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.



Adjoining a new isolated point without changing the space
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.


