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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.

Rachel
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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 02:05
  • 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 02:14
  • @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 06:39

2 Answers2

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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 $y \in P_n \setminus U(x,B)$.

Brian M. Scott
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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
  • This argument is similar to my objection to the proof in part 6) of example 103 in Steen & Seebach. I don't understand how Fn(xn) can be a subset of U since Fn(xn) is set of all points that are the same as xn at the finite set of indices in Fn, whereas U contains P0 - the set of all points in which each integer apart from 0 appears at most once. A point in Fn(xn) could have any number at indices not in Fn. – Ben Jun 13 '21 at 15:27
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    @Ben: There is a slight error in that section, but it’s easily repaired. We cannot ensure that $|F_n|=n$ for each $n$. Rather, there is a strictly increasing sequence $\langle k_n:n\in\Bbb N\rangle$ of positive integers such that $F_n={\alpha_j:j=1,\ldots,k_n}$ for each $n$. Now suppose that we already have $x^n\in P_0$ and $F_{n-1}\subseteq A$. $P_0\subseteq U$, and $U$ is open, so $x^n$ has a basic open nbhd $H(x^n)\subseteq U$, and the only real problem is seeing why we can choose $H$ so that $H\supseteq F_{n-1}$. If you look ahead to the choice of $x^{n+1}$, you’ll see that $x^n$ was ... – Brian M. Scott Jun 13 '21 at 20:44
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    ... chosen so that $x_{\alpha_j}^n=j$ whenever $\alpha_j\in F_{n-1}$, and $x_\alpha^n=0$ otherwise. Thus, if $\alpha_j\in F_{n-1}\cap H$, then $x_{\alpha_j}^n=j$, and if $\alpha\in H\setminus F_{n-1}$ then $x_\alpha^n=0$. That is, $H(x^n)$ and $F_{n-1}(x^n)$ agree on all indices that they have in common, so $$(F_{n-1}\cup H)(x^n)=F_{n-1}(x^n)\cap H(x^n)\subseteq U,.$$ We could therefore simply let $F_n=F_{n-1}\cup H$, except that we want $k_n>k_{n-1}$, which won’t be the case if it just happens that $H\subseteq F_{n-1}$. However, we can always assume that ... – Brian M. Scott Jun 13 '21 at 20:50
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    ... $H\setminus F_{n-1}\ne\varnothing$, since adding an extra index in $A\setminus F_{n-1}$ to $H$ just shrinks the set $H(x^n)$ and keeps the nbhd inside $U$. Now let $k_n=|F_n|$, enumerate $F_n\setminus F_{n-1}$ as ${\alpha_{k_{n-1}+1},\ldots,\alpha_{k_n}}$, and define $x^{n+1}\in P_0$ by setting $x_{\alpha_j}^{n+1}=j$ for $j=1,\ldots,k_{n+1}$ and $x_\alpha^{n+1}=0$ for $\alpha\in A\setminus F_{n+1}$. Now the argument in the book works, though the authors should have made the recursive construction clearer: as it stands, you have to read ahead to see what $x^n$ looks like. – Brian M. Scott Jun 13 '21 at 20:56
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    @Ben: Now for your actual objection. It’s perfectly true that $F_n(x^n)$ contains points not in $P_0$, so it is not a subset of $P_0$, but that doesn’t matter: we only need it to be a subset of $U$, and by definition every point in $P_0$ has a basic open nbhd contained in $U$. $P_0$ is not open, so we cannot expect its points to have open nbhds contained in $P_0$ itself. – Brian M. Scott Jun 13 '21 at 20:58
  • @BrianM.Scott: OK, I think I get it; when you say $H(x^n)$ is a basic open nbhd, do you mean that is it a member of a local base for $x^n$? and so by the statement in 2) in the book, all points in that nbhd are fixed to the same values at just a finite number of coordinates? That is something I didn't get either; in 2) why should it be that "$\pi_{\alpha}(B_i)=\mathbb{Z^+}$ for all but finitely many $\alpha$" is that proved in another example? – Ben Jun 13 '21 at 22:41
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    @Ben: Yes: it is specifically the basic open set $$\bigcap_{\alpha\in H}\pi_\alpha^{-1}[{x_\alpha^n}],.$$ The points in this set are those that agree with $x^n$ on all of the (finitely many) indices in $H$. In (2) the authors are assuming that the sets $B_i$ are standard basic open sets in the product; by definition these are sets of the form $\prod_{\alpha\in A}U_\alpha$ such that $U_\alpha=Z_\alpha^+$ for all but finitely many $\alpha\in A$. – Brian M. Scott Jun 14 '21 at 21:43
  • @BrianM.Scott OK, so I don't see where they state that assumption explicitly, but on page 8 they define a Tychonoff topology to be the coarsest topology such that the projection functions $\pi_{\alpha}$ are continuous. This means that in a Tychonoff topology open sets are generated by unions and finite intersections of all sets which fix a single coordinate, which means open sets can only fix a finite number of coordinates. And I guess they assume a Tychonoff topology for example 103, since they stated that in example 102 which is similar. – Ben Jun 14 '21 at 22:57
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    @Ben: A basic open set in the product can fix only finitely many coordinates; arbitrary open sets are more complicated, since they can be any union of basic open sets. The topology on a product of topological spaces is always the Tikhonov topology unless some different topology is explicitly specified. In fact, the Tikhonov topology is often called simply the product topology. – Brian M. Scott Jun 15 '21 at 20:04
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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.

Henno Brandsma
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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