Past exam of the options II course of the University of Oxford / 2013 / E4

A topological space is compact if every open cover of it has a finite subcover. That is, if $X={\bigcup }_{i\in I}{U}_i$ with all $U_i$ open, finitely many of those same open sets already cover $X$.

If the intersection were empty, the open complements $X\setminus F_i$ would cover the compact space. A finite subcover would give an empty finite intersection of the corresponding $F_i$. But the sets are nested, so any finite intersection is the set with largest index, which is nonempty. This contradiction provesThis is the nested-closed-set form of the finite intersection property for compact spaces.

Suppose no $F_N$ is contained in $U$. Then $K_i=F_i\cap (X\setminus U)$ are nested nonempty closed subsets of the compact space. Part (ii) gives a point in their intersection. Such a point lies both in ${\bigcap }_i{F}_i$ and outside $U$, contradicting the hypothesis. Thus some $F_N\subseteq U$, and nesting givesCompactness upgrades containment of the limiting intersection to eventual containment of the whole nested sets.

For each $b\in B$, use the supplied Hausdorff condition fact to choose disjoint open sets $U_b,V_b$ with $A\subseteq U_b$ and $b\in V_b$. Compactness of $B$ gives finitely many $b_1,\dots ,b_r$ whose $V_{b_j}$ cover $B$. SetBoth are open; $U$ contains $A$ and $V$ contains $B$. They are disjoint because membership in $U\cap V$ would put a point in some $U_{b_j}\cap V_{b_j}$. If one compact set is empty, use the empty set and $X$ directly. Thus disjoint compact subspaces of a Hausdorff space admit disjoint open neighborhoods.

A topological space is connected if it cannot be written as the union of two disjoint nonempty open subsets. Equivalently, its only subsets that are both open and closed are the empty set and the whole space. Under this definition the empty space is connected.

If $A$ is disconnected, a separation writes it as $A=A_1\cup A_2$ with $A_1,A_2$ disjoint, nonempty and open in its subspace topology. By definition there are ambient open sets $U,V$ with $A_1=U\cap A$ and $A_2=V\cap A$. These give exactly the four conditions stated.

Conversely, those conditions make $U\cap A,V\cap A$ two nonempty disjoint relatively open sets whose union is $A$. They are therefore a separation. This proves the equivalence. The ambient sets may overlap outside $A$, which is why the condition here is $U\cap V\cap A=\varnothing$.

If $A$ is disconnected, take a separation $A=A_1\cup A_2$. Each piece is also relatively closed, being the complement of the other. They are compact as closed subsets of the compact space $A$.

Part (b), using the ambient Hausdorff property, gives disjoint ambient open neighborhoods of $A_1$ and $A_2$. They meet $A$, cover it, and are disjoint everywhere, as required. Conversely, such ambient open sets give a separation on intersecting with $A$, by part (ii). Hence compactness and the Hausdorff property permit fully disjoint ambient neighborhoods.

Put $K={\bigcap }_i{F}_i$. If some $F_i$ is empty, so is $K$, which is connected. Otherwise the compact sets are closed in the Hausdorff ambient space, and can all be considered as nested nonempty closed subsets of the compact space $F_0$. Part (a)(ii) gives $K\ne \varnothing$, and $K$ is compact.

Suppose $K$ were disconnected. Part (c)(iii) gives disjoint ambient open sets $U,V$ covering $K$, both meeting it. Part (a)(iii), applied within $F_0$ to $U\cup V$, gives $F_N\subseteq U\cup V$ for some $N$. Both $U$ and $V$ meet $F_N$ because they meet $K\subseteq F_N$. Their intersections with $F_N$ would therefore disconnect $F_N$, a contradiction. Thus