Suppose $X$ is a $T_1$ space with an infinite set of isolated points. Show that if $X^\sharp = X \cup \lbrace \infty \rbrace$ is obtained by adding a single new isolated point, then $X$ and $X^\sharp$ are homeomorphic.

I am almost embarrased to raise this, which seems obvious. The proof must be simple, but it eludes me for now. Maybe it is an exercise in some textbook. You can clearly establish a 1-1 equivalence between the isolated points of $X$ and those of $X^\sharp$. But it is not clear how this equivalence would extend to the closure of the isolated points.

The theorem is easy when $X$ is compact $T_2$ and $cl(D) = \beta(D)$, where $D$ is the set of isolated points.


In my answer to https://mathoverflow.net/questions/26414 I descibed a somewhat simpler-looking example than Nik's, but proving that it works may be harder. Take two copies of $\beta\mathbb N$ and glue each non-isolated point of one copy to the corresponding point of the other copy. Any way of "absorbing" a new isolated point into the two copies of $\mathbb N$ forces a relative shift of those two copies, which forces corresponding shifts of the non-isolated points, which in turn conflicts with the gluing. The perhaps surprising thing about this example is that, if you add two isolated points, the result is (easily) homeomorphic to the original.

  • $\begingroup$ Yeah, that's cleaner. $\endgroup$
    – Nik Weaver
    Aug 7 '12 at 20:50

Well, I think this is false. Start with a family of $2^{2^{\aleph_0}}$ mutually non-homeomorphic connected spaces, and attach them to the non-isolated points of $\beta {\bf N}$. (I.e., start with the disjoint union of $\beta {\bf N}$ and the other spaces, and factor out an equivalence relation which identifies each point of $\beta {\bf N} - {\bf N}$ with a point of one of the other spaces.) Any homeomorphism between $X$ and $X^\sharp$ has to take isolated points to isolated points; taking closures, it takes $\beta{\bf N}$ onto itself; and by connectedness it takes each of the extra spaces onto itself. So it has to fix each point of $\beta {\bf N} - {\bf N}$. Now the question is whether a bijection between ${\bf N}$ and ${\bf N}$ minus a point can fix $\beta {\bf N} - {\bf N}$ pointwise. The answer is no because iterating the map, starting on the missing point, yields a sequence within ${\bf N}$ that gets shifted by the map, and it is easy to see that this shift does not fix the ultrafilters supported on that sequence.


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