# Is the basis of $\mathbb{R}_d \times \mathbb{R}_d$ = basis of $\mathbb{R}_d \times$ basis of $\mathbb{R}_d$?

I have the following question:

Let $\mathbb{R}_d$ denote $\mathbb{R}$ with the discrete topology. $\mathcal{B}(X)$ denotes the base or basis of the topology of $X$. Then show that $\mathcal{B}(\mathbb{R}_d) \times\mathcal{B}(\mathbb{R}_d) \neq \mathcal{B}(\mathbb{R}_d \times \mathbb{R}_d).$

The $\mathcal{B}(\mathbb{R}_d)$ is the family of singletons and each singleton is both open and closed.

Another posting regarding the product of discrete topologies , Product Topology of Discrete Sets, seems to make sense but I can't reconcile that explanation with what I am asked to prove.

If we consider the set $S = \{ (x,y) : x = y \}$ , then $S$ is closed in $\mathcal{B}(\mathbb{R}_d \times \mathbb{R}_d)$. But $S$ is not closed in $\mathcal{B}(\mathbb{R}_d) \times\mathcal{B}(\mathbb{R}_d)$. I do not understand why that is.

Any clarification would be appreciated.

Edit: The problem above is a modification of a problem from Folland's Real Analysis Text (Chapter 7, Exercise 29):

If $X$ is a set of cardinality > $\mathfrak{c}$ with the discrete topology, hen $\mathcal{B}_{X \times x}\neq \mathcal{B}_X \otimes \mathcal{B}_X$. In fact, $D = \{ (x,y) : x=y \}$ is closed but not in $\mathcal{B}_X \otimes \mathcal{B}_X$ (Use Exercise 5 in Section 1.2 and Proposition 1.23). If $D \in \mathcal{B}_X \otimes \mathcal{B}_X$ then $D \in \mathcal{M}$ where $\mathcal{M}$ is a sub-$\sigma$-algebra of $\mathcal{B}_X \otimes \mathcal{B}_X$ generated by a countable family of rectangles, hence $D \in \mathcal{N} \otimes \mathcal{N}$ where $\mathcal{N}$ is a countably generated sub-$\sigma$-algebra of $\mathcal{B}_X$. Then $\{x \} \in D_x \in \mathcal{N}$ for all $x$ but card$(\mathcal{N}) \leq \mathfrak{c}$. The same reasoning applies if $X$ is replaced by its one-point compactification.

I will re-read the chapter again. I am missing something fundamental.

• "Is a basis ...". There could be many distinct bases... – Eric Towers Nov 21 '17 at 4:02
• I see nothing at your link claiming that $S$ is not closed in the topology with basis $\mathcal{B}(\mathbb{R}_d) \times \mathcal{B}(\mathbb{R}_d)$. Where does the claim that $S$ is not closed originate? – Eric Towers Nov 21 '17 at 4:18
• @Eric Towers, I was referencing, perhaps incorrectly, an exercise from Folland's Real Analysis Text. It's Exercise 29 on page 231 of the second edition. – testguy Nov 21 '17 at 4:26
• @Eric TowersI can't discern the difference between $\mathcal{B}(\mathbb{R}_d) \times\mathcal{B}(\mathbb{R}_d)$ and $\mathcal{B}(\mathbb{R}_d \times \mathbb{R}_d)$. That's my problem. – testguy Nov 21 '17 at 4:31
• It's not about the base but about the Borel sets of those spaces! – Henno Brandsma Nov 21 '17 at 22:38

We take for a basis of the discrete topology the collection of singleton subsets, $\mathcal{B}(\mathbb{R}_d) = \{ \{x\} : x \in \mathbb{R}\}$ and $\mathcal{B}(\mathbb{R}_d \times \mathbb{R}_d) = \{ \{(x,y)\} : x \in \mathbb{R}, y \in \mathbb{R} \}$.

Let $x \in \mathbb{R}$. Then $\{x\} \in \mathcal{B}(\mathbb{R}_d)$ and $\{x\} \times \{x\} \in \mathcal{B}(\mathbb{R}_d) \times \mathcal{B}(\mathbb{R}_d)$.

But, although $\{(x,x)\} \in \mathcal{B}(\mathbb{R}_d \times \mathbb{R}_d)$, $\{(x,x)\} \not \in \mathcal{B}(\mathbb{R}_d) \times \mathcal{B}(\mathbb{R}_d)$. (Why? $\{x\} \times \{x\} = (\{x\},\{x\}) \neq \{(x,x)\}$.)

