4th Geometry Festival, Budapest

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E. 5-4)). 5-2) fact that homeomorphic spaces S continuous functions was noted. 5-l), yet C(UT,I_I) is always the same as C(T,F$. Still, there is a result going in the converse direction. If C(S,R) is isomorphic to C(T,R), then the characters of the algebras are also identifiable, hence so are vS and UT - at least in the sense that they are in 1-1 correspondence. But this is rather weak. Are they homeomorphic? Is the 1-1 correspondence alluded to above a homeomorphism? As it happens, it is. Part of the vehicle we use to show it is the following result which provides for a topological identification of UT and H.

UFU (PI If T is Lindelaf (CU) ucu (P) possesses a countable subcover (CUn or, n T = @. Thus p 4 UT. V equivalently, (nun) Though completely regular Hausdorff Lindel6f spaces are replete, the converse is false. To see this, let T denote with the topology generated by the half-open intervals (a,bl. T is completely regular, Hausdorff and LindelEf (see below), hence replete, hence so is TXT. TxT is not Lindel6f however, since the closed subspace {(t,-t) It f TI is discrete and nondenumerable. Proof that T is Lindel6f.

5-4) RELATIVE PSEUMCOME'ACTNESS. Hausdorff space. cl E C UT. ) Proof We show that E is not relatively pseudocompact if and only if cl E a UT. B Suppose that p E cl E-UT, so that there is some x F C(T,R) such that B n . Thus for each n C 3 there is a point t E E at which x is ) G(p) = -. Conversely if E is not relatively pseudocompact, then a positive function x c C(T,€J) exists which is unbounded on E. Thus there is a sequence (t ) of points from E such that x(t ) ) n. Consequently at any n n .

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