If J is uncountable, then $R^J$ is not normal.

Question

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.

Answer 1

I think that this link will explain all. It does all the steps, in quite similar notation as well.

Answer 2

$\newcommand{\set}[1]{\left\{#1\right\}}$

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 $$B_2=\set{s_1,\ldots,s_{n_1},s_{n_1+1},\ldots,s_{n_2}},$$ 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 $U\cap V\neq\varnothing$.

Answer 3

Solution given by @ashpool is really good indeed. Yet, I have modified a bit for smoothness as follows.

$U(x,B) \text{ is an open subset of } X, \text{ and } x\in U(x,B).\\ \text{ If } U \text{ is a neighborhood of } \ x\in X, \text{ then there exists a finite subset } B\subseteq J \text{ such that } U(x,B)\subseteq U.\\ \text{ Hence, the sets of the form } U(x,B) \text{ form a basis of } X.\\ \text{ Note that if } B_1\subseteq B_2, \text{ then } U(x,B_2)\subseteq U(x,B_1).\\ \text{ For each } n\in\mathbb{Z}_+, \text{ let }P_n \text{ be the set of } x\in X \text{ which is injective on } x^{-1}\left(\mathbb{Z}_+-\left\{n\right\}\right)=J-x^{-1}(n).\\ \text{ The constant function } J\rightarrow\{n\} \text{ is an element of }P_n, \text{ so } P_n\neq\emptyset. \text{ Let } x\in P_n.\\ \text{ Then } x^{-1}\left(\mathbb{Z}_+-\left\{n\right\}\right) \text{ is countable, so its complement } x^{-1}(\left\{n\right\}) \text{ is uncountable [since }\\ J \text{ is uncountable].}\\ \text{ If } m\neq n, \text{ then } n\in\mathbb{Z}_+-\left\{m\right\}, \text{ hence }x^{-1}({n})\subseteq x^{-1}(\mathbb{Z}_+-\left\{m\right\}),\\ \text{ so } x^{-1}(\mathbb{Z}_+-\left\{m\right\}) \text{ is uncountable, hence } x\notin P_m. \text{ Hence, } P_n\cap P_m=\emptyset.\\ \text{ Each } P_n \text{ is a closed subset of } X.\\ \text{ To see this, let } x\in X-P_n.\\ \text{ Then } x \text{ is not injective on } x^{-1}\left(\mathbb{Z}_+-\left\{n\right\}\right), \text{ so } x(s_1)=x(s_2)\neq n \text{ for some distinct } s_1,s_2\in J.\\ \text{ If } y\in U(x,{s_1,s_2}), \text{ then } y(s_1)=y(s_2)\neq n, \text{ so } y \text{ is not injective on } y^{-1}(\mathbb{Z}_+-\left\{n\right\}), \text{ so } y\in X-P_n, \text{ i.e., } U\left(x,\left\{s_1,s_2\right\}\right)\subset X-P_n.\\ \text{ Hence, } X-P_n \text{ is open.}\\ \text{ Let } U \text{ and } V \text{ be open subsets of } X \text{ containing } P_1 \text{ and } P_2, \text{ respectively.}\\ \text{ We will show that } U\cap V\neq\emptyset, \text{ proving that } X \text{ is not quasinormal.}\\ \text{ Let } x_1\in X \text{ be the constant function } J\rightarrow\{1\}. \text{ Then } x_1\in P_1\subseteq U, \text{ so there exists a finite nonempty subset } B_1\subseteq J \text{ such that } U(x_1,B_1)\subseteq U. \\ \text{ Let } B_1={\alpha_1,\ldots,\alpha_{n_1}},\text{ where } n_1\geq1.\\ \text{ Let } x_2\in X \text{ be the function defined as follows: }\\ x_2(\alpha_i)=i\ \ \mathrm{for}\ \ 1\le i\le n_1,x_2(\alpha)=1\ \mathrm{if} \ \alpha∉B_1.\\ \text{ Then } x_2\in P_1\subseteq U, \text{ so there exists a finite subset } B_2\subseteq J \text{ such that } B_1⫋B2 \text{ and } U(x_2,B_2)\subseteq U.\\ \text{ Let}B_2={\alpha_1,\ldots,\alpha_{n_1},\alpha_{n_1+1},\ldots,\alpha_{n_2}}, \text{ where } n_2>n_1.\\ \text{ Then we have } x_2\left(\alpha_i\right)=i\ \mathrm{for}\ \ \alpha_i\in\ B_1,\ \ x_2\left(\alpha_i\right)=1\ \mathrm{\mathrm{if}}\ \ \alpha_i\in B_2-B_1. $ $\text{ Let } x_3\in X \text{ be the function defined as follows:}\\ x_3(\alpha_i)=i\ \ \mathrm{for\ }\ \ 1\le i\le n_2,x_3(\alpha)=1\ \mathrm{if\ }\ \ \alpha\notin B_2.\\ \text{ Then } x_3\in P_1\subseteq U, \text{ so there exists a finite subset } B_3\subseteq J \text{ such that } B_2⫋B_3 \text{ and } U(x_3,B_3)\subseteq U.\\ \text{ Let } B_3={\alpha_1,\ldots,\alpha_{n_1},\alpha_{n_1+1},\ldots,\alpha_{n_2},\alpha_{n_2+1},\ldots,\alpha_{n_3}}, \text{ where } n_3>n_2.\\ \text{ Then we have } x_3\left(\alpha_i\right)=i\ \mathrm{for}\ \ \alpha_i\in\ B_2,\ \ x_2\left(\alpha_i\right)=1\ \mathrm{\mathrm{if}}\ \ \alpha_i\in B_3-B_2. $ $\text{ Continuing in this way, we obtain a sequence } \left\{U\left(x_i,B_i\right)\right\}_{i=1}^\infty\ \text{ such that } U(x_i,B_i)\subseteq U.\\ \text{ Let }A:={\alpha_1,\alpha_2,\alpha_3,\ldots}=\bigcup_{i=1}^{\infty}B_i, \text{ and } y\in X \text{ the function defined by}\\ y\left(\alpha_j\right)=j\ \ \mathrm{for\ all}\ \ \alpha_j\in A,\ \ y(\alpha)=2\ \ \mathrm{if}\ \ \alpha\notin A.\\ $ $\text{ Then } y\in P_2\subseteq V, \text{ so there exists a finite subset } B\subseteq J \text{ such that } U(y,B)\subseteq V.\\ \text{ We can choose } i\in\mathbb{Z}_+ \text{ large enough so that } A\cap B\subseteq B_i.\\ \text{ Then we may define a point } z\in\ X \text{ by }\\ z\left(\alpha_i\right)=i\ \text{ whenever } \alpha_i\in\ B_i, z\left(\alpha_i\right)=1\ \text{ whenever } \mathrm{\ }\alpha_i\in\ B_{i+1}-B_i\ ,\ \text{ and } \ z\left(\alpha\right)=2\ \text{ otherwise.}\\ \text{ Then } z\left(\alpha\right)=y\left(\alpha\right) \text{ if } \alpha\in\ B\cap\ B_i, \text{ and otherwise if } \alpha\in\ B, z\left(\alpha\right)=y\left(\alpha\right)=2;\\ \text{ thus } z\in\ U\left(y,B\right)\subset\ V.\\ \text{ Furthermore, } z\left(\alpha_j\right)=j=x_{i+1}(\alpha_j) \text{ if } \alpha_j\in\ B_i \text{ and } z\left(\alpha_j\right)=1 \text{ if } \alpha_j\in\ B_{i+1}-B_i.\\ \text{ Therefore } z\in\ U\left(x_{i+1},B_{i+1}\right)\subset\ U.\\ \text{So, } z\in\ U\cap\ V \text{ as desired. Hence, we conclude that } U\cap\ V\neq\emptyset. $

