finite intersection compactness counter example for $(0,1)$ 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?
 A: 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.
