14.2.2013 0:49, quasi wrote: > Kaba wrote: > >> Let X be a locally Euclidean Hausdorff space. Show that if X is >> compact, then X is second countable. >> ... >> Any hints? > > Since X is locally Euclidean, each x in X is contained in > an open set U_x homeomorphic to some Euclidean space, hence > U_x, regarded as a subspace of X, has a countable base, > B_x say.
> It follows that V is a union of open sets from B. > > Thus, B is a base for X, hence X is second countable.
My first thought was to take preimages of the basis sets in R^n, to get a basis for each U_x, and then take the union of these preimage sets under x. However, this does not work since the homeomorphism distorts the sets into ever new open sets. Of course, compactness rescues here, as your proof shows. Thanks.