# How to construct a Bernstein set and what are their applications?

Bernstein Set: A subset of the real line that meets every uncountable closed subset of the real line but that contains none of them. It's from wiki.

My question is this: How to construct a Bernstein set? And what's its application in mathematics?

One uses the axiom of choice in some form to construct a Bernstein set. The most straightforward way is by transfinite recursion. There are $2^\omega$ uncountable closed subsets of $\Bbb R$, so we can list them as $\{F_\xi:\xi<2^\omega\}$, and it can be proved that each of them has cardinality $2^\omega$.

Now suppose that $\eta<2^\omega$, and for each $\xi<\eta$ you’ve chosen points $x_\xi,y_\xi\in F_\xi$ so that all of these points are distinct. Let $X_\eta=\{x_\xi:\xi<\eta\}$ and $Y_\eta=\{y_\xi:\xi<\eta\}$. Then $|X_\eta\cup Y_\eta|<2^\omega$, so $F_\eta\setminus(X_\eta\cup Y_\eta)$ is infinite, and we can choose distinct $x_\eta,y_\eta\in F_\eta\setminus(X_\eta\cup Y_\eta)$ to continue the construction.

Now let $X=\bigcup_{\xi<2^\omega}X_\xi$ and $Y=\bigcup_{\xi<2^\omega}Y_\xi$; by construction $X$ and $Y$ are disjoint sets meeting each uncountable closed subset of $\Bbb R$, so both are Bernstein sets.

Added: In my experiences they are most useful as a tool for constructing (counter)examples. This post in Dan Ma’s Topology Blog is a good example of such use.

– Paul
Jul 12, 2012 at 1:00
• Does your construction imply that X U Y = R? Dec 7, 2015 at 0:39
• @user114634: No, it doesn't. Dec 7, 2015 at 0:46

Notice that the usual construction of the Bernstein set can also yield $\mathfrak c$ disjoint Bernstein sets in an arbitrary uncountable Polish space by working diagonally, similarly to how you construct a bijection between $\mathbf N$ and $\mathbf N^2$: you take one point from the first closed set into the first Bernstein set, two points from what remains of the second closed set to first two Bernstein sets etc, resulting in a sequence of distjoint sets, each of which intersects all but possibly some $\kappa<\mathfrak c$ uncountable closed sets. Then you notice that every (uncountable) closed set contains $\mathfrak c$ other (uncountable) closed subsets, so each of those actually intersects ALL uncountable closed sets.

In many ways, Bernstein sets are "as pathological as possible" for a subset of a Polish space, and as such are often good source of examples of pathological behavior.

Bernstein sets are not only non-measurable (and extremely so: the outer measure of a Bernstein set with respect to any continuous Borel measure is full, while the inner is zero!), but also do not have Baire property.

They are also quite interesting as topological measure spaces. Every compact subset of the Bernstein set is countable, so it's impossible to define a (nontrivial) continuous Radon measure on Bernstein set, which makes them sort of the opposite of Polish spaces -- in a Polish space, any finite Borel measure is Radon.

• "The usual construction of the Bernstein set can also yield c disjoint Bernstein sets in an arbitrary uncountable Polish space by working diagonally." It's nice:) Vote me!
– Paul
Jul 12, 2012 at 2:11

This is exactly the Proposition 3.1 in Handbook of Game Theory, Volume 1, Chapter 3, Jan Mycielski.

There is a payoff set such that the Gale-Stewart game is not determined.

Proof: Let the payoff set be the Bernstein set. Given an arbitary strategy, the set of all possible outcomes generated by it is a perfect set(see Brian M. Scott's answer), so for any strategy of one player, there is pair of opponent's strategies that generate two outcomes that are in the payoff set and its complement respectively.

Since a Bernstein set is nonmeasurable, you can't do it without some form of the Axiom of Choice: there's no "explicit" construction.

An example of an application: show that every measurable set of positive measure contains a nonmeasurable set.

Here's one: If $\omega_1$ injects into reals then there is an uncountable set of reals without perfect subset. Hence the statement that every set of reals has the perfect set property implies that $\omega_1$ is inaccessible in $L$ (constructible universe).

Here's an "application": Let P be a partition of reals induced by the equivalence relation $x \sim y$ iff $x - y$ is rational. A choice set for P is called a Vitali set. Show that for each real $m > 0$ there is a Vitali set of outer measure $m$.

See this blog post of Dan Ma for a detailed construction of Bernstein set.

A Bernstein set $$B$$ intersect every closed uncountable sets and contains none of them. In other words the set $$B$$ and it's set theoretic complement $$B^c$$ intersect every closed uncountable sets.

Intersecting every closed uncountable sets can be replaced by intersecting every perfect sets( why?)

$$A\subset \Bbb{R}$$ is called saturated non measurable set iff$$m_i(A) =0=m_i(A^c)$$

Where $$m_i(A)$$ is the inner measure of $$A$$ and $$A^c=\Bbb{R}\setminus A$$

See here

1. Bernstein set is a saturated non measurable set.

2. Complement of a Bernstein is also Bernstein.

3. Bernstein set doesn't exists in ZF alone, one need some form of choice! (see here)

4. Bernstein set doesn't have the Baire Property i.e one can't express Bernstein set as a symmetric difference of an open set and a meager set (see here).

5. Any measurable subset of a Bernstein set must be a null set (lebesgue outer measure $$0$$). See here

6. Any subset of Bernstein set having the Baire property must be of meager.

Two applications of Bernstein sets:

Theorem $$1$$: Any set of positive measure contains a set that fails to be measurable.

Proof: Let $$A\in\mathcal{L}(\Bbb{R})$$ and $$m(A) >0$$.

Consider $$A\cap B$$ and $$A\cap B^c$$ where $$B$$ is a Bernstein set.

Then at least one of the sets $$A\cap B$$ or $$A\cap B^c$$ must be non measurable otherwise both the sets $$A\cap B$$ and $$A\cap B^c$$ being the measurable subsets of Bernstein set must have measure $$0$$ but it would contradict that $$m(A) >0$$.

Theorem $$2$$: Any non meager (second category) subset of $$\Bbb{R}$$ contains a set that fails to have the property of Baire.

Proof: Let $$A\subset \Bbb{R}$$ be a non meager set.

Consider $$A\cap B$$ and $$A\cap B^c$$ where $$B$$ is a Bernstein set.

Then at least one of the sets $$A\cap B$$ or $$A\cap B^c$$ doesn't have the Baire property otherwise both the sets $$A\cap B$$ and $$A\cap B^c$$ being the BP subsets of Bernstein set must be meager but it would contradict that $$A$$ is non meager.