If J is uncountable, then $R^J$ is not normal.
Let X = $(Z+)^J$; it
will suffice to show that X is not normal, since X is a closed subspace of $R^J$ . We
use functional notation for the elements of X, so that the typical element of X is
a function x : J $\rightarrow$ Z+.
(a) If x $\in$ X and if В is a finite subset of J, let U(x, B) denote the set consisting
of all those elements у of X such that у ($\alpha$) = x($\alpha$) for $\alpha\in$ B. Show the sets
U(x, B) are a basis for X.
(b) Define $P_n$ to be the subset of X consisting of those x such that on the set
J — $x^{-1}$(n), the map x is injective. Show that $P_1$ and $P_2$ are closed and
disjoint.
(c) Suppose U and V are open sets containing $P_1$ and $P_2$, respectively. Given a
sequence $\alpha_1$,$\alpha_2$ ,… of distinct elements of J, and a sequence
0 =
< $n_0$ < $n_1$ <…
of integers, for each i > 1 let us set
$B_i$ ={ $\alpha_1$,$\alpha_2$,…,$\alpha_{n_i}$}
and define $x_i$ $\in$ X by the equations
$x_i$($\alpha_j$) = J for 1 =< j =< $n_{i-1}$ .
$x_i(\alpha)$ = 1 for all other values of $\alpha$.
Show that one can choose the sequences $\alpha_j$ and $n_j$ so that for each i, one
has the inclusion
U($x_i$,$B_i$)$\subset$U.
[Hint: To begin, note that $x_1(\alpha)$ = 1 for all $\alpha$; now choose $B_1$ so that
U($x_1$,$ B_1$) $\subset$ U]
this is an exercise from munkres topology ive done part a & b but i have no idea for part c . can anyone help me with c ?
of course there is some similar questions about this problem but they werent any help.
thank u.
Best Answer
$\newcommand{\set}[1]{\left\{#1\right\}}$
I found the article in the link unnecessarily complicated, so I wrote a more readable proof below: (Why do I type essentially the same proof again? Well, because it is fun. Isn't it why we do mathematics?)
$U(x,B)$ is an open subset of $X$, and $x\in U(x,B)$. If $U$ is a neighborhood of $x\in X$, then there exists a finite subset $B\subseteq J$ such that $U(x,B)\subseteq U$. Hence, the sets of the form $U(x,B)$ form a basis of $X$. Note that if $B_1\subseteq B_2$, then $U(x,B_2)\subseteq U(x,B_1)$.
For each $n\in\mathbb{Z}_+$, let $P_n$ be the set of $x\in X$ which is injective on $x^{-1}(\mathbb{Z}_+\backslash\set{n})$. The constant function $J\to\set{n}$ is an element of $P_n$, so $P_n\neq\varnothing$.
Let $x\in P_n$. Then $x^{-1}(\mathbb{Z}_+\backslash\set{n})$ is countable, so its complement $x^{-1}(\set{n})$ is uncountable. If $m\in\mathbb{Z}_+\backslash\set{n}$, then $$x^{-1}(\set{n})\subseteq x^{-1}(\mathbb{Z}_+\backslash\set{m}),$$ so $x^{-1}(\mathbb{Z}_+\backslash\set{m})$ is uncountable, hence $x\not\in P_m$, i.e., $P_n\cap P_m=\varnothing$.
Each $P_n$ is a closed subset of $X$. To see this, let $x\in X\backslash P_n$. Then $x$ is not injective on $x^{-1}(\mathbb{Z}_+\backslash\set{n})$, so $x(s_1)=x(s_2)\neq n$ for some distinct $s_1,s_2\in J$. If $y\in U(x,\set{s_1,s_2})$, then $y(s_1)=y(s_2)\neq n$, so $y$ is not injective on $y^{-1}(\mathbb{Z}_+\backslash\set{n})$, so $y\not\in P_n$. Hence, $U(x,\set{s_1,s_2})\cap P_n=\varnothing$.
Let $U$ and $V$ be open subsets of $X$ containing $P_1$ and $P_2$, respectively. We will show that $U\cap V\neq\varnothing$, proving that $X$ is not quasinormal.
Let $x_1\in X$ be the constant function $J\to\set{1}$. Then $x_1\in P_1\subseteq U$, so there exists a finite nonempty\ subset $B_1\subseteq J$ such that $U(x_1,B_1)\subseteq U$. Let $$B_1=\set{s_1,\ldots,s_{n_1}},$$ where $n_1\geq1$.
Let $x_2\in X$ be the function defined as follows: $$\begin{cases} x_2(s_i)=i\quad&\mbox{for}\ \ 1\leq i\leq n_1,\\ x_2(s)=1\quad&\mbox{if}\ \ s\not\in B_1. \end{cases}$$ Then $x_2\in P_1\subseteq U$, so there exists a finite subset $B_2\subseteq J$ such that $B_1\varsubsetneqq B_2$ and $U(x_2,B_2)\subseteq U$. Let $$B_1=\set{s_1,\ldots,s_{n_1},s_{n_1+1},\ldots,s_{n_2}},$$ where $n_2> n_1$.
Let $x_3\in X$ be the function defined as follows: $$\begin{cases} x_3(s_i)=i\quad&\mbox{for}\ \ 1\leq i\leq n_2,\\ x_3(s)=1\quad&\mbox{if}\ \ s\not\in B_2. \end{cases}$$ Then $x_3\in P_1\subseteq U$, so there exists a finite subset $B_3\subseteq J$ such that $B_2\varsubsetneqq B_3$ and $U(x_3,B_3)\subseteq U$. Let $$B_3=\set{s_1,\ldots,s_{n_1},s_{n_1+1},\ldots,s_{n_2},s_{n_2+1},\ldots,s_{n_3}},$$ where $n_3> n_2$.
Continuing in this way, we obtain a sequence $\set{U(x_i,B_i)}_{i=1}^\infty$ such that $U(x_i,B_i)\subseteq U$. Let $$B\mathrel{\mathop:}=\set{s_1,s_2,s_3,\ldots}=\bigcup_{i=1}^\infty B_i,$$ and $y\in X$ the function defined by $$\begin{cases} y(s_i)=i\quad&\mbox{for all}\ \ i\geq1,\\ y(s)=2\quad&\mbox{if}\ \ s\not\in B. \end{cases}$$ Then $y\in P_2\subseteq V$, so there exists a finite subset $C\subseteq J$ such that $U(y,C)\subseteq V$. We can choose $i\in\mathbb{Z}_+$ large enough so that $B\cap C\subseteq B_i$. If $1\leq j\leq n_i$, then $$x_{i+1}(s_j)=y(s_j)=j,$$ so $x_{i+1}|_{B_i}=y|_{B_i}$. Since $$B_{i+1}\cap C\subseteq B\cap C\subseteq B_i,$$ it follows that $$x_{i+1}|_{B_{i+1}\cap C}=y|_{B_{i+1}\cap C}.$$ Hence, there exists $z\in X$ such that $$z\in U(x_{i+1},B_{i+1})\cap U(y,C).$$ Since $U(x_{i+1},B_{i+1})\subseteq U$ and $U(y,C)\subseteq V$, we conclude that $U\cap V\neq\varnothing$.