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Feb 16, 2014 at 14:47 vote accept Greg Zitelli
Feb 16, 2014 at 10:43 comment added alpha The confusion in your question has been cleared up by Qiaochu Yuan but the following remark might interest you. The compactifications of $X$ correspond to the $C^\ast$ subalgebras of $C^b(X)$ ($X$ need not be locally compact). Interesting special cases: the bounded uniformly continuous functions (for uniform spaces), the algebra obtained by adding a unit to $C_0(X)$ ($X$ locally compact), the almost periodic functions ($X$ a locally compact group) and many more. The corresponding spectra are, in these cases, the Samuel, the one-point and the Bohr compactification.
Feb 16, 2014 at 10:01 review Close votes
Feb 16, 2014 at 20:37
Feb 16, 2014 at 4:58 comment added Qiaochu Yuan The Stone-Čech compactification is defined for any topological space, with no hypotheses whatsoever. The map $X \to \beta X$ is an embedding iff $X$ is Tychonoff, which is still a weaker condition than $X$ being locally compact Hausdorff (this is the condition you need for the existence of the one-point compactification).
Feb 16, 2014 at 4:45 answer added Qiaochu Yuan timeline score: 4
Feb 16, 2014 at 0:47 comment added Paolo Ghiggini Are you sure that the Stone–Čech compactification of $\mathbb{R}$ is the one-point compactification? This seems to contradict the universal property stated here: en.wikipedia.org/wiki/… For example, not all continuous functions from $\mathbb{R}$ to $[0,1]$ factor through a map from $S^1$.
Feb 16, 2014 at 0:46 comment added Branimir Ćaćić Maybe I'm misreading you, but the Stone-Čech compactification of $\mathbb{R}$ can't possibly be the one-point compactification, since $C_b(\mathbb{R})$ isn't isomorphic to $C_0(\mathbb{R})^+ \cong C_0((0,1))^+ \cong C(S^1)$.
Feb 16, 2014 at 0:20 history asked Greg Zitelli CC BY-SA 3.0