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I am trying to understand the finite intersection definition of compactness, so we have the definitions:

Let $F$ be a collection of sets. Then $F$ is said to have the Finite Intersection Property if for every finite collection of sets from $F$, ${F_1,F_2,...,F_n}⊆F$, we have that $⋂_{i=1}^n F_i ≠∅$

and

Let X be a topological space. Then X is compact if and only if for every collection of closed sets $\mathcal{F}$ from X, we have that if $\mathcal{F}$ has the finite intersection property then $⋂_{F∈\mathcal{F}} F≠∅$

I want to form a counterexample, ie a set that contains a family of closed sets with the finite intersection property that is not compact. I think this family $\mathcal{F}$ must be an infinite set, since any finite set would automatically have the intersection be non-empty from the finite intersection property. My teacher said to consider the set $(0,1) \subset \mathbb{R}$ which we know is not compact. He said to consider the family of sets $(0,\frac{1}{n})$, which I understand violates what we want (infinite intersection is empty but finite is not). However, my question is, aren't these sets not closed? Maybe he meant $(0,\frac{1}{n}]$ which I think would be closed under our original set $(0, 1)$ even if not closed in $\mathbb{R}$, similar to the examples discussed here: Difference between closed, bounded and compact sets

If so, does the "closed sets" part of the definition refer to closed under the set we are concerned with, ie $(0,1)$, and not the general space, ie $\mathbb{R}$?

TLDR: Does anyone know a relatively simple example for a family of closed sets $\mathcal{F}$ from a space $X$ that satisfies the finite intersection property, but $X$ is not compact? Specifically for proving $(0,1)$ is not compact?

pongdini
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    You are right. $(0,\frac 1n)$ is not closed in $(0,1)$ but $(0,\frac 1n]$ is. – Kavi Rama Murthy Sep 06 '22 at 07:17
  • When you play with topology space $(X, \tau)$ and say some set is closed, you always mean closed with respect to $\tau$. If not, you have to specify it explicitly. Theorems including different topologies are usually not so easy. So his example does refer to $(0, \frac 1 n]$ being closed in $(0, 1)$. An equivalent example in $\mathbb R$ would be $(-\infty, -n]$, or $[n, \infty)$. – Esgeriath Sep 06 '22 at 07:21

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He certainly meant $\mathcal F = \{ F_n = (0,1/n] \mid n \ge 2\}$. The $F_n$ are closed in $(0,1)$, $\mathcal F$ has the finite intersection property, but $\bigcap_{n \ge 2}F_n = \emptyset$.

Of course "collection of closed sets $\mathcal F$ from $X$" means that each $F \in \mathcal F$ is closed in $X$. This has nothing to do with a possibly bigger ambient space $Y \supset X$ in which $X$ has the subspace topology. If $X$ is not closed in $Y$, then there is no reason to believe that closed subsets of $X$ are closed in $Y$. Look at $X = (0,1)$ and $Y = \mathbb R$.

I guess your teacher introduced the concept of compactness by the standard "open cover property" (each open cover of $X$ has a finite subcover) and then proved the statement in your question. The idea is that we have a $1$-$1$-correspondence between open and closed subsets of $X$ given by forming the complement $M \mapsto X \setminus M$, which has the property that $\bigcup M_\alpha \mapsto (X \setminus \bigcup M_\alpha) = \bigcap (X \setminus M_\alpha)$. Thus open covers of $X$ corresspond to families of closed subsets with empty intersection, and open covers having finite subcovers coorespond to families of closed subsets with empty intersection having a finite subfamily with empty intersection.

This coorespondence allows you to find counterexamples. In $(0,1)$ the sets $U_n = (1/n,1)$ with $n \ge 2$ form an open cover having no finite subcover, thus their complements $F_n = (0,1) \setminus U_n$ will do.

Paul Frost
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