1
$\begingroup$

I have two questions about ultrafilter on Stone-Čech compatification of natural number .

I would like to know:

(a) Why The set $ \omega$ corresponds to the set of principal ultrafilters? And the set $\beta\omega - \omega$ to the set of free ultrafilters?

(b) How are the elements $\beta\omega - \omega$ of ? Are they isolated points? Why?

$\endgroup$
3
$\begingroup$

No point of $\beta\omega\setminus\omega$ is isolated: $\beta\omega$ is a compactification of $\omega$, so $\omega$ is dense in $\beta\omega$, and therefore every non-empty open set in $\beta\omega$ contains points of $\omega$. (And if all points of $\beta\omega\setminus\omega$ were isolated, then $\beta\omega$ would be an infinite discrete space and therefore not compact.)

When we say that a compact space $K$ is a compactification of a space $X$, we mean that $K$ has a dense subset that is homeomorphic to $X$. For each $n\in\omega$ let $p_n$ be the principal ultrafilter over $n$; then $\{p_n:n\in\omega\}$, the set of principal ultrafilters in $\beta\omega$, is a discrete subset of $\beta\omega$ that is dense in $\beta\omega$. In other words, $\{p_n:n\in\omega\}$ is a dense subset of $\beta\omega$ that is homeomorphic to $\omega$, and by definition $\beta\omega$ is therefore a compactification of $\omega$. The set $\omega$ corresponds to the set of principal ultrafilters because the natural embedding of $\omega$ as a dense subset of $\beta\omega$ is the map $\omega\to\beta\omega:n\mapsto p_n$ that takes each $n\in\omega$ to the corresponding principal filter on $\omega$.

$\endgroup$
  • $\begingroup$ I would like to know more about the space $\beta \omega$. Where can I find more about it? Any exposition or article will do. $\endgroup$ – Idonknow Apr 19 '18 at 23:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.