I am reading Rudin's "Fourier analysis on groups" and doing a review of Topology by reading his appendix. He describes one point compactification like this: Given any topological space $S$, build $S_{\infty}=S\cup\{\infty\}$ and topologize it by calling a subset $A\subseteq S_\infty$ open if either $A$ is open in $S$ or if the complement of $A$ is a compact subset of $S$.

Now, I can't prove that this gives a new topology - if $A$ is an open set that contains $\infty$ and $B$ is an open set which does not, I can't see how to prove $A\cap B$ is open. If the complement of $A$ was closed, in addition to being compact, that would solve it; and indeed, the wikipedia article demands explicitly that the complement be closed. If the space was Hausdorff than a compact set would automatically be closed, but this is not assumed.

So, has Rudin omitted a condition, or can we prove this is a topology even without the "closed" condition?

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    $\begingroup$ I don't think it makes a topology without the closed condition. Consider real line with two $0$, which is not Huasdorff, and set $A$ the real line with one $0$, and $B$ any open set containing the two $0$. $\endgroup$ – Minghao Liu May 18 '14 at 11:38

The author omitted the Hausdorff assumption; probably had it on the back of his mind, but not in written text. Since the Hausdorff topological spaces are prevalent in analysis, the assumption becomes one of the things that "go without saying".

Another example is Conway's book on functional analysis, where the author gives some exercises involving topological spaces before revealing in Chapter IV that "[t]he attitude that has been adopted in this book is that all topological spaces are Hausdorff".

Here's the simplest possible counterexample: let $X$ be two-element set $\{a,b\}$ with antidiscrete topology $\mathcal T = \{\varnothing, \{a,b\}\}$. All subsets of $X$ are compact, since we only have finitely many open sets in the topology. Adjoining $\infty$ leads, according to the definition, to "topology" $$\{\varnothing, \{a,b\}, \{\infty\}, \{a,\infty\}, \{b,\infty\}, \{a,b,\infty\} \}$$ which is not a topology, as $\{a,b\}\cap \{a,\infty\} = \{a\}$ is not there.

  • $\begingroup$ Actually Rudin takes great care to differentiate the Hausdorff case from the non-Hausdorff one (he defines one-point compactification for all spaces, and then describes what happens in the special, locally compact Hausdorff space). Still, thanks for the counterexample. $\endgroup$ – Gadi A May 19 '14 at 7:27

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