Urysohn's Lemma
== Lemma == Let $T = \struct {S, \tau}$ be a $T_4$ topological space. Let $A, B \subseteq S$ be closed sets of $T$ such that $A \cap B = \O$. Then there exists an Urysohn function for $A$ and $B$. == Proof == Let $T = \struct {S, \tau}$ be a $T_4$ space. Let $A, B \subseteq S$ be closed sets of $T$ such that $A \cap B = \O$. Let $P = \Q \cap \closedint 0 1$ where $\closedint 0 1$ is the closed unit interval. $\Q$ is countable, therefore so is $P$. === Creation of Domain === We are going to construct a set $\Bbb U \subseteq \tau$ of open sets with $P$ as an indexing set: :$\Bbb U = \set {U_i: i \in P}$ such that: :$\forall p, q \in P: p < q \implies {U_p}^- \subseteq U_q$ where ${U_p}^-$ denotes the set closure of $U_p$. We define $U_p$ by induction, as follows. List the elements of $P$ in the form of an infinite sequence $\sequence z$. Let $z_0 = 1, z_1 = 0$. In general, let $P_n$ denote the set consisting of the first $n$ elements of $\sequence z$. Let $\map \PP n$ be the proposition: :$U_p$ is defined for all $p \in P_n$, and: ::$(1): \quad \forall p, q \in P_n: p < q \implies {U_p}^- \subseteq U_q$ ==== Basis for the Induction ==== As $S$ is a $T_4$ space we have that: :$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U_1, V \in \tau: A \subseteq U_1, B \subseteq V$  We have that $A \subseteq U_1$ is a closed set of $S$. We define $U_1 = S \setminus B$ where $S \setminus B$ denotes the complement of $B$ in $S$. As $S$ is $T_4$, we can choose an open set $U_0 \in \tau$ such that $A \subseteq U_0$ and ${U_0}^- \subseteq U_1$. Thus $\map \PP 1$ is shown to hold. This is our basis for the induction. ==== Induction Hypothesis ==== This is our induction hypothesis: Let $\map \PP k$ be the proposition: :$U_p$ is defined for all $p \in P_k$, and: :: $(1): \quad \forall p, q \in P_k: p < q \implies {U_p}^- \subseteq U_q$ We want to show that if $\map \PP k$ holds, then: :$U_p$ is defined for all $p \in P_{k + 1}$, and: :: $(1): \quad \forall p, q \in P_{k + 1}: p < q \implies {U_p}^- \subseteq U_q$ ==== Induction Step ==== This is our induction step: Let $r = z_{k + 1}$ be the next element in $\sequence z$. Consider $P_{k + 1} = P_k \cup \set r$. It is a finite subset of the closed unit interval $\closedint 0 1$. We consider the usual $<$ ordering on $P_{k + 1}$, which is a subset of $\closedint 0 1$ which in turn is a subset of $\R$.  From Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements, $P_{k + 1}$ has both a minimal element $m$ and a maximal element $M$. From Predecessor and Successor of Finite Toset, every element other than $m$ and $M$ has an immediate predecessor and immediate successor. We already know that $z_1 = 0$ is the minimal element and $z_0 = 1$ is the maximal element of $P_{k + 1}$. So $r$ must be neither of these. Thus: :$r$ has an immediate predecessor $p$ :$r$ has an immediate successor $q$ in $P_{k + 1}$. The sets $U_p$ and $U_q$ are already defined by the inductive hypothesis. As $T$ is a $T_4$ space, there exists an open set $U_r \subseteq \tau$ such that: :${U_p}^- \subseteq U_r$ :${U_r}^- \subseteq U_q$ We now show that $(1)$ holds for every pair of elements in $P_{k + 1}$. If both elements are in $P_n$, then $(1)$ is true by the inductive hypothesis. If one is $r$ and the other is $s \in P_k$, then: :$s < p \implies {U_s}^- \subseteq {U_p}^- \subseteq U_r$ and: :$s \ge q \implies U_r \subseteq {U_q}^- \subseteq {U_s}^-$ Thus $(1)$ holds for every pair of elements in $P_{k + 1}$. Therefore by induction, $U_p$ is defined for all $p \in P$. We have defined $U_p$ for all rational numbers in $\closedint 0 1$. We now extend this definition to every rational $p$ by defining: :$U_p = \begin{cases} \O & : p < 0 \\ S & : p > 1 \end{cases}$ It is easily checked that $(1)$ still holds. === Definition of Function === Let $x \in S$. Define $\map \Q x = \set {p: x \in U_p}$. This set contains no rational number less than $0$ and contains every rational number greater than $1$ by the definition of $U_p$ for $p < 0$ and $p > 1$. Thus $\map \Q x$ is bounded below and its infimum is an element in $\closedint 0 1$. We define: :$\map f x = \map \inf {\map \Q x}$ Now we need to show that $f$ satisfies the conditions of Urysohn's Lemma. Let $x \in A$. Then $x \in U_p$ for all $p \ge 0$, so $\map \Q x$ equals the set $\Q_+$ of all non-negative rationals. Hence $\map f x = 0$. Let $x \in B$. Then $x \notin U_p$ for $p \le 1$ so $\map \Q x$ equals the set of all the rationals greater than $1$. Hence $\map f x = 1$. === Continuity of Function === To show that $f$ is continuous, we first prove two smaller results: :$(a): \quad x \in {U_r}^- \implies \map f x \le r$ We have: $x \in {U_r}^- \implies \forall s > r: x \in U_s$ so $\map \Q x$ contains all rationals greater than $r$. Thus $\map f x \le r$ by definition of $f$. {{qed|lemma}} :$(b): \quad x \notin U_r \implies \map f x \ge r$ We have: $x \notin U_r \implies \forall s < r: x \notin U_s$ so $\map \Q x$ contains no rational less than $r$. Thus $\map f x \ge r$. {{qed|lemma}} Let $x_0 \in S$ and let $\openint c d$ be an open real interval containing $\map f x$. We will find a neighborhood $U$ of $x_0$ such that $\map f U \subseteq \openint c d$. Choose $p, q \in \Q$ such that: :$c < p < \map f {x_0} < q < d$ Let $U = U_q \setminus {U_p}^-$. Then since $\map f {x_0} < q$, we have that $(b)$ implies vacuously that $x \in U_q$. Since $\map f {x_0} > p$, $(a)$ implies that $x_0 \notin U_p$. Hence $x_0 \in U$ and thus $U$ is a neighborhood of $x_0$ because $U$ is open. Finally, let $x \in U$. Then $x \in U_q \subseteq {U_q}^-$, so $\map f x \le q$ by $(a)$. Also, $x \notin {U_p}^-$ so $x \notin U_p$ and $\map f x \ge p$ by $(b)$. Thus: $\map f x \in \closedint p q \subseteq \openint c d$ Therefore $f$ is continuous.  == Also see == * Urysohn's Lemma Converse {{Namedfor|Pavel Samuilovich Urysohn|cat = Urysohn}} == Sources == * {{BookReference|Counterexamples in Topology|1978|Lynn Arthur Steen|author2 = J. Arthur Seebach, Jr.|ed = 2nd|edpage = Second Edition|prev = Definition:Urysohn Function|next = Urysohn's Lemma Converse}}: Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces * {{planetmath|url = proofofurysohnslemma|title = Urysohn's Lemma}}