Problem of the Week #117 - August 25th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Many topological spaces have countable bases, but no $T_1$ space has a locally finite basis unless it is discrete. Prove this fact.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Euge. You can find his solution below.

[sp]Let $X$ be a $T_1$ space. If $X$ is discrete, then the collection of singletons of $X$ is a locally finite basis for $X$.

Conversely, suppose $X$ has a locally finite basis, $\mathcal{B}$. Take an arbitrary element $x\in X$. Consider the subcollection $\Sigma := \{B\in \mathcal{B}\;|\; B \ni x\}$. Since $\mathcal{B}$ is locally finite, for some neighbourhood $G$ of $x$, $G \cap B = \emptyset$ for all but finitely many $B\in \mathcal{B}$. Thus $x \notin B$ for all but finitely many $B\in \mathcal{B}$. Consequently, $\Sigma$ is finite.

Enumerate the elements of $\Sigma$ as $B_1,\, B_2,\ldots, \, B_n$. Let $U := \cap_{i = 1}^n B_i$. Then $x\in U $. If there was an $a\in U $ different from $x$, then as $X$ is $T_1$ and $\mathcal{B}$ is a basis, there would be a $B\in\Sigma$ not containing $a$. This contradiction shows that $U = \{x\}$. Since the intersection of finitely many open sets is open, it follows that $\{x\}$ is open. Hence, $X$ is discrete.[/sp]