# Disjoint subsets in a compactification.

Let $X$ be the plane with the usual Euclidean metric, and let $cX$ be the compactification corresponding to the algebra of all bounded, uniformly continuous functions on $X$.

I'm having trouble thinking of an example of two disjoint closed subsets of $X$ whose closures in $cX$ are not disjoint.

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How is this compactification defined? What text or book or paper are you working from? –  Henno Brandsma Jan 30 '12 at 14:22
In class, we were shown that the set of all compactifications on a completely regular space $X$ has a one-to-one correspondence with the set of all closed subalgebras $A$ of $C_{b}(X)$, the set of continuous bounded functions on $X$, that separate points and closed sets in $X$. To construct the compactification of $X$, we considered the set $F \subset C(X, [0,1]$ that separates points and closed sets. After enumerating $F$ as $\{ f_{a} : a \in A\}$, we took the diagonal product of $F$ to get an embedding from $X$ into $[0,1]^A$. –  josh Jan 30 '12 at 14:56
Ok, and this has (IIRC) the property that every function from $A$ can be extended to $cX$. And if we take all bounded functions we get $\beta X$. What is the reverse correspondence, BTW? –  Henno Brandsma Jan 30 '12 at 15:05
Try the cross of axes and the graph of $\frac{1}{x}$ for $x \neq 0$, maybe. I have a hunch that might work. –  Henno Brandsma Jan 30 '12 at 15:14
Consider for some such $f$, $x_n = (n, \frac{1}{n})$ in said graph, and $y_n = (n,0)$ in the cross. There is a common subsequence of them such that both the image subsequences converge, and uniform continuity then implies the limits are the same. –  Henno Brandsma Jan 30 '12 at 16:01