Ultrafilter on Stone-Čech compatification of natural number

Question

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?

Answer 1

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 $\omega$.