By J. Bernstein

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Then X is a locally compact space, but Y is not locally compact subspace of X. To show that the completeness is not hereditary, put X = [0, 1] (the unit interval as a subspace of R1 ) and Y = [0, 1) as a subspace of X. Then X is a complete metric space with respect to the usual metric, but Y is not a complete subspace, because any sequence in Y , convergent to 1 is a Cauchy sequence in Y which does not converge in Y . In the case of 32 completeness it is to be noted that the Cauchyness depends on the metric.

In (X, τ1 , τ2 ), τ1 is deﬁned to be completely regular with respect to τ2 if for each τ1 closed set C and each point x ∈ /C there is a real valued function f on X into [0, 1] such that f (x) = 0, f (C) = 1, and f is τ1 upper semi-continuous and τ2 lower semi-continuous. Furthermore, (X, τ1 , τ2 ) is a-6 Several topologies on one set strong (weak) pairwise completely regular if τ1 is completely regular with respect to τ2 and (or) τ2 is completely regular with respect to τ1 . Bitopological normality was deﬁned by Kelly [6] as follows: (X, τ1 , τ2 ) is pairwise normal if for each τ1 closed set A and τ2 closed set B disjoint from A there is a τ1 open set V containing B and a τ2 open set U disjoint from V containing A.

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