Vtyjkyuk

I've got a problem to show something. Actually, we consider $X$ a topological space , and we want to show :

$X$ compact $iff forall (F_i)_i in I $ collection of closed set which is closed by finite intersection : $bigcap_i in I F_i neq emptyset$

It's okay for the direction $implies$. But for the other direction, I'm stuck. Actually, this is what I did :

Let $(O_i)_i in I in (mathcalO_X)^I$ such that : $X = bigcup_i in I O_i$. If there is $i_0 in I$ such that $U_i_0 = X$, then $X = U_i_0$ and it's okay. Otherwise, we have :

$X = (bigcap_i in I U_i^c)^c$, and we consider the collection : $(U_i^c)_i in I$ which is a collection of closed subset. Then, I tried different things, I mainly try to introduce a set as : $mathcalA = ; J$ finite and $ (U_j^c)_j in J$ closed by finite intersection $$ or other set like this in order to find a $J subset I$, $J$ a finite set, which verify : $bigcup_j in J U_j = X$. But I didn't find.

If someone could help me, it will be very appreciated !

Thank you !

For the reverse direction I would recommend arguing by contrapositive.

Claim: If $X$ is not compact, then there exists a family of closed sets $U_alpha_alpha in I$ that is closed under finite intersections, such that $bigcaplimits_alphain I U_alpha = varnothing.$

Proof: Suppose $X$ is not compact. Then, there exists an open cover $V_alpha_alpha in I$ for $X$ which does not admit a finite subcover. We will assume without loss of generality that for all $alpha in I$, $V_alpha neq varnothing$. As such, we may for all $alpha in I$ define $V_alpha = U_alpha^C$ so that $mathscrG=U_alpha_alpha in I subseteq mathscrP(X)$ is a family of nonempty, closed sets. Fix a finite subset $J subset I$. Since $V_alpha_alpha in I$ does not admit a finite subcover,
$$varnothing neq bigcuplimits_alpha in J V_alpha subset X implies varnothing neq bigg(bigcuplimits_alpha in J V_alphabigg)^C subset X. $$
But from our definition of $U_alpha_alpha in I$ and de Morgan's Law, it follows that
$$bigcaplimits_alpha in JU_alpha=bigcaplimits_alphain JV_alpha^C=bigg(bigcuplimits_alpha in J V_alphabigg)^C neq varnothing.$$
Since $J subset I$ was an arbitrary finite set, we conclude that $U_alpha_alpha in I$ has the finite intersection property. Now constructing the set
$$mathscrA:=J in mathscrP(I):textrmJ finite$$ and letting $I'$ be an indexing set for $mathscrA$ so that $mathscrA=J_beta_beta in I'$, we in turn construct the family of sets $mathscrH=W_beta_beta in I'$ such that for each $beta in I'$, $W_beta=bigcaplimits_alpha in J_betaU_alpha$. Further defining $mathscrI=I cup I'$, and $mathscrF=mathscrG cup mathscrH$, we see that $mathscrF =K_alpha_alpha in mathscrIsubseteq mathscrP(X)$ is a family of closed sets which is closed under finite intersections, with $$bigcaplimits_alpha in mathscrIK_alpha=bigcaplimits_alpha in IU_alpha.$$
However, since the $V_alpha_alpha in I$ are an open cover for $X$ by assumption, we may again invoke the definition of $U_alpha_alpha in I$ to show that
$$bigcaplimits_alpha in mathscrIK_alpha=bigcap_alpha in IU_alpha=bigcap_alpha in IV_alpha^C=bigg(bigcuplimits_alpha in I V_alphabigg)^C =X^C=varnothing,$$
which is what we promised to show.

Regards!
-Dan