Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does anyone know how to prove that $\mathbb{R}^\mathbb{R}$ (with the product topology) does not fulfill the $T_4$ axiom?

It would be sufficient to have an uncountable subset $A \subseteq \mathbb{R}^\mathbb{R}$ which is closed and discrete as subspace (because this is impossible for any separable $T_4$-space), but I do not know if such a subspace exists.

share|improve this question
    
What about the collection of "tuples" $\{A_r\}$ for $r\in \mathbb{R}$ with the $x$ component of $A_x$ equal to 1, and all other components equal to zero? (think of $A_x$ as an uncountable tuple of numbers, indexed by $\mathbb{R}$) –  gabbering Sep 5 '12 at 23:21
1  
@Sebastian: This set is not closed, since it has the 0-Tuple as limit point. –  Dune Sep 5 '12 at 23:28
    
Yes, hmm I'll have to think about it some more –  gabbering Sep 5 '12 at 23:34
4  
There's exercise 32.9 in Munkres outlining an argument due to A.H. Stone: Part (a)-(c) and Part (d). –  t.b. Sep 5 '12 at 23:57
1  
@StevenStadnicki Why should the topology of $\omega_1$ enter into this? The products $X^Y$ and $X^Z$ are homeomorphic if $Y$ and $Z$ have the same cardinality. –  Miha Habič Sep 6 '12 at 1:07

1 Answer 1

up vote 7 down vote accepted

It suffices to show that the product of $|\Bbb R|$ copies of $\Bbb N$ is not normal, since this product is a closed subspace of $\Bbb R^{\Bbb R}$. The following argument is expanded from hints to Exercise 3.1.H(a) in Engelking, General Topology; I’ve included it because it works by exhibiting a large closed discrete subset of the product, as suggested by Dune. However, it’s actually true that $\Bbb N^{\omega_1}$ is non-normal, even if $\omega_1<2^\omega$; this is an old result of A.H. Stone. At the end I’ve appended a brief sketch of the argument.

Let $I=[0,1]$, and let $X={^I\Bbb N}$ with the product topology. For each $t\in I$ define

$$f_t:I\to\Bbb N:x\mapsto\begin{cases} 0,&\text{if }x=t\\ k,&\text{if }k\in\Bbb Z^+\text{ and }\frac1{k+1}<|x-t|\le\frac1k\;. \end{cases}$$

Define $$h:I\to X:x\mapsto\big\langle f_t(x):t\in I\big\rangle\;,$$

and let $D=\operatorname{ran}h$. Clearly $|D|=|I|=2^\omega$, and I’ll show that $D$ is a closed, discrete subset of $X$. Since $X$ is separable by the Hewitt-Marczewski-Pondiczery theorem, it will then follow from Jones’s lemma that $X$ is not normal.

It’s easy to see that $D$ is discrete. Fix $y=h(u)=\langle f_t(u):t\in I\rangle\in D$. Let $$B=\Big\{\langle x_t:t\in I\rangle\in X:x_u=0\Big\}\;;$$ $f_u(u)=0$, so $B$ is an open nbhd of $y$ in $X$. Moreover, $f_u(t)\ne 0$ when $t\in I\setminus\{u\}$, so $B\cap D=\{y\}$.

Showing that $D$ is closed in $X$ takes more work.

Suppose that $y=\langle y_t:t\in I\rangle\in X$ is such that for all $t\in I$, $y_t\ne 0$; then $y\notin\operatorname{cl}_XD$.

Proof: For each finite $F\subseteq I$ let $B_F=\{x\in X:\forall t\in F(x_t=y_t)\}$, and let $\mathscr{B}$ be the collection of all such $B_F$; $\mathscr{B}$ is a local base at $y$ in $X$. For $B_F\in\mathscr{B}$ we have

$$\begin{align*} h^{-1}[B_F]&=\{u\in I:h(u)\in B_F\}\\ &=\{u\in I:\forall t\in F(f_t(u)=y_t)\}\\ &=\left\{u\in I:\forall t\in F\left(\frac1{y_t+1}<|u-t|\le\frac1{y_t}\right)\right\}\\ &=\bigcap_{t\in F}R_t\;, \end{align*}$$

where for each $t\in I$ we define

$$\begin{align*} R_t&=\left\{u\in I:\frac1{y_t+1}<|u-t|\le\frac1{y_t}\right\}\\ &=I\cap\left(\left[t-\frac1{y_t},t-\frac1{y_t+1}\right)\cup\left(t+\frac1{y_t+1},t+\frac1{y_t}\right]\right)\;. \end{align*}$$

For each $t\in I$ let $$V_t=I\cap\left(t-\frac1{y_t+1},t+\frac1{y_t+1}\right)\;;$$ $V_t$ is an open nbhd of $t$ in the usual topology on $I$ that is disjoint from $R_t$. $I$ is compact in the usual topology, so there is a finite $F\subseteq I$ such that $\{V_t:y\in F\}$ covers $I$. But then $$h^{-1}[B_F]=\bigcap_{t\in F}R_t\subseteq I\setminus\bigcup_{t\in F}V_t=\varnothing\;,$$ and $B_F$ is an open nbhd of $y$ disjoint from $D$. $\dashv$

Now suppose that $y=\langle y_t:t\in I\rangle\in X\setminus D$ is such that $y_s=0$ for some $s\in I$; $y\ne h(s)$, so there is a $t\in I\setminus\{s\}$ such that $y_t\ne f_t(s)$. Moreover, if $u\in I\setminus\{x\}$, then $f_u(s)\ne 0$, so $$B_{s,t}=\Big\{\langle x_u:u\in I\rangle\in X:x_s=0\text{ and }x_t=y_t\Big\}$$ is an open nbhd of $y$ disjoint from $D$.

It follows that $D$ is closed in $X$ and hence that $X$ is not normal.


Stone’s result, that $X=\Bbb N^{\omega_1}$ is not normal, is proved by exhibiting two disjoint closed sets in $X$ that cannot be separated by disjoint open sets. Two sets that work are

$$H_0=\Big\{\langle n_\xi:\xi<\omega_1\rangle\in X:\forall m\in\Bbb N\setminus\{0\}\big(|\{\xi<\omega_1:n_\xi=m\}|\le 1\big)\Big\}$$ and

$$H_1=\Big\{\langle n_\xi:\xi<\omega_1\rangle\in X:\forall m\in\Bbb N\setminus\{1\}\big(|\{\xi<\omega_1:n_\xi=m\}|\le 1\big)\Big\}\;.$$

They’re clearly disjoint, since every point of $H_i$ has all but countably many coordinates equal to $i$ for $i=0,1$, and it’s easy to show that they’re closed.

If $X$ were normal, there would be open sets $U_0$ and $U_1$ such that $H_0\subseteq U_0$, $H_1\subseteq U_1$, and $\operatorname{cl}U_0\cap\operatorname{cl}U_1=\varnothing$. $X$ is separable by the Hewitt-Marczewski-Pondiczery theorem, so let $D$ be a countable dense subset of $X$. Let $D_0=U_0\cap D$ and $D_1=U_1\cap D$. For each $x\in D_0$ let $B(x)$ be a product basic open nbhd of $x$ contained in $U_0$, and for each $x\in D_1$ let $B(x)$ be a product basic open nbhd of $x$ contained in $U_1$. Let $V_0=\bigcup_{x\in D_0}B(x)$ and $V_1=\bigcup_{x\in D_1}B(x)$; then $\operatorname{cl}V_0=\operatorname{cl}U_0$ and $\operatorname{cl}V_1=\operatorname{cl}U_1$. Each of the sets $B(x)$ for $x\in D_0\cup D_1$ depends on only finitely many coordinates, so $V_0$ and $V_1$ depend on only countably many coordinates, and therefore $\operatorname{cl}U_0$ and $\operatorname{cl}U_1$ depend on only countably many coordinates.

Let $C\subseteq\omega_1$ be the union of the countable sets of coordinates on which $\operatorname{cl}U_0$ and $\operatorname{cl}U_1$ depend. Let $\varphi:C\to\Bbb N$ be any injection, and define $x\in H_0$ and $y\in H_1$ by

$$x_\xi=\begin{cases}\varphi(\xi),&\text{if }\xi\in C\\0,&\text{otherwise}\end{cases}$$

and

$$u_\xi=\begin{cases}\varphi(\xi),&\text{if }\xi\in C\\1,&\text{otherwise}\;.\end{cases}$$

Suppose that $z\in X$ is such that $z\upharpoonright C=\varphi$. Then $z$ agrees with $x$ on $C$, and $x\in\operatorname{cl}U_0$, so $z\in\operatorname{cl}U_0$. But by the same reasoning $z\in\operatorname{cl}U_1$. This contradiction shows that $X$ is not normal.

share|improve this answer
    
Thank you very much for this great answer! Do we get in the same way, that an uncountable product of a non-compact Hausdorff space is not normal? I mean, does every such space contain $\mathbb{Z}_+$ as closed subspace (up to homeomorphism)? –  Dune Sep 6 '12 at 11:01
    
I see, having an infinite discrete closed subspace is equivalent to not being countably compact (assuming $T_1$). Therefore the above proof generalizes to: an uncountable product of non-countable-compact Hausdorff spaces is not normal. The case of non-compact Hausdorff spaces, which is also true according to Wikipedia, must be proved in a different way. –  Dune Sep 6 '12 at 12:17
    
I wanted to direct your attention to this related question (already raised by @Dune in these comments) which I first thought might be answered using the ideas here, but I admit that even after some thinking I still don't see how. –  t.b. Sep 9 '12 at 2:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.