend the mapping (r(x)) to a continuous mapping (y=R(x)) of the bicompactification (\beta G) onto (\beta H). The function (\tilde\rho(x_1,x_2)=\rho^*(R(x_1), R(x_2))) is defined on the set (R^{-1}(H)\supseteq G). It is not difficult to verify that the function (\tilde\rho(x_1,x_2)) defines a continuous generalized metric on (R^{-1}(H)). If (x_1,x_2\in G), then

[
\tilde\rho(x_1,x_2)=\rho^(R(x_1),R(x_2))=\rho^(r(x_1),r(x_2))=\rho^*(y_1,y_2)=\rho(x_1,x_2).
]

Thus, (\tilde\rho(x_1,x_2)) is an extension of the generalized metric (\rho(x_1,x_2)) to (R^{-1}(H)). By hypothesis, the generalized metric (\rho(x_1,x_2)) cannot be extended to (x_0\cup G), hence (x_0\notin R^{-1}(H)), and consequently (y_0=r(x_0)\notin H).

We now make use of a theorem of Katětov ({}^{(5)}), from which it follows directly, under the hypotheses of Theorem 2, that metric spaces of cardinality not exceeding (m) are functionally closed. By construction, the cardinality of the metric space (H) does not exceed the cardinality of the space (G), and consequently (H) is functionally closed. By Lemma 1 there exists a continuous function (\varphi(y)), defined on (H), which cannot be extended continuously to (y_0\cup H); but then, by Lemma 3, the function (\varphi(R(x))), continuous on (G), cannot be extended continuously to (x_0\cup G).

Theorem 2 is proved.

Received
7 II 1958

REFERENCES

({}^{1}) E. Hewitt, Trans. Am. Math. Soc., 64, No. 1, 45 (1948).
({}^{2}) P. S. Alexandrov, Uspekhi Mat. Nauk, 2, No. 1, 5 (1947).
({}^{3}) G. I. Kats, DAN, 99, No. 6, 897 (1954).
({}^{4}) A. A. Kubenskii, DAN, 117, No. 5, 748 (1957).
({}^{5}) M. Katětov, Fund. Math., 38, 73 (1951).