Normal space and Tietze/Urysohn

**CSM** · June 8th 2011, 06:50 AM

I’m having a really hard time with this one. On a conceptual level and on a technical level. I’ve got my three topology books here, but I can’t seem to make anything out of it. Can somebody enlighten me?

Let $T$ be a topological space. And let $T$ be normal and $\{C_1, C_2,…,\}$ a countable family of closed subsets of $T$ . Suppose that every point of $T$ has an neighbourhood $U$ such that $U\cap C_i\not=\emptyset$ for at most one $C_i$ . (Note in perticular that $C_i\cap C_j=\emptyset$ when $i\not= j$ .)

Prove: there are open sets $U_1,U_2,…$ such that $C_i\subset U_i$ for each $i$ and such that $\overline{U_i}\cap\overline{U_j}=\emptyset$ when $i\not= j$
(Use Tietze Extension Theorem or Urysohn’s lemma)

**Opalg** · June 9th 2011, 01:21 AM

Originally Posted by CSM

I’m having a really hard time with this one. On a conceptual level and on a technical level. I’ve got my three topology books here, but I can’t seem to make anything out of it. Can somebody enlighten me?

Let $T$ be a topological space. And let $T$ be normal and $\{C_1, C_2,…,\}$ a countable family of closed subsets of $T$ . Suppose that every point of $T$ has an neighbourhood $U$ such that $U\cap C_i\not=\emptyset$ for at most one $C_i$ . (Note in perticular that $C_i\cap C_j=\emptyset$ when $i\not= j$ .)

Prove: there are open sets $U_1,U_2,…$ such that $C_i\subset U_i$ for each $i$ and such that $\overline{U_i}\cap\overline{U_j}=\emptyset$ when $i\not= j$
(Use Tietze Extension Theorem or Urysohn’s lemma)

Each $x\in C_1$ has an open neighbourhood $U_x$ with $U_x\cap C_n = \varnothing$ for all n>1. Let $V_1 = \textstyle\bigcup_{x\in C_1} U_x$ , and use Urysohn or Tietze to get an open set $U_1$ with $C_1\subseteq U_1\subseteq\overline{U_1}\subseteq V_1.$

Now use induction. Suppose that open sets $U_1,\ldots,U_n$ with disjoint closures have been chosen so that $C_k\subseteq U_k$ and $\overline{U_k}\cap C_r = \varnothing$ for $1\leqslant k\leqslant n$ and for all $r>n.$ Show that each $x\in C_{n+1}$ has an open neighbourhood $U_x$ disjoint from $C_r$ for all r>n, and also disjoint from $\overline{U_k} (1\leqslant k\leqslant n).$ As for the case n=1, define open sets

$C_{n+1}\subseteq U_{n+1}\subseteq\overline{U_{n+1}}\subseteq V_{n+1} = \bigcup_{x\in C_{n+1}} U_x,$

and check that $U_{n+1}$ has the right properties to continue the induction.

**CSM** · June 11th 2011, 05:26 AM

Originally Posted by Opalg

Each $x\in C_1$ has an open neighbourhood $U_x$ with $U_x\cap C_n = \varnothing$ for all n>1. Let $V_1 = \textstyle\bigcup_{x\in C_1} U_x$ , and use Urysohn or Tietze to get an open set $U_1$ with $C_1\subseteq U_1\subseteq\overline{U_1}\subseteq V_1.$

Now use induction. Suppose that open sets $U_1,\ldots,U_n$ with disjoint closures have been chosen so that $C_k\subseteq U_k$ and $\overline{U_k}\cap C_r = \varnothing$ for $1\leqslant k\leqslant n$ and for all $r>n.$ Show that each $x\in C_{n+1}$ has an open neighbourhood $U_x$ disjoint from $C_r$ for all r>n, and also disjoint from $\overline{U_k} (1\leqslant k\leqslant n).$ As for the case n=1, define open sets

$C_{n+1}\subseteq U_{n+1}\subseteq\overline{U_{n+1}}\subseteq V_{n+1} = \bigcup_{x\in C_{n+1}} U_x,$

and check that $U_{n+1}$ has the right properties to continue the induction.

Thank you very much for explaining me.

But I can't grasp it totally.
In you first sentence you tell me to use Urysohn/Tietze to get an open set $U_1$ with $C_1\subseteq U_1\subseteq\overline{U_1}\subseteq V_1.$
How is Tietze/Urysohn used there?

BTW. I've got a lemma + proof in my book that says: A topological space $X$ is normal if and only if for each closed subset $E$ of $X$ and each open set $W$ containing $E$ there exists an open set $U$ containing $E$ such that $\overline{U}\subset W$ .
Isn't that just what I need for getting that set $U_1$

I thought I had to do something with functions and inverse images.... by the hint...?

I really appreciate the time you take to help me.

**CSM** · June 11th 2011, 05:39 AM

I'd really like to understand the way you tried to solve this, nevertheless I'll show another attempt at solving it (helped by others):

$\bullet$ Let $C=\bigcup_n C_n$ .
Then $C$ is closed. Because if you take a $x$ not in $C$ , we'll find a neighbourhood $O_x$ of $x$ intersecting at most one $C_n$ , say $C_k$ . Now $O_x\backslash C_k$ is open, it contains $x$ (for $x$ is not in $C_k$ because it is not in $C$ ) and misses all $C_n$ en thus $C$ ).

$\bullet$ Each $C_n$ is still closed in $C$ , (...), and for defining a continuous function, we only need to define a continuous function $f_n$ on every $C_n$ and combine them ( $f$ will be continuous again).

$\bullet$ Define $f_n$ on $C_n$ as being constant $n$ . Extend $f$ to $X$ (Tietze) and use the $f^{-1}(n-\frac{1}{3},n+\frac{1}{3})$ , $n\in N$ as the required $U_n$ .

- $f(C_1)=1, f(C_2)=2,...$ etc. So $f^{-1}(\frac{2}{3},1\frac{1}{3})=C_1$ and $f^{-1}(1\frac{2}{3},2\frac{1}{3})=C_2$ so it easy to show that their closures are disjoint.

**Opalg** · June 11th 2011, 08:33 AM

That looks like a very good way to do the question.

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