= Solution
Write $K=\bigcap_{i\in I}K_i$. Each $K_i$ is <closed> in $X$, by the <compact subset of a Hausdorff space> theorem. Therefore $K$ is <closed>, and is <compact> because $X$ is <compact>. The nonempty finite <intersections> give the <finite intersection property>, so part (a)(ii) also yields $K\ne\varnothing$.
Suppose $K$ were not <connected>. Part (c) would give <disjoint> <open sets> $U,V$ containing $K$ in their <union> and each meeting $K$. Let
$$
C=X\setminus(U\cup V).
$$
This is a <closed subset> of the <compact space> $X$, and $C\cap\bigcap_{i\in I}K_i=\varnothing$. Apply the contrapositive of the <finite intersection characterization of compactness> to the family of <closed sets> $\{C\}\cup\{K_i:i\in I\}$. Some finite subfamily has empty <intersection>. Since every finite <intersection> of just the $K_i$ is nonempty, that subfamily must include $C$. Thus for some finite $J\subseteq I$,
$$
C\cap\bigcap_{j\in J}K_j=\varnothing,
\qquad\text{or equivalently}\qquad
K_J:=\bigcap_{j\in J}K_j\subseteq U\cup V.
$$
But $K\subseteq K_J$ and $K$ meets both $U$ and $V$. Therefore $K_J\cap U$ and $K_J\cap V$ are nonempty <disjoint> relatively <open sets> covering $K_J$, contradicting its assumed <connectedness>. This proves
$$
\boxed{\bigcap_{i\in I}K_i\text{ is nonempty, compact and connected}.}
$$
This is <connectedness of an intersection from connected finite intersections>. If $I=\varnothing$, its <intersection> is $X$; the stated hypothesis for the finite <set> $J=\varnothing$ already makes $X$ nonempty and <connected>, so that case is included.
Back to article page
