# One-point compactification in non-Hausdorff spaces

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?

• 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 $0$. – Minghao Liu May 18 '14 at 11:38

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.