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15 votes
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Proof that the Cardinality of Borel Sets on $\mathbb R$ is $c$ without using the ordinals .

Your idea is sound, but it requires more work. As pointed out, you have only described so far a very small subcollection of the Borel sets. Instead, show that you can associate to each Borel set a ...
Andrés E. Caicedo's user avatar
15 votes
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Why do we work on the Borel sigma algebra and not on the Lebesgue sigma algebra?

I think the comments already mentioned striking points but here is an easy example that shows pathologies that can happen: Consider the inclusion-like map $$ f\colon \Bbb R \to \Bbb R^n, x\mapsto (x, ...
GhostAmarth's user avatar
  • 2,403
14 votes
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What is the significance of a Borel $\sigma$-algebra?

The Borel-$\sigma$-algebra is important, because some very natural measures simply cannot be defined on all the subsets of $\mathbb{R}^d$. Let me show you the standard example: take $d=1$. We would ...
stochastic-conch's user avatar
13 votes
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If $A\times A$ is Borel must $A$ be Borel?

Define $f:X\to X\times X$ by $f(x)=(x,x)$. Then $f$ is continuous, and $f^{-1}(A\times A)=A$. If $A\times A$ is Borel, it follows that $A$ is Borel.
Eric Wofsey's user avatar
13 votes
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Are all convex sets Borel sets?

The answer is no (except in the trivial case $n=1$), by a counting argument. Let $C$ be the closed unit ball centered at the origin, and let $O$ be the open unit ball centered at the origin. Any set $...
Noah Schweber's user avatar
12 votes
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Image of Borel set under continuous and injective map

We want to show that $$\mathscr{B} (\mathbb{R}^n) \subset \mathscr{A}:= \{A \subset \mathbb{R}^n : f(A) \in \mathscr{B}(\mathbb{R}^n) \}$$ By the definition of Borel $\sigma$-algebra it is enough to ...
Yanko's user avatar
  • 13.8k
10 votes
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Is there a set $A \subset [0,1]$ such that both $A$ and $[0,1] \setminus A$ intersect every positive-measure set?

As shown in this question there are $\mathfrak c$ Borel sets, which can be extended to the fact that there are $\mathfrak c$ Borel sets of positive measure. Each one with positive measure must have ...
Ross Millikan's user avatar
10 votes
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What is a Borel set?

A Borel set is actually a simple concept. Any set that you can form from open sets or their complements (i.e., closed sets) using a countable number of intersections or unions is a Borel set. It ...
Apoorv's user avatar
  • 375
10 votes
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Nontrivial Borel Sets

The relevant topic here is descriptive set theory. The standard texts on the subject are Moschovakis and Kechris; I tend to prefer the latter, especially as a first introduction unless you're already ...
Noah Schweber's user avatar
10 votes

Is a subset of a Borel set of measure zero always Borel?

If $A$ is a subset of $\Bbb R$ of measure $0$, then all its subsets, $2^{|A|}$ many of them, are measure $0$ (and so Lebesgue measurable, Borel or not). Any subspace of a separable metric space is ...
Henno Brandsma's user avatar
10 votes

Why are all sets we can construct Borel sets?

"Concrete" isn't a word with a precisely defined meaning here. The author is using the word "concrete" to indicate that Borel descriptions are more complicated than computable or ...
Mitchell Spector's user avatar
9 votes
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For what spaces does every countable Borel cover have a finite subcover?

This is an extraordinarily strong property: it is equivalent to $X$ having only finitely many open sets. Indeed, it is easy to see that if $X$ has finitely many open sets, then it has only finitely ...
Eric Wofsey's user avatar
9 votes
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Does every uncountable Borel subset of $\mathbb R$ contains a perfect subset´╝č

Yes - this is called the perfect set property. The analytic (= continuous image of Borel) sets also have this property; for more complicated sets (e.g. complements of analytic sets), the perfect set ...
Noah Schweber's user avatar
9 votes
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$\sigma$-algebra generated by open balls

No. See, for example, this question here or this question on MO. Separability, or equivalently second countability, is sufficient to guarantee the ball sigma-algebra is the same as the Borel sigma-...
user10354138's user avatar
  • 33.3k
8 votes

translation and dilation invariance of borel sets

Proofs of things about $\sigma$-algebras often begin by saying "Let $A=$..." and proceed by showing that $A$ is a $\sigma$-algebra. Here: Let $A$ be the collection of all sets $E$ such that every ...
David C. Ullrich's user avatar
8 votes

Is the sum (difference) of Borel set with itself a Borel set?

