# proof of Urysohn’s lemma

First we construct a family ${U}_{p}$ of open sets of $X$ indexed by the
rationals such that if $$, then $\overline{{U}_{p}}\subseteq {U}_{q}$. These
are the sets we will use to define our continuous function^{}.

Let $P=\mathbb{Q}\cap [0,1]$. Since $P$ is countable^{}, we can use
induction^{} (or recursive definition if you prefer) to define the sets
${U}_{p}$. List the elements of $P$ is an infinite^{} sequence in some way;
let us assume that $1$ and $0$ are the first two elements of this
sequence. Now, define ${U}_{1}=X\backslash D$ (the complement of $D$ in
$X$). Since $C$ is a closed set^{} of $X$ contained in ${U}_{1}$, by
normality^{} of $X$ we can choose an open set ${U}_{0}$ such that $C\subseteq {U}_{0}$ and $\overline{{U}_{0}}\subseteq {U}_{1}$.

In general, let ${P}_{n}$ denote the set consisting of the first $n$ rationals in our sequence. Suppose that ${U}_{p}$ is defined for all $p\in {P}_{n}$ and

$$ | (1) |

Let $r$ be the next rational number in the sequence. Consider ${P}_{n+1}={P}_{n}\cup \{r\}$. It is a finite subset of $[0,1]$ so it inherits
the usual ordering $$ of $\mathbb{R}$. In such a set, every element
(other than the smallest or largest) has an immediate predecessor and
successor^{}. We know that $0$ is the smallest element and $1$ the
largest of ${P}_{n+1}$ so $r$ cannot be either of these. Thus $r$ has
an immediate predecessor $p$ and an immediate successor $q$ in
${P}_{n+1}$. The sets ${U}_{p}$ and ${U}_{q}$ are already defined by the
inductive hypothesis so using the normality of $X$, there exists an
open set ${U}_{r}$ of $X$ such that

$$\overline{{U}_{p}}\subseteq {U}_{r}\text{and}\overline{{U}_{r}}\subseteq {U}_{q}.$$ |

We now show that (1) holds for every pair of elements in ${P}_{n+1}$. If both elements are in ${P}_{n}$, then (1) is true by the inductive hypothesis. If one is $r$ and the other $s\in {P}_{n}$, then if $s\le p$ we have

$$\overline{{U}_{s}}\subseteq \overline{{U}_{p}}\subseteq {U}_{r}$$ |

and if $s\ge q$ we have

$$\overline{{U}_{r}}\subseteq {U}_{q}\subseteq {U}_{s}.$$ |

Thus (1) holds for ever pair of elements in ${P}_{n+1}$ and therefore by induction, ${U}_{p}$ is defined for all $p\in P$.

We have defined ${U}_{p}$ for all rationals in $[0,1]$. Extend this definition to every rational $p\in \mathbb{R}$ by defining

$$ |

Then it is easy to check that (1) still holds.

Now, given $x\in X$, define $\mathbb{Q}(x)=\{p:x\in {U}_{p}\}$. This
set contains no number less than $0$ and contains every number greater
than $1$ by the definition of ${U}_{p}$ for $$ and $p>1$. Thus
$\mathbb{Q}(x)$ is bounded below and its infimum^{} is an element in
$[0,1]$. Define

$$f(x)=\text{inf}\mathbb{Q}(x).$$ |

Finally we show that this function $f$ we have defined satisfies the conditions of lemma. If $x\in C$, then $x\in {U}_{p}$ for all $p\ge 0$ so $\mathbb{Q}(x)$ equals the set of all nonnegative rationals and $f(x)=0$. If $x\in D$, then $x\notin {U}_{p}$ for $p\le 1$ so $\mathbb{Q}(x)$ equals all the rationals greater than 1 and $f(x)=1$.

To show that $f$ is continuous, we first prove two smaller results:

(a) $x\in \overline{{U}_{r}}\Rightarrow f(x)\le r$

*Proof*. If $x\in \overline{{U}_{r}}$, then $x\in {U}_{s}$ for all $s>r$
so $\mathbb{Q}(x)$ contains all rationals greater than $r$. Thus $f(x)\le r$ by definition of $f$.

(b) $x\notin {U}_{r}\Rightarrow f(x)\ge r$.

*Proof*. If $x\notin {U}_{r}$, then $x\notin {U}_{s}$ for all $$
so $\mathbb{Q}(x)$ contains no rational less than $r$. Thus $f(x)\ge r$.

Let ${x}_{0}\in X$ and let $(c,d)$ be an open interval of $\mathbb{R}$
containing $f(x)$. We will find a neighborhood^{} $U$ of ${x}_{0}$ such that
$f(U)\subseteq (c,d)$. Choose $p,q\in \mathbb{Q}$ such that

$$ |

Let $U={U}_{q}\backslash \overline{{U}_{p}}$. Then since $$, (b) implies that $x\in {U}_{q}$ and since $f({x}_{0})>p$, (a) implies that ${x}_{0}\notin \overline{{U}_{p}}$. Hence ${x}_{0}\in U$.

Finally, let $x\in U$. Then $x\in {U}_{q}\subseteq \overline{{U}_{q}}$, so $f(x)\le q$ by (a). Also, $x\notin \overline{{U}_{p}}$ so $x\notin {U}_{p}$ and $f(x)\ge p$ by (b). Thus

$$f(x)\in [p,q]\subseteq (c,d)$$ |

as desired. Therefore $f$ is continuous and we are done.

Title | proof of Urysohn’s lemma |
---|---|

Canonical name | ProofOfUrysohnsLemma |

Date of creation | 2013-03-22 13:09:23 |

Last modified on | 2013-03-22 13:09:23 |

Owner | scanez (1021) |

Last modified by | scanez (1021) |

Numerical id | 4 |

Author | scanez (1021) |

Entry type | Proof |

Classification | msc 54D15 |