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My understanding of Borel set is the collection of all open/close intervals on $\mathbb{R}$. So if we create a map from Borel set to $\mathbb{R}$, namely map all the open/close intervals centred at the same real number to that number in $\mathbb{R}$, we can see that it is a many-to-one mapping. Therefore, the cardinality of Borel cannot be smaller than that of $\mathbb{R}$. However, a measure theory lecturer says the cardinality of Borel is less than or equal to $\mathbb{R}$, depending on the continuum hypothesis. Could someone help to clarify?

Sam
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    There are Borel sets that aren't like what you described, and an enormous amount at that. The continuum hypothesis is essentially irrelevant for the cardinality of Borel sets, but AC isn't. If AC holds, then the cardinality of Borel sets is $\lvert \Bbb R\rvert$. The idea is that with transfinite induction you can write the set of Borel sets as a union of $\aleph_1$-many sets of cardinality $\lvert\Bbb R\rvert$. There are models of ZF+not C where $\Bbb R$ is the union of a countable family of countable sets, and therefore in those models the set of Borel sets is the whole $\mathcal P(\Bbb R)$. – Sassatelli Giulio Aug 26 '22 at 11:12
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    @Sam: The Borel sets are the sigma algebra generated by the topology, so they certainly include the open sets and the closed sets, but of course there are many other sets as well. – MPW Aug 26 '22 at 11:13
  • @SassatelliGiulio the definition I mentioned is given in the lecture notes I used. But is there any issue with my thinking (in terms of mapping)? It seems to me with that thinking Borel set can’t be smaller than real number – Sam Aug 26 '22 at 17:37
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    @Sam So I guess that for you $(0,1)\cup (2,3]$ isn't Borel? Anyways, call $B(\Bbb R)$ the set of Borel sets with the actual definition and $K\subseteq B(\Bbb R)$ the set of the intervals that are either open or closed. Proving that $|\Bbb R|\le |K|$ is easy. You can do it with surjectivity of the partially defined function you've described, or more simply with the injection $x\mapsto [x,x+1]$. For that matter, $|K|\le |\Bbb R|$ is easy too because you can inject $K$ into the disjoint union of a few copies of $\Bbb R^2$ and $\Bbb R$. What needs some sophistication is $| B(\Bbb R)|\le |\Bbb R|$. – Sassatelli Giulio Aug 26 '22 at 19:02
  • The answer is here https://math.stackexchange.com/questions/70880/cardinality-of-borel-sigma-algebra (and doesn't depend on the continuum hypothesis as far as I can tell). – Rob Arthan Aug 26 '22 at 22:03
  • @RobArthan the 'countable' part already got me confused. Here I quote my lecture notes: a. Let $\Omega = \mathbb{R}$ and let $\mathcal{O}$ be the set of all open sets in $\mathbb{R}$. Then $\sigma(\mathcal{O})$ is called the Borel sets, denoted by $\mathcal{B}$. How does this definition imply Borel set is countable? Even just all the interval with the same size centred at all real numbers are already uncountably infinite? – Sam Aug 27 '22 at 12:05
  • @SassatelliGiulio - Thank you for the correction. I've deleted the false comment. – Paul Sinclair Aug 27 '22 at 17:22
  • I don't understand your comment. The set of Borel sets is not countable. – Rob Arthan Aug 28 '22 at 00:57
  • @RobArthan in https://mathworld.wolfram.com/BorelSet.html it says 'Roughly speaking, Borel sets are the sets that can be constructed from open or closed sets by repeatedly taking countable unions and intersections.' Sounds like Borel set is countable – Sam Aug 28 '22 at 08:09
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    There are uncountably many open sets, so the construction starts with an uncountable set and keeps adding to it. – Rob Arthan Aug 28 '22 at 12:17

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