Welcome to our community

Be a part of something great, join today!

Problem of the Week #71 - October 7th, 2013

Status
Not open for further replies.
  • Thread starter
  • Moderator
  • #1

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Here's this week's problem.

-----

Problem: Let $X$ be a completely regular topological space, but not compact. Determine whether the Stone–Čech compactification $\beta(X)$ is metrizable or not.

-----

Hint:
First determine if $\beta(X)$ is limit point compact.


Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Moderator
  • #2

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
No one answered this week's question. You can find my solution below.

We first prove that $\beta(X)$ is not limit point compact.

Proof
: Suppose that $\beta(X)$ is limit point compact. Then $X$ must have a limit point in $\beta(X)-X$, i.e. $x_n\in X$ converges to $y\in\beta(X)-X$. We show that $y$ is a limit point of two sequences that have no points in common (hence, showing $\beta(X)$ is not limit point compact).

First, we see to construct a subsequence in which no two points are the same. Let
\[n_k = \left\{\begin{array}{ll} 1, & k=1\\ \min\{n>n_{k-1}\,|\,x_n\notin \{x_{n_1},x_{n_2},\ldots,x_{n_{k-1}}\}\}, & k>1\end{array}\right.\]
Since $x_n$ converges to $y$, the subsequence $x_{n_k}$ converges to $y$ as well. Also, no two points in $x_{n_k}$ are the same. For now, let $x_{n_k}=x_N$. Let $A=\{x_1,x_3,\ldots\}$ be the set of odd points, and let $B=\{x_2,x_4,\ldots\}$ be the set of even points of the sequence $x_N$. We now show that $\overline{A}=A\cup\{y\}$ and $\overline{B}=B\cup\{y\}$.

Clearly, any neighborhood of $y$ contains a point of $A$, thus $y\in\overline{A}$. Thus, $A\subset A\cup\{y\}\subset\overline{A}$. In order to prove our claim, it suffices to show that the complement of $A\cup\{y\}$ is open. Let $z$ be a point in the complement. Since $z$ is not the limit of the sequence $\left(x_{2N+1}\right)$ ($y$ is the limit), we can construct a neighborhood around $z$ such that it doesn't contain $y$ and it only contains finitely many points in $\left(x_{2N+1}\right)$. Since $z\notin A$, we can remove these finitely many points; this gives us a neighborhood that is disjoint from $A\cup\{y\}$. Therefore, $A\cup\{y\}$ is closed and $\overline{A}=A\cup\{y\}$. By a similar argument for $B$, we see that $\overline{B}=B\cup\{y\}$. Therefore, $\overline{A}\cap\overline{B}=\{y\}\neq\emptyset$.

We now show that $\overline{A}\cap\overline{B}=\emptyset$, which leads to the contradiction we seek. It should be clear that $A\cap B=\emptyset$ since $x_N$ does not contain any identical terms. They are closed subsets of $X$ for $\text{Cl}_X A=X\cap\overline{A}=X\cap(A\cup\{y\})=A$ (similar result for $B$). Now, by Urysohn's Lemma of normal spaces, there exists a continuous function $f:X\rightarrow[0,1]$ such that $A\subset f^{-1}(0)$ and $B\subset f^{-1}(1)$. We now apply the universal property of $\beta(X)$: since $f:X\rightarrow[0,1]$ is an open map into a compact Hausdorff space, there there exists a unique function $\beta f: \beta(X)\rightarrow[0,1]$. Now, $\overline{A}\subset\beta f^{-1}(0)$ and $\overline{B}\subset\beta f ^{-1}(1)\implies \overline{A}\cap\overline{B}=\emptyset$. Therefore, $y\in\beta(X)-X$ can't be a limit point of any sequence in $X$. Therefore $\beta(X)$ is not limit point compact.$\hspace{.25in}\blacksquare$

Since $\beta(X)$ is not limit point compact, we can now show that $\beta(X)$ is not metrizable.

Proof
: Suppose that $X$ is normal and not compact. We see that $X\subset\beta(X)$ since $\beta(X)$ is compact while $X$ is not. Since no point in $\beta(X)-X=\overline{X}-X$ is the limit of any sequence of points in $X$, $\beta(X)$ fails to satisfy the sequence lemma. Therefore, $\beta(X)$ is not first countable, which implies $\beta(X)$ is not metrizable.$\hspace{.25in}\blacksquare$
 
Status
Not open for further replies.