The Eighth Prague Topological Symposium August 18-24, 1996 Economical University Prague, Czech Republic

Normality-type Properties in \exp(X) by A.P. Kombarov (Moscow State University )

All spaces are assumed to be T₁ . Recall that a space is said to be pseudonormal if every countable closed subset has arbitrarily small closed neighborhoods . A space is called to have property D [4] if every countable closed discrete set has arbitrarily small closed neighborhoods. A regular space is said to have property wD [4] if every infinite closed discrete set has an infinite subset which has arbitrarily small closed neighborhoods. Clearly, normal implies pseudonormal implies D implies wD . Let Q be a topological property. A space X is said to be point-Q ( \ \nabla -Q ), if for every x in X a subspace X\setminus{x} ( \ the subset X²\backslash\Delta , where \Delta={(x,x):x in X} is a diagonal) has the property Q . The exponential space \exp(X) is the set of all non-empty closed subsets of X with Vietoris (finite) topology.

Theorem 1. If \exp(X) is a point-D space, then X is a hereditarily separable perfectly normal countably compact space.

Corollary 1. If \exp(\exp(X)) or \exp(X× X) is point-D , then X is a metrizable compact space.

Corollary 2. If \exp(X) is hereditarily pseudonormal, then X is a hereditarily separable perfectly normal countably compact space.

We note here that a space X is a metrizable compact space if \exp(X) is hereditarily normal [3] or is regular hereditarily countably paracompact [5]. Every regular countably paracompact space is pseudonormal. Thus the next problem seems to be natural.

Problem 1. Is X a metrizable compact space, if \exp(X) is hereditarily pseudonormal?

Theorem 2. Let X be a compact space. Then \exp (X) is point-wD if and only if X is hereditarily separable and perfectly normal.

Theorem 3. If \exp(X) is \nabla -normal, then X is a hereditarily separable perfectly normal compact space.

Problem 2 Is X a metrizable compact space, if \exp(X) is \nabla -normal?

If \exp(X) is regular and perfect (=every closed set is a G_\delta ), then X is a metrizable compact space [7]. The next theorem 4 is an extension of this theorem.

Theorem 4. If \exp(X) is Hausdorff and \omega -perfect (=every separable closed set is a G_\delta ) , then X is a metrizable compact space.

A space X is said to be weakly normal [1, 2], if for every two disjoint closed subsets A and B of X there exists a continuous mapping f of X into R^\omega such that images of A and B are disjoint. Clearly, every normal space is weakly normal. If there exists a one-to-one continuous mapping of X onto a separable metrizable space, then the space X is weakly normal. So the Niemytzkij plane or the square of the Sorgenfrey line can serve as examples of weakly normal spaces which are not normal [1]. Can one use weak normality instead of normality in Velicko's theorem [8] (and in Coban's theorem[3]): if \exp(X) is ( hereditarily ) normal , then X is a( metrizable) compact space? It is easy to see that this is not the case. Indeed there exists a one-to-one continuous mapping of \exp(\omega) onto D^\omega , so \exp(\omega) is hereditary weakly normal, but the space \omega is not compact.

Theorem 5. If X is a countably compact space and if \exp(X) is weakly normal, then X is compact.

Theorem 6. If X is a countably compact space and if \exp(X) is hereditarily weakly normal, then X is a perfectly normal hereditarily separable compact space.

Corollary 4. If X is countably compact and \exp(\exp(X)) or \exp(X× X) is hereditarily weakly normal, then X is a metrizable compact space.

Problem 3. Is a compact space X metrizable, if \exp(X) is hereditarily weakly normal ?

And the final theorem 7 is a slight generalization of Noble's theorem from [6].

Theorem 7. All powers of a T₁ -space X are weakly normal if and only if X is compact T₂ .

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