R subset and Borel set

Which is the simplest example of a subset of R that is not a Borel set?

Ps: I asked my probability professor how I can see if a subset of R is a Borel set or not? He answered that in the case we will use the set are always Borel set and in fact he said that all subset of R are Borel set. So at this point I was very interested: why I can't imagine a subset of R that is not a Borel set? So I asked him an example of a subset of R and he answered that he will not do that because it will only confuse me and because is not important. So I'm asking here!

Ps2: I don't have any background of topology, geometry and so on. I know analysis (I-II) and linear algebra (I-II) so keep it simple please.

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math.stackexchange.com/q/137277 – user47854 Nov 1 '12 at 22:05

Our imagination is limited. That much is a fact. We can expand it by training our minds, and slowly understand the properties of new objects to the point we are convinced that we can imagine them.

The fact that you don't have much background in abstract mathematics means that it is unlikely that you can come up with such set on your own.

Why? Well, when we think about sets of real numbers we think about intervals, this is because intervals are nice. They are defined by two endpoints and that's that. We could maybe think about a finite union of intervals, which is again defined by a few (read: finitely many) points and that's that.

However can you imagine the rationals? Can you imagine the rationals and discern them, in your imagination, from the irrationals or the real numbers? Can you imagine the irrational numbers? Can you discern this set from the entire real numbers? It takes some experience to understand those set to begin with, let alone to imagine them, and even more time to come up with similar examples on your own.

The rational and irrational numbers are both very simple sets in terms of "description", namely if we look at the Borel sets, they are generated from the intervals by unions, complements, and intersections. We can assign each set some sort of complexity rank -- how many steps we need to do before we can generate it from the intervals themselves. Both the rationals and irrationals are quite simple. The Borel sets themselves are vastly more complicated, and in fact you cannot even imagine Borel sets which are very complicated. At least not without understanding a lot more about this.

All that is fine, but one can easily use the way we define the Borel sets to show that in fact there are only $2^{\aleph_0}$ Borel sets. Cantor's theorem, however, tells us that there are $2^{2^{\aleph_0}}$ sets of real numbers and that this is a much larger number of sets. In particular almost all sets of real numbers are not Borel.

To your specific question, it is difficult to give an example of a concrete non-Borel set, in particular because we need to use something called the axiom of choice in order to prove the existence of such sets. The axiom of choice is a useful tool in modern mathematics whose power is in its non-constructiveness, namely we can use it to prove that a certain object exists without demonstrating an example for such object. For example, we can prove that there are non-Borel sets, but we cannot really point out at a set and say it is a non-Borel set. While there are examples of sets which are not Borel sets, such as the one linked by jkl in the comments (see: Constructing a subset not in $\mathcal{B}(\mathbb{R})$ explicitly), we still have to use the axiom of choice to prove that this set is not a Borel set.

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I would like to give an answer that is similar to Asaf's but a bit less pessimistic.

One can define the notion of a Borel code, which is a recipe for building up a set of reals from intervals under the operations of countable unions, countable intersections, and complementation. Every Borel code specifies a particular Borel set.

A bit of elementary set theory shows that there are only as many Borel codes as there are real numbers, so we can list all the Borel codes as $(c_x : x \in \mathbb{R})$. Let $B_x$ be the Borel set coded by $c_x$. Define $A = \{x \in \mathbb{R} : x \notin B_x\}$. Then the set $A$ is not equal to the set $B_x$ for any real $x$ because $x$ itself is in one but not the other. (This is Cantor's diagonal argument.)

Therefore we have a set $A$ that has no Borel code. Whether it is simple or not depends on whether the coding system is simple or not. I would say that conceptually it is simple but the details are rather tedious.

Because $A$ has no Borel code, it has no good reason to be Borel. It could still be Borel for a silly reason. For example, if $\mathbb{R}$ were a union of countably many countable sets, then every set would be Borel for a silly reason. To rule this out, we need the Axiom of Choice (really just the Axiom of Countable Choice.)

Using the Axiom of Choice, we can show that the collection of sets with Borel codes is closed under countable unions and intersections. Given a countable sequence of sets $A_1$, $A_2$, $A_3$,... with Borel codes, to get a Borel code for their intersection or union, one first has to choose a code $c_i$ for each set $A_i$, which requires the Axiom of Choice because each $A_i$ will have many Borel codes to choose from. Then the sequence of codes $(c_i : i<\omega)$ gives you a code for the intersection or union.

Because the collection of sets with Borel codes is also closed under complementation and contains every interval, it must contain all the Borel sets. So the set $A$ we constructed, having no Borel code, is not Borel.

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I wasn't pessimistic, when I get pessimistic -- you'll know. – Asaf Karagila Nov 2 '12 at 7:08
@AsafKaragila ha :) – Trevor Wilson Nov 2 '12 at 17:15