Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2010 / A

A T0 space distinguishes any two distinct points by an open set containing one and not the other. A Tychonoff space, also called a completely regular Hausdorff space, is a $T_1$ space in which a point outside a closed set can be separated from it by a continuous function: after rescaling, $f(x)=0$ and $f{|}_C=1$ with values in $[0,1]$.

The product topology has the subbasis of a topologyA map into the product is continuous exactly when each coordinate map is continuous. One direction follows by composition with the continuous product projections; the other follows by taking inverse images of the displayed subbasic sets.

Define the diagonal evaluation map $\Delta (x)=(f_{\lambda }(x){)}_{\lambda \in \Lambda }$. Its coordinates are continuous, so the map is continuous by continuity into a product space. Point separation makes it injective. For a basic open set $B=f_{\lambda }^{-1}(U)$,Every open subset of $X$ is a union of such $B$, so its image is open in $\Delta (X)$. The inverse on the image is continuous, and $\boxed{\Delta \text{ is a topological embedding}}$. If $X$ is $T_0$, any two distinct points are distinguished by an open set and then by one basic neighborhood within it. Equality of all coordinates would prevent that distinction. Thus the basis hypothesis alone implies point separation in a $T_0$ space. The unsubscripted $f$ in the printed separation hypothesis is understood as the chosen coordinate $f_{\lambda }$.

For the Sierpiński space $S=\{0,1\}$ with opens $\varnothing ,\{1\},S$, each open $U\subseteq X$ gives a continuous characteristic map ${\chi }_U:X\to S$ with ${\chi }_U^{-1}(\{1\})=U$. These inverse images form the whole topology, hence a basis. In a $T_0$ space they separate points, proving the embedding into $S^{\tau }$.

For a Tychonoff space, use all continuous maps $X\to [0,1]$. Given $x\in U$ with $U$ open, complete regularity supplies $f(x)=0$ and $f=1$ on $X\setminus U$. Then $f^{-1}([0,1/2))$ is an open neighborhood of $x$ contained in $U$. These inverse images form a basis. The diagonal theorem gives an embedding into a product of closed unit intervals.

A three-point space works even without any separation axiom. TakeFor each open $U$ of an arbitrary space, let $g_U=2$ on $U$ and $0$ elsewhere. It is continuous. For each point $a$, let $h_a=1$ at $a$ and $0$ elsewhere. It is also continuous, because its image lies in the subspace $\{0,1\}$ of $Y$, which has the indiscrete topology; equivalently $h_a^{-1}(\{2\})=\varnothing$. The $g_U$ supply every open set as an inverse image, and the $h_a$ distinguish every pair of points. Using the diagonal evaluation map for their combined family gives a universal three-point space for product embeddings:

Finally embed the given completely regular Hausdorff space into a cube $K=[0,1{]}^I$. The Tychonoff theorem makes $K$ compact; it is Hausdorff and hence normal. The image of the compact subset $C$ is compact, so closed in $K$. The Tietze extension theorem says that a continuous bounded real-valued function on a closed subset of a normal space extends continuously to the whole space. The given function on $C$ is bounded by compactness, so apply this theorem to its transported function on the image of $C$ in $K$. Restrict the extension to the image of $X$ and pull it back along the embedding. This yields $\boxed{F:X\to \mathbb{R}\text{ continuous},F{|}_C=f}$. This proves that compact subsets of Tychonoff spaces are C-embedded, using normality of the compact cube.

For $y\in Y$, every neighborhood of $y$ in the subspace is of the form $V\cap Y$ with $V$ an ambient open neighborhood. Since $A\subseteq Y$, this meets $A$ exactly when $V$ meets $A$. Thus

A Hausdorff compactification is a topological embedding $h:X\to K$ with dense image in a compact Hausdorff space $K$. Define the domination preorder byIt is reflexive by the identity map and transitive by composition. Any comparison map is onto: its image is compact and hence closed in $K$, and contains the dense set $h(X)$. Two compactifications are equivalent if a homeomorphism $u:K\to K^{\prime }$ satisfies $u\circ h=h^{\prime }$.

Such a homeomorphism and its inverse give domination both ways. Conversely, if comparison maps exist in both directions, each composite agrees with the identity on the dense embedded copy of $X$. Two continuous maps to a Hausdorff space agreeing on a dense set agree everywhere. The comparison maps are therefore inverse homeomorphisms. Thus equivalence is exactly mutual domination.

The Stone-Čech extension property says that every continuous map from $X$ into any compact Hausdorff space extends continuously to the compactification. The extension is unique by density. For two compactifications with this property, extend each embedding through the other compactification. This gives comparison maps both ways, hence equivalence by the preceding argument.

For the perfect-map characterization by Stone-Čech remainders, the universal property first gives a unique continuous $F:\beta X\to \beta Y$ extending the composite $X\stackrel{f}{\to }Y\hookrightarrow \beta Y$. The required remainder condition is equivalent toOne inclusion is already forced by the extension property; the other forbids any point of the remainder mapping into $Y$.

Suppose first that $f$ is a perfect map. Fix $y\in Y$ and let $K=f^{-1}(y)$, a compact subset of $X$ and hence a closed subset of $\beta X$. If $p\in F^{-1}(y)\setminus K$, normality of $\beta X$ supplies an open $U$ with $K\subseteq U$ and $p\notin \overline{U}$. Because $f$ is closed,is an open neighborhood of $y$ and $f^{-1}(V)\subseteq U\cap X$. Choose an open $V^{\prime }$ of $\beta Y$ with $V^{\prime }\cap Y=V$. The open set $F^{-1}(V^{\prime })$ contains $p$, and its intersection with the dense subset $X$ lies in $U$. Every point of that open set is consequently in $\overline{U}$, contradicting the choice of $p$. Therefore $F^{-1}(y)=K\subseteq X$ for every $y\in Y$, proving the remainder condition. This argument also handles an empty fiber.

Conversely suppose the remainder condition holds. Every fiber $f^{-1}(y)=F^{-1}(y)$ is a closed subset of $\beta X$, hence compact, and lies in $X$. To show that $f$ is closed, let $A$ be closed in $X$ and set $D={\overline{A}}^{\beta X}$. The initial closure identity gives $D\cap X=A$. Compactness makes $F(D)$ closed in $\beta Y$, and the remainder condition givesIndeed any preimage in $D$ of a point of $Y$ must belong to $X$ and hence to $A$. Thus $f(A)$ is closed in $Y$. Together with compact fibers this proves

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.