# “Proof” that $\mathbb{R}^J$ is not normal when $J$ is uncountable

In 'Topology' by Munkres, he leaves as an exercise to prove that $\mathbb{R}^J$ is not normal under the product topology when $J$ is uncountable. The proof outlined as exercise 32.9 is the same one given in the 'Counterexamples in Topology' book, space #103, Uncountable Products of $\mathbb{Z}^+$. My question is a two-fer:

1) Is what I've outlined below an alternative proof to parts (c) and (d) in his problem as stated? Or have I gone wrong at some step? I am skeptical, as my "proof" is quite a bit simpler than what is given in the outline in the book, and it is mentioned a few times that the proof is quite difficult. Perhaps I am missing the point, which is to see what Stone did in his proof with his particular sequences and sets. Which leads me to my second question...

2) When simpler proofs do exist, is it tradition in mathematics to honor the original proofs as a matter of principle, publishing them instead of more streamlined versions?

Proof that $\mathbb{R}^J$ is not normal:

The proof begins by examining a closed subspace of $\mathbb{R}^J$, in this case $X = (\mathbb{Z}_+)^J$, and showing that it is not normal. The elements in $X$ are written using the map notation, $x:J \rightarrow \mathbb{Z}_+$.

First, in part (a), it's shown that if $x \in X$ and $B$ is a finite subset of $J$, then the sets $U(x, B) = \{ y \in X : y(\alpha) = x(\alpha), \alpha \in B\}$ form a basis for $X$.

Second, in part (b), it's shown that if $P_n$ is the subset of $X$ consisting of $x \in X$ s.t. $x$ is injective on $J - x^{-1}(n)$, that $P_n$ is closed. Moreover, if $n \neq m$, then $P_n \cap P_m = \emptyset$. The proof of this utilizes the basis elements in (a).

Here's where my proof differs from that of the literature. Consider a set $P_n$ for some $n \in \mathbb{Z}^+$, and suppose that $U$ is an open set containing $P_n$. We will show that $U = X$, and so $X$ cannot be normal, since two closed sets $P_n$ and $P_m$ cannot be separated by disjoint open sets.

Choose some $\alpha' \in J$, and define a set of sequences $x_i \in X$ as follows, where $i \in \mathbb{Z}^+$: Let $x_i(\alpha) = i$ when $\alpha = \alpha'$, and $x_i(\alpha) = n$ otherwise. Then each sequence $x_i \in P_n$, since $x_i(\alpha) = n$ on all but at most a singleton in $J$, namely $\alpha'$. Thus, $x_i \in U$ for each $i \in \mathbb{Z}^+$. As such, the coordinate corresponding to $\alpha'$ in the product of open sets that make up $U$ must be equal to $\mathbb{Z}^+$. Since $\alpha'$ was arbitrary, every coordinate in the product $U$ must be equal to $\mathbb{Z}^+$, and so $U = X$.

Thus $X$ cannot be normal, as two disjoint closed sets, $P_n$ and $P_m$ where $n \neq m$, cannot be separated by disjoint open sets.

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Here is the exercise from Munkres (since Google doesn't have scans): Parts (a)-(c), Part (d). – t.b. Sep 18 '11 at 2:05
Concerning your question 2): no, definitely not. If better proofs are found and become reasonably well-known, they sooner or later replace the original ones. – t.b. Sep 18 '11 at 2:14
@Rachel: You seem to be assuming that $U$ must be of the form $\prod_i U_i$ with $U_i \subset \mathbb Z^+$. This is not the case. In general, $U$ will be an arbitrary union of such sets (in particular one may not be able to write $U$ as a product). – Sam Sep 18 '11 at 6:39

Even without seeing just where the problem is, you can see that there must be one: if $m \ne n$, then $X\setminus P_m$ is an open nbhd of $P_n$ properly contained in $X$, so your conclusion that there can be no such set is false.

The flaw in your argument is that the open nbhd $U$ of $P_n$ need not be a basic open nbhd: it need not be a product of open sets. In fact, no set of the form $U(x,B)$ can contain $P_n$. To see this, let $B_0 = \{\alpha\in B:x(\alpha)=n\}$, and let $B_1 = B \setminus B_0$. If $B_1 \ne \varnothing$, define $y \in X$ by $y(\alpha) = n$ for all $\alpha \in J$; clearly $y \in P_n \setminus U(x,B)$. If $B_1 = \varnothing$, fix $\alpha_0 \in B_0$, and define $y \in X$ by $$y(\alpha) = \begin{cases}n+1,&\alpha = \alpha_0\\ n,&\text{otherwise}; \end{cases}$$ clearly we again have $y \in P_n \setminus U(x,B)$.

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Ah, I see now! And yes, you're right, the complement of $P_m$ contains $P_n$ and is open, so that does throw a wrench in the "proof". So, it seems that what I have actually shown is that if $U$ is a basis element in the product topology containing $P_m$, then it is $X$. – Rachel Sep 18 '11 at 13:03
indeed, but open sets are unions of basic open sets, and for different points in $U$ we use different open basic neighborhoods in it... – Henno Brandsma Sep 18 '11 at 13:28

You cannot prove what you claim: $P_1$ and $P_2$ are closed and disjoint so clearly $X \setminus P_2$ is an open superset of $P_1$, unequal to $X$.

The mistake in the proof is that you show: for every coordinate $\alpha$ and every $m$ in $\mathbb{Z}^{+}$ there is a basic neighborhood $U(\alpha,m)$ that it is a subset of $U$ and $m \in p_{\alpha}[U(\alpha, m)]$. It does not follow from that that $U = X$ from that. It only shows that $U$ projects onto each factor space, but this is not enough. Consider the diagonal in a finite product of discrete spaces, for an example of that: the diagonal is open, and projects onto each factor in the full set, yet is "thin" as well, and not equal to the full product.

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I think I see what you're saying here about the diagonal set. If U is the union of each diagonal in $\mathbb{Z}^+$, then each coordinate projects onto $\mathbb{Z}^+$, but U contains only the constant sequences, hardly $X$ at all. – Rachel Sep 18 '11 at 13:09