1. ## Ultrafilters Topology

Let $X$ be a topological space. let $X'$ be the collection of all the ultrafilters on $X$.
we say that a set $S\subseteq X'$ is open and belongs to the base of $X'$ if

there is an open $U\subseteq X$, such that $U'\in S$ $\Leftrightarrow$ $U\in U'$.

Let $N'$ be the space of ultrafilters (filters on $N$ with the discrete topology) as explained above.
Prove that if a sequence $U_{i \ i\in N}$ converge in $N' \ \ U_i=U_j$ for every large enough $i,j$.

any help is appreciated...

2. Originally Posted by aharonidan
Let $X$ be a topological space. let $X'$ be the collection of all the ultrafilters on $X$.
we say that a set $S\subseteq X'$ is open and belongs to the base of $X'$ if

there is an open $U\subseteq X$, such that $U'\in S$ $\Leftrightarrow$ $U\in U'$.

Let $N'$ be the space of ultrafilters (filters on $N$ with the discrete topology) as explained above.
Prove that if a sequence $U_{i \ i\in N}$ converge in $N' \ \ U_i=U_j$ for every large enough $i,j$.

any help is appreciated...
Maybe I'm being deceived, but isn't this just a consequence that for any discrete space $X$ the only convergent sequences are eventually constant?

3. Originally Posted by aharonidan
Let $X$ be a topological space. let $X'$ be the collection of all the ultrafilters on $X$.
we say that a set $S\subseteq X'$ is open and belongs to the base of $X'$ if

there is an open $U\subseteq X$, such that $U'\in S$ $\Leftrightarrow$ $U\in U'$.

Let $N'$ be the space of ultrafilters (filters on $N$ with the discrete topology) as explained above.
Prove that if a sequence $U_{i \ i\in N}$ converge in $N' \ \ U_i=U_j$ for every large enough $i,j$.

any help is appreciated...
Another way to phrase this problem is to say that every convergent sequence in the Stone–Čech compactification of N is eventually constant. I believe that this is a standard result in the theory of the Stone–Čech compactification, but I do not know of a proof. You might try looking here for ideas and further references.