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Jan 10, 2015 at 14:27 comment added Nik Weaver @LisaK: the generalization is this. If $X$ contains such a configuration, then identify it with ${\bf N}$ and define $f$ as in my other answer. Let $f$ be zero off the configuration. A cluster point of a set $A$ is a point $x$ every neighborhood of which has infinite intersection with $A$.
Oct 5, 2013 at 7:08 vote accept CommunityBot
Sep 30, 2013 at 22:59 comment added Nik Weaver Incidentally, this condition is weaker than "does not contain a countably infinite set of isolated points that has a limit point". For instance, let $X$ be $\mathbb{N}$ together with one point of $\beta \mathbb{N}$. This $X$ does contain a copy of $\mathbb{N}$ that has a cluster point, but every infinite subset of $\mathbb{N}$ can be split into two infinite pieces, only one of which has a cluster point.
Sep 30, 2013 at 19:19 history answered Nik Weaver CC BY-SA 3.0