𝐵 paracompact does not imply 𝐵^{𝐼} paracompact

As is well known the category of paracompact spaces is important in algebraic topology and the theory of fiber spaces. The following question arises naturally. If a space (Hausdorff space) B is paracompact, is the space of paths BI (I= [0, 1]), with, of course, the compact-open topology, also paracompact? The following simple example answers the question in the negative. Let X denote the set of real numbers with the half-open interval topology [1]. This now well-known space has the following properties: regular, Lindel6f (hence paracompact [2], hence normal [3]) and totally disconnected. It is also known that XXX is not normal [1 ] (hence not paracompact). Since XI and X are homeomorphic, XI is paracompact so a slight adjustment must be made to provide the counter-example. Let C(X) denote the cone over X, i.e., in XXI identify XX {I } to a point, thus obtaining C(X). Then, if p: XXI -*C(X) is the identification map, C(X) is topologized by employing the weakest topology which renders p continuous. Since XXI is Lindelof and regular, it follows that C(X) is Lindelof and regular, hence paracompact. What we will show now is that C(X)I is not paracompact. The idea is the following: X appears in C(X) as a closed subset, namely the base of the cone. Therefore XXX appears in C(X) X C(X) as a closed subset and hence C(X) X C(X) is not paracompact. Thus, if we can imbed C(X) X C(X) in C(X)I as a closed subset, it will follow that C(X)I is not paracompact. We leave to the reader the simple proofs of the following lemmas.