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We can show that $\mathbb{R}$ and $\mathbb{R}^2$ or ($\mathbb{R}^n$) have same cardinality using the following one-to-one and onto mapping:

Say x = (0.123456789....)

Then f(x) = {(0.13579...),(0.2468..)}

My question is can we claim that the Borel sigma algebra in $\mathbb{R}$ has a correspondent sigma algebra (if it is, is it THE Borel sigma algebra in $\mathbb{R}^2$?) by mapping each Borel set D to $\mathbb{R}^2$ by f(D)? Why or why not could you give the argument??

Similarly (I am more certain of this) When we have N^n and N using this mapping could we claim that the sigma algebra in N has a correspondent sigma algebra in N^n? Does the statement hold for N^infinity and N as well, if N^infinity and N have the same cardinality ( i am not sure about my last statement)?

Thank you!

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This is not really related to the core question, but note that the function you mention is not a bijection (think about numbers having more than one expansion). – Ittay Weiss May 29 '13 at 4:59
up vote 4 down vote accepted

The Borel sigma-algebra of $\mathbb{R}^2$ is defined as that generated by the subsets of $\mathbb{R}^2$ that are open in the product topology. It is almost, but not quite, correct to say that this is equal to the collection of pointwise images of Borel subsets of $\mathbb{R}$ under your a correspondence given by splitting and interleaving decimal expansions.

The problem, as Ittay pointed out in a comment, is that this correspondence is not quite a bijection between $\mathbb{R}$ and $\mathbb{R}^2$. Nevertheless, by changing the argument a bit, one can still show that $\mathbb{R}$ and $\mathbb{R}^2$ are Borel-isomorphic. One way to do this is to get a Borel injection from $\mathbb{R}^2$ to $\mathbb{R}$ and then use the Borel version of the Cantor–Bernstein–Schroeder Theorem.

Indeed, a Borel injection $f: \mathbb{R}^2 \to \mathbb{R}$ can be defined as follows: Given $(x,y) \in \mathbb{R}^2$, let $x_0 x_1 x_2 \ldots$ and $y_0y_1y_2 \ldots$ be the unique decimal expansions of $x$ and $y$ respectively with the property that the digits are not eventually all 9's from some point on. Then let $f(x,y)$ be the real number whose decimal expansion is $x_0y_0x_1y_1 \ldots.$ (For simplicity of notation I am ignoring the issue of where the decimal point goes, but this is easy to deal with.) Unfortunately the range of this function is missing uncountably many real numbers (such as those whose even digits are eventually all 9's and whose odd digits are not) and the easiest way that I see to deal with this issue is to use the Cantor–Bernstein–Schroeder Theorem, although it may be overkill.

The sets $\mathbb{R}$ and $\mathbb{R}^2$ with these Borel structures (and indeed $\mathbb{R}^n$ for any positive integer $n$) are examples of standard Borel spaces, which are the objects one gets from Polish topological spaces by forgetting the topology and remembering only the Borel structure. The Borel isomorphism we established above proves a special case of a theorem of Kuratowski, which says that every uncountable Polish space is Borel-isomorphic to $\mathbb{R}$, so up to isomorphism there is only one uncountable standard Borel space.

The sets $\mathbb{N}$ and $\mathbb{N}^n$ with the discrete Borel structure (every subset is a union of countable many singletons and is therefore Borel in the usual topology) are Borel-isomorphic to one another. They are examples of the countably infinite standard Borel space, which is also unique up to isomorphism.

The set $\mathbb{N}^\infty$ (also known as $\mathbb{N}^\mathbb{N}$) of integer sequences with the Borel algebra generated by the product topology is an uncountable standard Borel space—it is Borel-isomorphic to $\mathbb{R}$ rather than to $\mathbb{N}$.

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Hi Trevor thank you for the reply, could you clarify your sentence "They are examples of the countably infinite standard Borel space, which is also unique up to isomorphism". How is it related to set N with discrete Borel structure? Also how do you "modify" the function, i mean explicitly? – Salih Ucan May 29 '13 at 8:15
@SalihUcan By that sentence I mean that all countably infinite standard Borel spaces are isomorphic to each other (equivalently, they are all isomorphic to $\mathbb{N}$ with the discrete Borel structure.) In the countable case every set is Borel, so any bijection is a Borel isomorphism, and what I said follows from the fact that $\mathbb{N}^n$ and $\mathbb{N}$ have the same cardinality. – Trevor Wilson May 29 '13 at 15:07
The reason the non-uniqueness of decimal expansions is not a problem is that it only happens on a countable set, and if $X$ is a Borel space and $S$ is a countable set, then $X\setminus S$ is Borel-isomorphic to $X$. I will add more about this later. – Trevor Wilson May 29 '13 at 15:09
@SalihUcan I had to modify the argument a bit more than I thought. If you have any questions about the latest edit, please let me know. – Trevor Wilson May 29 '13 at 22:27

The fact that there are $1-1$ mapping between two sets says almost nothing about the measure of the two sets. Consider a silly example of the Cantor set, you can show relatively easily that it has cardinality $c$ while having zero measure.

The $f$ you defined from $\mathbb{R}^{1}\rightarrow \mathbb{R}^{2}$ is measurable (because it is continuous), and hence $f(D)$ does generate a $\sigma$-algebra in $\mathbb{D}^{2}$. But this $\sigma$-algebra is "twisted" in some sense. For example, the image of $(0.4,0.5)$ in the plane includes points "outside" of $(0.4,0),(0.5,0)$ like $(0.4,0.9)$ from $0.49$ or $(0.49,0.95)$ from $(0.4995)$. It is intuitively not the right choice for us to define a $\sigma$-algebra on the plane, where the open sets arise from open balls or open retangles.

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Does your example show that the Borel structure induced by $f$ is different from the usual one? – Trevor Wilson May 29 '13 at 4:53
No, but I think it showed aesthetically it is not preferable over other definitions. – Bombyx mori May 29 '13 at 4:56
Okay, I just wanted to make sure the OP didn't get the idea that you were saying it was not equivalent to the usual definition. – Trevor Wilson May 29 '13 at 4:57
On second thought, I think your post is wrong. You say "this $\sigma$-algebra is 'twisted' in some sense." How can that be, when it is the same as the usual sigma-algebra? – Trevor Wilson May 29 '13 at 5:00
@TrevorWilson: I hope so - could you add some details to the claim "there is only one uncountable standard Borel space"? I have never seen this fact in literature. The wiki article is nice but did not give a proof. – Bombyx mori May 29 '13 at 5:00

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