> [!theorem]- Set Construction > > ![[urysohn_sets_goal.png]] > > Let $(X, \topo)$ be a [[Normal Space|normal space]], and $A, B \subset X$ be [[Closed Set|closed sets]]. Let > $ > \Delta = \bracs{\frac{k}{2^n}: n \in \nat, 0 < k < 2^n} > $ > be the collection of dyadic rational numbers in $(0, 1)$, then there is a family of [[Open Set|open sets]] > $ > \cf = \bracs{U_{r}: r \in \Delta} > $ > such that $A \subset U_r \subset B^c$ for all $r \in \Delta$, and that the [[Topological Closure|closure]] $\ol{U_r} \subset U_{s}$ for all $r < s$. > > *Proof*. Denote $U_0 = A$ and $U_1 = B^c$, and let $\cf_{0} = \bracs{U_0, U_1}$, then $A \subset U_r \subset B^c$ for all $U_r \in \cf_0$, and $\ol{U_0} = A \subset B^c = U_1$. Moreover, all sets in $\cf_0$ except $U_0$ is open. > > ![[urysohn_sets_steps.png]] > > Let $n \in \nat$, and suppose that $\cf_{n - 1}$ satisfies the above criteria (except $U_0$ being closed), and write > $ > \cf_{n - 1} = \bracs{U_{\frac{k}{2^{n - 1}}}: 0 \le k \le 2^{n - 1}} = \bracs{U_{\frac{2k}{2^{n}}}: 0 \le k \le 2^{n - 1}} > $ > Let $k \in [0, 2^n]$ be an odd number, then $k + 1$ and $k - 1$ are even, $U_{k - 1}, U_{k + 1} \in \cf_{n - 1}$ are already defined. Since $\ol{U_{k - 1}} \subset U_{k + 1}$, $\ol{U_{k - 1}}$ and $U_{k + 1}^c$ are disjoint closed sets. As $X$ is normal, > $ > \exists U \in \topo: \ol{U_{k - 1}} \subset U \subset \ol{U} \subset (U_{k + 1}^{c})^c = U_{k + 1} > $ > Let $U_k = U$ be such an open set, then $A \subset U_k \subset B^c$. Moreover, $\ol{U_k} \subset \ol{U_{k + 1}} \subset U_s$ for all odd $s > k$, and $\ol{U_k} \subset \ol{U_{k + 1}} \subset {U_{s - 1}} \subset {U_s}$ for all even $s > k + 1$. Define > $ > \cf_{n + 1} = \bracs{U_{\frac{k}{2^n}}: 0 \le k \le 2^{n - 1}} > $ > then $\cf_{n + 1}$ also satisfies the desired criteria. > > Take $\cf = \bigcup_{n \in \nat}\cf_{n} \setminus \bracs{U_0, U_1}$, then every $U_r \in \cf$ satisfies $A \subset U_r \subset B^c$, and with the exclusion of $U_0$, every set in $\cf$ is open. > > For any $U_r, U_s$, there exists $\cf_{n}$ such that $U_r, U_s \in \cf_n$. In which case, if $r < s$, $\ol{U_r} \subset U_s$. Moreover, for any dyadic number $r = \frac{k}{2^{n}}$, $U_r \in \cf_n \subset \cf$. Therefore $\cf$ has the desired properties. > > [!theoremb] Theorem > > Let $(X, \topo)$ be a normal topological space. If $A, B \subset X$ are disjoint closed sets, then there exists a [[Continuity|continuous]] function $f: X \to [0, 1]$ such that $f = 0$ on $A$ and $f = 1$ on $B$. > > As a result, every normal space is [[Completely Regular Space|completely regular]]. > > *Proof*. Let $\Delta = \bracs{\frac{k}{2^n}: n \in \nat, 0 < k < 2^n}$ be the collection of dyadic rational numbers in $(0, 1)$, then any number $x \in [0, 1]$ can be approximated by elements of $\Delta$. For any $r \in (0, 1)$, we can write $r = \sup_{s \in \Delta: s < r}s = \inf_{s \in \Delta: s > r}s$. > > > ![[urysohn_goal.png|400]] > > Let > $ > f: X \mapsto [0, 1] \quad x \mapsto \inf\bracs{r: x \in U_r} > $ > where we assign $x$ a value based on its stage of "interpolation" from $A$ to $B$. Then since $A \subset U_r \subset B$ for all $r \in \Delta$, $f(A) = \bracs{0}$ and $f(B) = \bracs{1}$. > > ### Continuity, with Base > > Let $\alpha \in [0, 1]$, then since $\alpha = \sup_{s \in \Delta: s < \alpha}s$, $f(x) < \alpha$ if and only if there exists $s \in \Delta: s < \alpha$ such that $f(x) < s$, which only happens when $x \in \bigcup_{s \in \Delta: s < \alpha}U_s$. Therefore > $ > f^{-1}((-\infty, \alpha)) = \bigcup_{s \in \Delta: s < \alpha}U_s > $ > As each $U_s$ is open, $\bigcup_{s < \alpha}U_s$ is open. > > Since $\alpha = \inf_{s > \alpha}s$, $f(x) > \alpha$ if and only if there exists $s > \alpha$ such that $f(x) > s$ and $x \in U_s^c$. If $x \in U_s^c$, then there exists $t \in \Delta \cap (\alpha, s)$ such that $U_t \subset \ol{U_t} \subset U_s$ and $x \in \ol{U_t}^c \supset U_s^c$. If $x \in \ol{U_t}^c$ for some $t$, then there exists $s \in \Delta \cap (\alpha, t)$ such that $U_s \subset U_t \subset \ol{U_t}$ and $x \in U_s^c \supset \ol{U_t}^c$. Therefore > $ > f^{-1}((\alpha, \infty)) = \bigcup_{s \in \Delta: s > \alpha}U_s^c = \bigcup_{t \in \Delta: t > \alpha}\ol{U_t}^c > $ > As each $\ol{U_t}^c$ is open, $\bigcup_{t > \alpha}\ol{U_t}^c$ is open. > > Since the open rays generate the standard topology on $\real$, and the preimage of those open rays are open, $f$ is continuous. > > > ### Continuity, Pointwise > > For any $x \in (0, 1)$ and a given radius, we can find a "interpolating" set where all values of $f$ between two boundaries are sufficiently close to $x$. > > ![[urysohn_bound.png|500]] > > Let $x \in X$, $r \in (0, 1)$ such that $f(x) = r$, and $\varepsilon > 0$ such that $B(x, \varepsilon) \subset (0, 1)$. Take $s \in \Delta \cap (r - \varepsilon, r)$ and $t \in \Delta \cap (r, r + \varepsilon)$. > > Since $s \in \Delta$, $f(y) > s > r - \varepsilon$ for all $y \in U_s^c \supset \ol{U_s}^c$. As $t \in \Delta$, $f(y) \le t < r + \varepsilon$ for all $y \in U_t$. > > ![[urysohn_open.png|400]] > > Let $U = \ol{U_s}^c \cap U_t$, then $U$ is open. Since $f(x) > s$, $x \not\in U_s$. Let $s' \in (s, r)$, then $f(x) > s'$ and $x \not\in U_r \supset \ol{U_s} \supset U_s$, which gives $x \in \ol{U_s}^c$. As $f(x) \le t$, $x \in U_t$. Therefore $U \in \cn(x)^o$ is a neighbourhood of $x$. > > Let $V \in \cn(f(x))$ be a neighbourhood of $x$. If $x \in (0, 1)$, then there exists $\varepsilon > 0$ such that $B(x, \varepsilon) \subset V, (0, 1)$. Let $U$ be as above, then $f^{-1}(V) \supset U$ is a neighbourhood of $x$. Therefore $f$ is continuous at $x$. > > If $f(x) = 0$, then take $U = U_t$ and $B(r, \varepsilon) = [0, \varepsilon)$, and $f^{-1}(V) \supset U_t = U$ is a neighbourhood of $x$. If $f(x) = 1$, take $U = \ol{U_s}^c$, and $B(r, \varepsilon) = (1 - \varepsilon, 1]$ and $f^{-1}(V) \supset \ol{U_s}^c = U$ is a neighbourhood of $x$. > > Since $f$ is continuous at every point in $X$, $f$ is continuous.