Answer 4

Many thanks to @ashpool for writing up the proof. I personally find it much easier to follow proofs when I understand the overall strategy the proof is pursuing, so I thought it might be worthwhile to write a bit about that. This is of course not to replace the detailed formal proof but to complement it for ease of understanding.

The proof starts out by showing that the neighbourhoods that constrain functions to take certain values at a finite number of points form a basis of the topology. Ideally, we would like this basis to contain only the neighbourhoods for “injective constraints”, i.e. those that prescribe different values at different points. If this subset of neighbourhoods also formed a basis, the proof would be quite straightforward: The constant function $x_1$ that maps all of $J$ to $1$ is in $P_1$, so $U$ must contain a neighbourhood of that function with a finite set of injective constraints. The function that fulfils those constraints and maps everthing else to $2$ is in $P_2$, and thus in $V$, so we’re done.

Unfortunately, the neighbourhoods with injective constraints don’t form a basis, and the proof is essentially about working around that. It starts off with a neighbourhood $B_1$ of $x_1$ that may have a mix of constraints, some injective with function values $\ne1$ and some non-injective with function values $1$. It disregards all those function values and instead constructs a function $x_2\in P_1$ that’s injective at all the constrained points. It then again takes a neighbourhood $B_2$ of that function. That neighbourhood may again impose a mix of constraints – but the key is that on the values that $B_1$ constrained, $B_2$ can impose only injective constraints (since it contains $x_2$, which is injective on those values).

The proof continues in this fashion, in each step disregarding any new function values imposed and just making sure that the next neighbourhood imposes at most injective constraints where the previous one imposed constraints, and that these are compatible with the ones from previous steps. Unless this process terminates after a finite number of steps (in which case we’ve found a neighbourhood in $U$ that imposes only injective constraints, so we can apply the straightforward proof in the second paragraph above), it constructs a countable set $B$ of points and a sequence of neighbourhoods $B_i$ such that a) all constraints of all the $B_i$ lie in $B$, b) for any finite subset of $B$, there is some $B_i$ that has only injective constraints on that subset and c) all these injective constraints are compatible. (Each $B_i$ may also impose further constraints on $B$ that may be incompatible with those of other $B_k$; what’s important is that each $B_i$ maintains compatibility on the set on which injectivity has already been enforced at that stage.)

At no point in this process (not even “in the limit”) do we have a neighbourhood that imposes only injective constraints (unless the process terminates); but the sequence $B_i$ as a whole in a sense serves as a factory of such neighbourhoods – as a whole it imposes injective constraints on all of $B$, and if we specify a finite subset of $B$, it gives us a neighbourhood with only injective constraints on that subset.

Now we take the function $y\in P_2$ that satisfies all the compatible constraints on $B$ of all the $B_i$ and maps everything else to $2$. It has a neighbourhood in $V$, which can only impose a finite number of constraints. Any injective constraints it imposes (in this case constraints other than to the function value $2$) must be on points in $B$, and they must impose the same function values as the compatible constraints of the $B_i$ (since $y$ has those function values), so our neighbourhood factory can give us a neighbourhood $B_i$ whose constraints are compatible with those of $V$. Neighbourhoods that impose compatible constraints intersect, so we’ve found two neighbourhoods, one in $U$ and one in $V$, that intersect, and hence contain points in $U\cap V$.