Stone Cech Compactification - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-18T14:15:59Z http://mathoverflow.net/feeds/question/106351 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106351/stone-cech-compactification Stone Cech Compactification unknown (google) 2012-09-04T14:46:23Z 2012-09-04T17:59:30Z <p>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! </p> http://mathoverflow.net/questions/106351/stone-cech-compactification/106371#106371 Answer by Joseph Van Name for Stone Cech Compactification Joseph Van Name 2012-09-04T17:59:30Z 2012-09-04T17:59:30Z <p>This space is not Frechet-Urysohn. More generally, in any normal space $X$ if <code>$(x_{n})_{n}$</code> 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 <code>$(x_{n})_{n}$</code> 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 <code>$x_{n}$</code> are unique. Then the set <code>$\{x_{n}|n\in\mathbb{N}\}$</code> is a closed discrete subspace of $X$. In particular, the sets <code>$A=\{x_{n}|n\textrm{ is even}\},B=\{x_{n}|n\textrm{ is odd}\}$</code> 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 <code>$^{\lim}_{n\longrightarrow\infty}f(x_{n})=^{\lim}_{n\longrightarrow\infty}f(x_{2n})=0$</code> and <code>$^{\lim}_{n\longrightarrow\infty}f(x_{n})= ^{\lim}_{n\longrightarrow\infty}f(x_{2n+1})=1$.</code> This is a contradiction.</p>