2
votes
2answers
84 views

Stone-Čech compactification using ultrafilters

Let $X = \omega \cup \{ x \}$ ne the Stone-Čech compactification of $\omega$. (I am viewing $X$ as a subspace of the set of ultrafilters over $\omega$). Let, $\mathcal A$, $\mathcal B$ be two disjoint ...
2
votes
1answer
62 views

A boolean algebra is complete if its stone space is extremally disconnected

I have the following proof, but I don't understand one of the steps: Theorem 4.4. A Boolean algebra is complete iff its Stone space is exlremally disconnected. Proof. Identify the given ...
3
votes
2answers
66 views

Čech-Stone compactification of $\mathbb N$ and ultrafilters on $\mathbb N$

I have found in the literature that the Čech-Stone compactification $\beta\mathbb N$ of $\mathbb N$ (or more generally, of any discrete topological space) can be identified with ultrafilters on ...
0
votes
2answers
48 views

What does it mean for an ultrafilter to have a limit?

I got this question from the construction of the Stone-Čech compactification using ultrafilters given in Wikipedia. There they say that if $F$ is an ultrafilter in a compact Hausdorff space $K$ then ...
-2
votes
1answer
43 views

the Stone–Čech compactification

$\beta X$ is the Stone–Čech compactification of $X$ and $\mathscr{B} = \{\widehat{A} : A \subset X \}$ is a base for a topology $\beta X$ where $‎\widehat{A} = \{ \mathcal{P} \in \beta X : A \in ...
0
votes
1answer
46 views

Question about Stone-Čech compactification

Assume $R=(\mathcal{U}_n)_n$ is a sequence of distinct ultrafilters on some set $X$. Since every Hausdorff space has an infinite discrete subspace, there is a subsequence $R=(\mathcal{V}_n)_n$ of ...
0
votes
1answer
98 views

Questions about the Stone-Čech compactification

A compactification of $X$ is a pair $( h, Y)$ where $Y$ is a compact space and $h\colon X ‎\to Y$ is an embedding such that $h(X)$ is dense in $Y$. According to definition of mentioned ...
0
votes
0answers
28 views

Let $(X,\tau)$ be a $T_B$-space…

A topological space is called $T_B$ if every compact subset is closed. (I):$Let (X,\tau)$ be a $T_B$-space which is not countably compact, $\{x_n :n \in \omega\}$ a set without accumulation points, ...
1
vote
0answers
45 views

A topological space is called $T_B$ if…

A topological space is called $T_B$ if every compact subset is closed. According to therem ( I, II , III), how does the below theorem proof?? Let $(X,\tau)$ be a $T_B$-space which is not ...
1
vote
0answers
84 views

Let $\mathcal{F}$ be a filter on $X$. $\mathcal{B} \subseteq P( X)$ is called a filter- base…

Let $\mathcal{F}$ be a filter on $X$. $\mathcal{B} \subseteq P( X)$ is called a filter- base satisfies bellow conditions: ( 1 ) : ‎$ ‎\mathcal{B}‎ ‎\neq‎ ‎\emptyset‎ $‏ ‎‎( 2 ) : ‎$ ‎\emptyset ...
2
votes
1answer
140 views

the Stone-Čech compactification by using ultrafilters

If $X$ is discrete, one can construct $\beta X$ as the set of all ultrafilters on $X$. But which kind of topology must we use in the above sentence? How can we define the Stone–Čech ...
1
vote
2answers
210 views

Stone-Čech compactification of a discrete space

I would like to know: Why is the Stone-Čech compactification of a discrete space the set of ultrafilters on that space?
1
vote
1answer
87 views

countably compact, KC minimal

** Lemma :Let $(X,\tau )$ be a KC-space which is not countably compact. Then X can be condensed onto a weaker KC-topology.** Proof: Let new topology $‎ ‎\tau‎^{‎\prime‎} = ‎\{U‎‎\in‎‎ ‎\tau:‎‎ ...
1
vote
1answer
101 views

Show that $X$ is separable.

I'm working through Kunen's Set Theory and I'm not sure how to proceed on part of one exercise. Let $X$ be compact Hausdorff and $\mathbb{O}_X$ be the poset of nonempty open sets of $X$ ordered by ...
3
votes
2answers
453 views

Showing the Universal Property of Stone-Čech Compactification Holds Using Ultrafilters

Let $X$ be a topological space, and $F$ an ultrafilter on $X$. We say that $x$ is an $F$-limit if every neighborhood of $x$ is in $F$. If $f: X \rightarrow Y$ is a map, then $f_{\ast}F = \{B ...
0
votes
1answer
55 views

Every minimal KC space is compact

A space $(X,\tau )$ is said to be minimal $KC$ if $(X,\tau)$ is $KC$ but no topology on $X$ which is strictly smaller than $\tau$ is $KC$. Theorem : Every minimal KC space is compact. Proof. ...
0
votes
1answer
87 views

topological space

Let $( X,\tau )$ be a $T_1$ topological space. Let $D = \{ d_n : n \in \omega \}$ be a countably infinite closed discrete subspace of $X$. Fix $P \in X$ and let $F \in \beta\omega- \omega$ be an ...
4
votes
1answer
109 views

Why does the z-ultrafilter Stone-Čech compactification construction have the universal property?

For the notions of $z$-filter, prime $z$-filters and $z$-ultrafilters see A Prime $\mathcal P$-filter is contained in a unique $\mathcal P$-ultrafilter? Let $X$ be a Tychonoff space. Let $BX$ be the ...
3
votes
0answers
46 views

Another question in relation to Tychonoff theorem

Let $X_i$ be compact topological spaces and let $X = \prod_{i \in I}X_i$ and let $\mathscr F$ be ultrafilter on $X$. Define $\mathscr F_i = \{Y \subseteq X_i : \pi_i^{-1}Y \in \mathscr F\}$. Here ...
4
votes
1answer
60 views

Compactification of a discrete space using ultrafilters.

I want to show for the collection of ultra filters on a (non-empty) set $A$, $U(A)$. That $U(A)$ is compact where the topology is derived from the base $U_B = \{F\in U(A)|B\in F\}$. Seeing as $A$ can ...
4
votes
3answers
79 views

Construction of Compactification.

Let $U(S)$ be the collection of ultrafilters on a non empty set $S$. And let $O(A) = \{U\in U(S)| A\in U\}$ for $A\subseteq S$. I am told to show that for $\tau = \{O(A)\subseteq U(S)| A\subseteq ...
4
votes
1answer
274 views

Tychonoff theorem (1/2)

I was trying to prove Tychonoff theorem. First I used (which I showed also): The following are equivalent (a) $X$ is compact (b) every filter of closed set $F$ on $X$ has non-empty ...
2
votes
1answer
149 views

Points in $\beta\omega$

Let $f$ be a continuous function on $\beta\omega$ with $f(x_0)=0$ for some non-principal ultrafilter $x_0\in \beta\omega$. Is there an integer (a principal ultrafilter) $n$ such that $f(n)=0$? EDIT: ...
7
votes
1answer
383 views

Addition on ultrafilters is non-commutative

I'm reading a book on ultrafilters and it tells me that showing addition on ultrafilters over the naturals (addition on the set of ultrafilters '$\beta \mathbb{N}$', defined in the standard way) is ...
3
votes
1answer
206 views

Set of ultrafilters $\beta \mathbb{N} - \mathbb{N}$ is not separable

I want to show the set of ultrafilters $\beta \mathbb{N} - \mathbb{N}$ (where $\beta \mathbb{N}$ is all ultrafilters on $ \mathbb{N}$) is not separable. I know we can take as a base of $\beta ...