Consequently, $\mathcal{B}(\mathbb{R}_d \times \mathbb{R}_d) \neq \mathcal{B}(\mathbb{R}_d) \times \mathcal{B}(\mathbb{R}_d)$

There's a basic "type mismatch" between elements of $\mathcal{B}(\mathbb{R}_d \times \mathbb{R}_d)$ and elements of $\mathcal{B}(\mathbb{R}_d) \times \mathcal{B}(\mathbb{R}_d)$. These can be tricky to notice, but once you do, they're hard to stop noticing...

• Thank you very much Eric! It makes sense. – testguy Nov 21 '17 at 4:35
• I don't know that I buy this reading of the question. it's perfectly normal in mathematics to overload notation based on the type of its arguments. I mean, using the usual definitions of natural numbers we have $1 = \{\{\}\}$ and $2 = \{\{\}, \{\{\}\}\}$, but nobody would argue that $$1 \times 2 = \{(\{\}, \{\}),(\{\}, \{\{\}\})\}$$ and is therefore not a number at all. Clearly anyone writing $\mathcal{B} \times \mathcal{B}$ intends to indicate the set $\{ b \times b' \; | \; b, b' \in \mathcal{B}\}$, not the literal Cartesian product of $\mathcal{B}$ with itself. – Daniel McLaury Nov 21 '17 at 5:00
• @Daniel, the question is a modification of a question from Folland's Real Analysis, Exercise 29 in Chapter 7. I understand your point. I will post the question from Folland's text. – testguy Nov 21 '17 at 13:53
• I posted the question from Folland's text. Thanks. I'll re-read the material again. – testguy Nov 21 '17 at 14:24

Prof. Brandsma pointed out that we are dealing with Borel sets not bases in the comments. Also, what I asked to prove is not true. My apologies.

If the space consisted of the set of ordinals less than or equal to the first uncountable ordinal $\omega_1$, with the order topology, then we can prove it. Exercise 7.28 on page 231 in Folland's Real Analysis states this. But $\mathbb{R}_d$ is not ordered.

There is a book by S. M. Srivastava on Borel sets: A Course on Borel Sets. This helped clear up my confusion as well as provide a solid proof.

Srivastava credits the proof to B.V Rao. Apparently Ulam first posed a question like this. B.V Rao provided the following proof\cite{sms}. First we identify the notation $\mathcal{B}(\mathbb{R}_d) = \mathscr{P}(\mathbb{R})$ and $\text{card}((\mathbb{R}_d))= \aleph_1 = \mathfrak{c}$ so as not to get confused. Also, $\text{card}(\omega_1) = \aleph_1.$ Under the Continuum Hypothesis, the following holds true.

Proposition: $\mathcal{P}(\omega_1) \otimes \mathcal{P}(\omega_1) = \mathcal{P}(\omega_1 \times \omega_1)$.

Proof. Let $A \subseteq \omega_1 \times \omega_1$. Write $A = B \cup C$, where $$B = A \bigcap \{(\alpha, \beta) \in \omega_1 \times \omega_1: \alpha \geq \beta\}$$ and $$C = A \bigcap \{(\alpha, \beta) \in \omega_1 \times \omega_1: \alpha \leq \beta\}$$ We shall show that $B$ is in the product $\sigma$-algebra. By symmetry, it will follow that $C$ is also in the product $\sigma$-algebra. The result then follows.

For each $\alpha < \omega_1$, $B_\alpha$ is countable, say $B_{\alpha} = \{\alpha_0,\alpha_1,\alpha_2, \dots \}$. Since the graph of every measurable function from a measurable space to a discrete measurable space whose cardinality is at most $\mathfrak{c}$ is measurable, $$G_n = \{ (\alpha, \alpha_n) : \alpha \in \omega_1 \}$$ is in the product of discrete $\sigma$-algebras. Now we note that $B = \bigcup_n G_n$.

If our space $X$ is Hausdorff and has cardinality greater than that of the continuum, then $\mathcal{B}(X \times X) \neq \mathcal{B}(X) \times \mathcal{B}(X).$ This is Exercise 7.29, page 231 in Folland. The proof is from Bogachev's Measure Theory Volume 2.