Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2008 / A / 2

For each point which belongs to at least one $U_t$, its membership-index set is a nonempty subset of $[0,1]$, so has an infimum in $[0,1]$. Assigning the value one to the other points therefore defines $f:X\to [0,1]$ everywhere.

For $0<r\le 1$,An index $t<r$ gives $f(x)\le t<r$. Conversely if the infimum is below $r$, some membership index must be below $r$. Thus this strict sublevel set is open.

For $0\le r<1$ we claimIf $x\notin \overline{U_t}$ and $t>r$, then $x$ cannot belong to any $U_s$ with $s\le t$, since $U_s\subseteq U_t$ for $s<t$. Hence $f(x)\ge t>r$. Conversely, if $f(x)>r$, density of $D$ supplies $r<t<u<f(x)$ with $t,u\in D$. Since $x\notin U_u$ and $\overline{U_t}\subseteq U_u$, one has $x\notin \overline{U_t}$. This proves the identity, and its right-hand side is open.

The remaining sublevel and superlevel sets, for thresholds outside these ranges, are either empty or all of $X$. Preimages of open rays are consequently open, and these rays generate the usual real topology. Therefore $f$ is continuous. The closure inclusion is what supplies openness of the strict superlevel sets; mere nesting would not suffice.

Urysohn's lemma states that, in a normal space, disjoint closed subsets $A,B$ can be separated by a continuous function $u:X\to [0,1]$ with $u=0$ on $A$ and $u=1$ on $B$.

First use the shrinking consequence of the separation property of a normal space: if a closed set $F$ lies in an open set $G$, there is an open $W$ with $F\subseteq W\subseteq \overline{W}\subseteq G$. To prove it, separate the disjoint closed sets $F$ and $X\setminus G$ by disjoint open neighborhoods $W,V$. Then $\overline{W}\subseteq X\setminus V\subseteq G$.

Let $D$ be the dyadic rationals in $[0,1]$. Set $U_1=X\setminus B$, and shrink around $A$ to choose $U_0$ with $A\subseteq U_0\subseteq \overline{U_0}\subseteq U_1$. At each subsequent dyadic stage, insert between each adjacent pair $s<t$ an open set $U_{(s+t)/2}$ such thatThe shrinking lemma applies because the first set is closed and contained in the last open set. Induction gives $\overline{U_s}\subseteq U_t$ for all dyadic $s<t$. Part (a) now constructs a continuous $u$. Each point of $A$ lies in $U_0$, so has value zero. Each point of $B$ lies in none of the $U_t$, so has value one. This proves the lemma.

For the embedding, let $I=C(X,[0,1])$, a set of continuous functions, and define the evaluation mapEvery coordinate is continuous, so $e$ is continuous in the product topology. The Hausdorff assumption makes singletons closed. For distinct points, Urysohn's lemma applied to their singletons gives a coordinate with different values, so $e$ is injective.

To check the inverse topology, take $x\in O$ with $O$ open. Apply the lemma to $\{x\}$ and $X\setminus O$, choosing a coordinate $u$ with $u(x)=1$ and $u=0$ outside $O$. ThenThis is the preimage of an open coordinate cylinder under $e$. Thus every original neighborhood is detected by coordinate neighborhoods, and $e^{-1}:e[X]\to X$ is continuous. The evaluation map is a topological embedding.

Let $\mathcal{B}$ be a topological basis consisting of clopen sets. For $U\in \mathcal{B}$ its characteristic function ${\chi }_U:X\to \{0,1\}$ is continuous, since the preimages of both singleton values are open. DefineThe coordinate criterion for product topology makes this map continuous. If $x\ne y$, the $T_0$ property gives an open set containing one and not the other. A topological basis element contained in that open set still distinguishes them, so their corresponding characteristic coordinates differ. Therefore $e$ is injective.

Finally, for each topological basis element $U$,The right-hand side is the preimage of an open cylinder in the product of discrete two-point topological spaces. These cylinders recover a topological basis for the topology of $X$, proving continuity of the inverse onto the image. Hence