Stone Cech Compactification

2012-09-04T14:46:23Z

Hello, I consider the stone cech compactification of a discrete space using ultrafilters. My question is, why is this space Frechet-Urysohn? (a topological space is Frechet-Urysohn if for every point $x \in \overline A \setminus A$, there is a sequence in A which converges to x) Thanks!

Answer by Joseph Van Name for Stone Cech Compactification

2012-09-04T17:59:30Z

This space is not Frechet-Urysohn. More generally, in any normal space $X$ if $(x_{n})_{n}$ is a sequence in $X$ that converges to a point $x\in\beta X$, then $x\in X$ (I think this is a problem in John Conway's Book on functional analysis). To prove this fact, suppose to the contrary that $(x_{n})_{n}$ is a sequence in $X$ that converges to a point $x\in\beta X\setminus X$. Without loss of generality, we may assume that all the points $x_{n}$ are unique. Then the set $\{x_{n}|n\in\mathbb{N}\}$ is a closed discrete subspace of $X$. In particular, the sets $A=\{x_{n}|n\textrm{ is even}\},B=\{x_{n}|n\textrm{ is odd}\}$ are disjoint closed subsets of $X$. Therefore there is a continuous function $f:X\rightarrow[0,1]$ such that $f$ maps $A$ to $0$ and $f$ maps $B$ to $1$. Furthermore, there is a unique extension of $f$ to a continuous mapping $\hat{f}:\beta X\rightarrow[0,1]$. However, in this case, we would have $^{\lim}_{n\longrightarrow\infty}f(x_{n})=^{\lim}_{n\longrightarrow\infty}f(x_{2n})=0$ and $^{\lim}_{n\longrightarrow\infty}f(x_{n})= ^{\lim}_{n\longrightarrow\infty}f(x_{2n+1})=1$. This is a contradiction.

Stone Cech Compactification - MathOverflow [closed]

Stone Cech Compactification

Answer by Joseph Van Name for Stone Cech Compactification