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Oct 12, 2011 at 15:55 comment added Bill Johnson Sure, post it on math.SE, Asaf. I don't have an account there.
Oct 12, 2011 at 10:42 comment added Asaf Karagila Bill: If you have a math.SE, could you post this on the linked question as well? If not, may I have your permission to post it as community wiki?
Oct 12, 2011 at 10:41 vote accept Asaf Karagila
Oct 12, 2011 at 10:27 comment added Emil Jeřábek @Asaf: The indicator function of $2B$ is continuous, and its extension to $\beta X$ gives a partition of $\beta X$ into two clopen subsets that are easily seen to be the closures of $2B$ and its complement, respectively. (In general, if $C$ is a clopen subset of $W$, then the closure of $C$ in $\beta W$ is clopen.) By a similar argument, the closure of $\frac12Y$ omits all points $-1/(2n+1)$, hence it does not contain any neighbourhood of its element $0$.
Oct 12, 2011 at 9:23 comment added Asaf Karagila Bill, the extension is onto $\beta Y$, I also fail to see why the closure of $2B$ is clopen, and why the closure of $\frac{1}{2}Y$ is not clopen in $\beta Y$.
Oct 11, 2011 at 23:33 history edited Bill Johnson CC BY-SA 3.0
Corrected typo
Oct 11, 2011 at 23:22 history answered Bill Johnson CC BY-SA 3.0