Although Hausdorff spaces aren't generally regular, a Hausdorff space that is also (say) locally compact will be regular, because any Hausdorff space is preregular.
So I’m trying to get my head around the proof of the Riesz Representation Theorem (I’ve mostly got it, I’m just trying to. $(\overline {B_n})$ is decreasing sequence of nonempty compact sets and hence their intersection is not empty: if it is empty then complements of $\overline {B_n}, n=2,3,,$ cover the compact set $\overline {B_1}$. Corollary 6 Let X be a compact Hausdor space. By the same construction, every locally compact Hausdorff space X is an open dense subspace of a compact Hausdorff space having at most one point more than X. Then for y ∈ Y, q − 1 ( y) is a closed subset of X but not necessarily compact. As part of doing this I was trying to figure out the role of the assumption that the space was a locally compact Hausdorff space: The proof generally seems to follow through with just normality (and maybe paracompactness? A topological space is a generalization of the notion of an object in three-dimensional space. (2) any compactification of a completely regular space that is not metrizable; in particular, the Stone-Cech compactification if the discrete space of integers. When X is also Hausdorff, the property of local compactness becomes much Although Hausdorff spaces are not, in general, regular, a Hausdorff space that is also (say) locally compact will be regular, because any Hausdorff space is preregular. (1) any topological (Tychonoff) product of an uncountable family of compact metric spaces. A subset Aof X is compact if and only if it is closed. Let X be a locally compact, second countable Hausdorff topological space and let Y be a Hausdorff quotient of X. This result also ).

The finite union of compact subspaces is compact.

Ordered compact spaces. A nonempty compact subset of the real numbers has a greatest element and a least element. A compact subspace of a metric space is closed and bounded. Thus from a certain point of view, it is really preregularity, rather than regularity, that matters in these situations.

space via the Gromov-Hausdorff ultrametric Zhengchao Wan Department of Mathematics The Ohio State University Columbus, Ohio 43210 wan.252@osu.edu Abstract We establish universality and ultra-homogeneity of (U,uGH), the collection of all compact ultrametric spaces endowed with the so-called Gromov-Hausdorff ultrametric. A space X is locally compact if for each x ∈ X, there exists an open neighborhood U of x with closure U¯ compact. A compact subspace of a Hausdorff space is closed. Hausdorff space, in mathematics, type of topological space named for the German mathematician Felix Hausdorff. Let q: X → Y denote the quotient map. Moreover, two compact subsets of a Hausdorff space can be separated by two open neighborhoods. Theorem 5.6 If Aand Bare disjoint compact subsets of a Hausdor space X, then there exist disjoint open sets Uand V in Xsuch that AˆUand BˆV. It consists of an abstract set of points along with a specified collection of subsets, called open sets, that satisfy three axioms: (1) the set itself and the empty set are open sets, (2) the intersection of a finite … The following results are left to the reader to prove. Thus from a certain point of view, it is really preregularity, rather than regularity, that matters in these situations. Hence there is a finite sub-cover.