1
$\begingroup$

Let X be a compact Hausdorff space, {$F_{n}$ | n $\in$ $\mathbb{N}$,} a descending collection of closed subsets of X; and O an open set containing $\cap$$F_{n}$. Show that $\exists$ N such that $F_{n} $$\subseteq$ O $\forall$ n $\ge$ N.

Ok I have made some progress: - I know all these closed sets are compact as X is compact and any closed set of a compact space is compact (thus every open cover of each closed set has a finite subcover) - I think maybe using the finite intersection property may be useful

I am looking for a way to approach this problem. How to set it up. Thank you

$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

If all sets $F_n\setminus O$ are non-empty, then it contardicts the finite intersection property.

$\endgroup$
9
  • $\begingroup$ How does compactness come into it? $\endgroup$
    – m.styles
    Nov 10, 2016 at 4:51
  • 1
    $\begingroup$ @m.styles Since each $F_n\setminus O$ is a closed subset of a compact Hausdorff space, it is compact. $\endgroup$
    – Alex Ravsky
    Nov 10, 2016 at 5:13
  • $\begingroup$ If all sets Fn∖O are non-empty does it contradict the finite intersection property of Fn for all n or Fn∖O for all n. This is the bit I am unsure of. $\endgroup$
    – m.styles
    Nov 14, 2016 at 6:01
  • $\begingroup$ @m.styles If all sets $F_n\setminus O$ are non-empty then the family $\{F_n\setminus O\}$ of closed sets of the compact space $X$ has the finite intersection property, but has the empty intersection. $\endgroup$
    – Alex Ravsky
    Nov 14, 2016 at 7:59
  • $\begingroup$ I now understand your argument but showing {Fn\O} has empty intersection is not straight forward to me? $\endgroup$
    – m.styles
    Nov 14, 2016 at 18:51
0
$\begingroup$

Suppose (by contradiction) that $(n_k)_{k\in \mathbb N}$ is a strictly increasing sequence in $\mathbb N$ such that $F_{n_k}\not \subset O$ for each $k.$ Choose $p_k\in F_{n_k}$ \ $O.$ Let $G_k=\{p_j:j\geq k\}.$ Then $$\emptyset \ne \overline {G_{k+1}}\subset \overline {G_k}\subset \overline {F_{n_k}}=F_{n_k}.$$ Let $G= \cap_{k\in \mathbb N}\overline {G_k}.$ We have $G\ne \emptyset.$ Take any $p\in G$. Then $p$ belongs to $\cap_{m\in \mathbb N}F_m,$ because for any $m$ we may take $n_k>m,$ and $p\in \cap_{j\geq k}\overline {G_{n_j}} \subset F_{n_k}\subset F_m.$ So $p\in O.$ But then $O$ is a nbhd of $p,$ with $p\in \overline {G_k}$ (for any $k$), which implies $G_k\cap O\ne \emptyset,$ a contradiction.

$\endgroup$
2
  • $\begingroup$ I don't see how this proves anything. I think using compact and Hausdorff property is essential? $\endgroup$
    – m.styles
    Nov 13, 2016 at 6:55
  • $\begingroup$ The negation of $\exists n(m>n\implies F_m\subset O)$ is that there exist infinitely many $n$ such that $F_n\not \subset O,$ which implies the existence of $(n_k)_k$ with $F_{n_k}\not \subset O.$ I showed this is untenable.... Each $\overline {G_k}$ is compact because it is closed in the compact $T_2$ space $X$. And $(\overline {G_k})_k$ is a nested sequence of non-empty compact sets in the compact $T_2$ space $X$ so $G$ is not empty. $\endgroup$
    – DanielWainfleet
    Nov 13, 2016 at 8:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .