# $\mathbb{R}^\mathbb{R}$ is not normal

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.

• 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 $\mathbb{R}$) – Moss Sep 5 '12 at 23:21
• @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 – Moss Sep 5 '12 at 23:34
• 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
• @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

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.

• 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