Choose disjoint perfect sets $P, Q \subseteq [0, 1]$ such that $P \cup Q$ is linearly independent over the field of rationals. So $+ \upharpoonright (P \times Q)$ is a homeomorphism from $P \times Q$ ...
hot_queen's user avatar
  • 7,157
8 votes
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Constructing sigma algebras in countably many steps

One can find proofs in many set theory texts for the stronger result that, for each countable ordinal level of the Borel hierarchy, there exist Borel sets not belonging to that level. Look in book ...
Dave L. Renfro's user avatar
7 votes
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Borel $\sigma$-algebra of subset of a topological space

For any subset $\Omega'$ it's fairly easy to see that $$\mathcal B(\Omega', \mathcal T') = \{ A \cap \Omega': A \in \mathcal B(\Omega, \mathcal T)\}$$ Proof: the right hand side is clearly a $\...
Henno Brandsma's user avatar
7 votes

On Complete Separable metrizable spaces, do the compact sets generate the Borel $\sigma$-algebra?

Any separable infinite-dimensional Banach space is a counterexample, for example $L_1$. The compact sets are meagre so they generate a sub-$\sigma$-algebra of the $\sigma$-algebra of sets that are ...
Colin McQuillan's user avatar
7 votes
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Is the Borel sigma algebra complete under some measure?

The Borel $\sigma$-algebra on any space is complete with respect to counting measure, since the only null set for counting measure is the empty set. However, the Borel $\sigma$-algebra on $\mathbb{R}$...
Eric Wofsey's user avatar
7 votes

Show that the set of real numbers that have a decimal expansion with the digit 5 appearing infinitely often is a Borel set.

Hint: The set $E_1$ of real numbers that have a decimal expansion where the first digit after the decimal point is $5$ is the union of $[n+ 0.5, n+0.6]$ and $[-n-0.6, -n-0.5]$ for $n = 0, 1, 2, \...
Robert Israel's user avatar
7 votes
Accepted

A set in the Borel $\sigma$-algebra over $[0,1]$ that isn't in the algebra generated by open sets

Claim: For any set $X$ obtainable the boundary is nowhere dense. Proof: Boundary of open set is nowhere dense. Now for the operations: 1) complement - since boundary is the same for a set and its ...
Max's user avatar
  • 14.3k
7 votes

Is every Borel set a countable union of intervals?

The Cantor set $C$ is an uncountable Borel set and it does not contain any non-trivial interval (i.e. a singleton). Hence $C$ can not be written an as countable union of intervals. However, its ...
Yiorgos S. Smyrlis's user avatar
6 votes
Accepted

Example of limit of a net of Borel functions not Borel

Let $A$ be a subset of $\mathbb R$ that is not Borel. I'll produce a net of Borel functions on $\mathbb R$ whose limit is the characteristic function of $A$. The index set $I$ for my net is the set ...
Andreas Blass's user avatar
6 votes
Accepted

Can we obtain the Borel $\sigma$-algebra on $[0;1]$ as a limit of finite algebra?

Yes there is a natural and useful construction of $\sigma$-algebras as a "limit". But not as the limit of a sequence, rather the limit of an uncountable collection, indexed by the countable ordinals (...
David C. Ullrich's user avatar
6 votes

What sets are obtained by adding $\aleph_1$ unions and intersections to the Borel algebra?

This is an excellent question! Note first that, just from cardinal arithmetic considerations, whether we obtain all sets is independent (and therefore so is whether we obtain all Lebesgue measurable ...
Andrés E. Caicedo's user avatar
6 votes
Accepted

What is the Sigma Algebra generated by Jordan measurable sets?

I found the answer in this journal paper. The Sigma algebra generated by the Jordan measurable sets is the collection of all sets which can be written as a union of a Borel set and a subset of a ...
Keshav Srinivasan's user avatar
6 votes

$[0,1)$ is in both $G_{\delta}$ and $F_{\sigma}$

$G_\delta$ is not the complement of $F_\sigma$. The complement of an $F_\sigma$ set is a $G_\delta$ set.
Robert Israel's user avatar
6 votes
Accepted

Is the interior of a Jordan curve a Borel set?

The image of a Jordan curve is a compact set, so that its complement is open. The interior of the Jordan curve is one of the two connected components of that complement, and therefore open as well. ...
Martin R's user avatar
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