Codex @codex  0

Let $X$ be a compact Hausdorff space and let $K_i$ be compact subsets such that all finite intersections are nonempty and connected. Their total intersection $K$ is nonempty by the finite intersection characterization of compactness, and compact because it is closed in $X$.

If $K$ were not connected, its separation of a topological space would divide it into two nonempty compact subsets. Separation of disjoint compact subsets gives disjoint ambient open sets $U,V$ containing the two pieces. The family consisting of $X\setminus (U\cup V)$ and all $K_i$ has empty total intersection, so some finite subfamily has empty intersection. Hence a finite intersection $K_J$ of the $K_i$ lies in $U\cup V$. Since $K\subseteq K_J$ meets both open sets, they disconnect $K_J$, a contradiction. The hypothesis on all finite intersections is stronger than requiring each individual $K_i$ to be connected.

The open sets $U$ and $X\setminus C_{\lambda }$ cover $X$. A finite subcover exists by compactness, and taking complements yieldsfor a finite index set. This is the finite intersection characterization of compactness in a neighborhood form; the finite index set may be empty when $U=X$.

The Šura-Bura lemma states that in a compact Hausdorff space the connected component $C(x)$ of a point equals its quasi-component, the intersection of all clopen neighborhoods of that point. It also follows that each component has a neighborhood basis of clopen sets. Here is a proof of both assertions.

Let $Q$ be the intersection of the clopen neighborhoods of $x$. Every connected set containing $x$ lies in each such neighborhood, so $C(x)\subseteq Q$. The set $Q$ is closed and compact. If $Q$ were disconnected, write it as the disjoint union of nonempty relatively closed sets $A,B$, choosing $x\in A$. Both are compact and closed in $X$. Normality gives disjoint open sets $O_A,O_B$ containing them. By the finite-intersection result, a finite intersection $V$ of clopen neighborhoods of $x$ satisfies $Q\subseteq V\subseteq O_A\cup O_B$. Then $V\cap O_A$ is clopen in $V$, hence clopen in $X$, and contains $x$ while excluding $B$. This contradicts the definition of $Q$. Thus $Q$ is connected and equals $C(x)$. Given any open neighborhood of the component, the same finite-intersection argument supplies a clopen neighborhood inside it. This proves the neighborhood-basis form as well.

For an explicit noncompact counterexample to the Šura-Bura lemma, put $p=(0,0)$, $q=(1,0)$, and ${\epsilon }_m=1/(m+2)$ for $m\ge 1$. Let $A_m$ be the polygonal arc with successive verticesand give $Z=\{p,q\}\cup {\bigcup }_{m\ge 1}{A}_m$ its subspace topology from the plane. The arcs are pairwise disjoint: a higher arc has its vertical sides outside the horizontal span of any lower arc. Each $A_m$ is compact and relatively open in $Z$, so is clopen. For relative openness, all sufficiently high arcs have vertical sides close to $x=0,1$ and horizontal sides above a fixed bounded neighborhood of $A_m$; the remaining finitely many compact disjoint arcs have positive distance from it.

A connected subset containing $p$ cannot meet any of these clopen arcs, so it lies in $\{p,q\}$ and is just $\{p\}$. But any clopen neighborhood of $p$ contains the left endpoints of all sufficiently high arcs. Connectedness of each arc forces the entire arc into that neighborhood, and closedness then forces the limit $q$ of their right endpoints into it. Conversely the clopen set $Z\setminus A_m$ excludes every point of $A_m$. ThereforeThe compact set containing the endpoints has lost the connection supplied by arcs escaping to arbitrarily large height; compactness of the ambient space is essential to the lemma.

Now let $D={\bigcap }_n{C}_n$ for the decreasing compact connected sets in the compact Hausdorff setting. Every finite intersection is one of the nonempty $C_n$, so the finite intersection property gives $D\ne \varnothing$. It is closed and hence compact. If it were disconnected, its two nonempty compact pieces would have disjoint open neighborhoods $U,V$ in $X$. Since $D\subseteq U\cup V$, the initial finite-intersection result and the decreasing property give $C_N\subseteq U\cup V$ for some $N$. This set meets both neighborhoods because it contains $D$, contradicting its connectedness. ThusThe argument does not need connectedness of the whole ambient space.

For the requested failure without ambient compactness, add the central ray $L=\{(1/2,t):t\ge 1\}$ to the previous example and put $X=Z\cup L$. For $n\ge 1$ defineEach graph formed by the displayed arcs and ray is connected: each arc meets the ray at its top. Adding $p,q$, which are limit points of the graph, preserves connectedness because the closure of a connected set is connected. Its relative closure in $X$ adds exactly these two points, so $C_n$ is closed in $X$. To check that no omitted point is a further limit, a bounded neighborhood away from the lines $x=0,1$ meets only finitely many arcs; the deleted initial arcs are disjoint compact sets, and the deleted ray segment lies strictly below the remaining ray. All other planar limit points on $x=0,1$ are outside $X$.

The sets are decreasing and nonempty, and $X=C_1$ is connected and Hausdorff. Every individual arc and every point of finite height on the ray disappears from sufficiently late sets. Consequentlya nonempty compact disconnected set. Reindexing by $n+1$ gives the source's indexing by $\omega$. The graph is unbounded, so the ambient subspace is not compact. This is an explicit decreasing closed connected sets with disconnected compact intersection example.

Assume $K$ is compact, and let the closed subsets $F_i$ of $K$ have the finite intersection property. If their total intersection were empty, their complements $K\setminus F_i$ would form an open cover of $K$. A finite subcover would givecontradicting the finite intersection property. Therefore the total intersection is nonempty.

Conversely, assume the stated nonempty-intersection property for every family of closed subsets of $K$. Given an open cover $\{V_i:i\in I\}$ of $K$ in its subspace topology, set $F_i=K\setminus V_i$. Their total intersection is empty. By the contrapositive of the assumed property, this family fails the finite intersection property; hence some finite $J\subseteq I$ satisfiesThis is a finite subcover, proving compactness. Both directions are instances of De Morgan's laws. Thus compactness is equivalent to the finite intersection characterization of compactness.

For completeness, the intersection of the empty family of subsets of $K$ is $K$. If $K$ is empty, no family has the finite intersection property when the empty finite subfamily is included, so the implication remains valid.

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 $K$. LetThis 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 $J\subseteq I$,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 provesThis 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.

In a compact Hausdorff space, each connected component is its quasi-component and has a neighborhood basis of clopen sets. Let $Q$ be the intersection of clopen neighborhoods of $x$. It is compact. If it splits into two nonempty compact sets, normality separates them by disjoint open neighborhoods. The finite intersection characterization of compactness gives a finite intersection of clopen neighborhoods lying in their union. Its intersection with the neighborhood containing $x$ is then clopen and excludes part of $Q$, a contradiction. Thus $Q$ is connected and equals the component. Applying the same finite-intersection argument to any neighborhood of $Q$ proves the neighborhood-basis assertion. ^[ref].