Prove there is a Bernstein set $B$ such that $B+B$ is also Bernstein

Question

Show that there exists a Bernstein set $B$ such that $B+B$ is also Bernstein.

I have tried to use the definition that neither $B$ nor its complement contain a perfect set.

Answer 1

Well, to start with, remember that a Bernstein set is built in stages: we list all the perfect sets as $\{P_\eta: \eta<\mathfrak{c}\}$ (where $\mathfrak{c}$ is the cardinality of the continuum), and we define a pair of sequences of sets of reals - $In_\eta$ and $Out_\eta$ - as follows:

• $In_0=Out_0=\emptyset$.

• At stage $\eta$, we pick distinct reals $r, s\in P_\eta\setminus (In_\eta\cup Out_\eta)$ and let $In_{\eta+1}=In_\eta\cup\{r\}$ and $Out_{\eta+1}=Out_\eta\cup\{s\}$. (By induction, both $In_\eta$ and $Out_\eta$ only have $\vert\eta\vert$-many elements, so such $r$ and $s$ exist.)

• For $\lambda$ a limit, we let $In_\lambda=\bigcup_{\eta<\lambda}In_\eta$ and $Out_\lambda=\bigcup_{\eta<\lambda}Out_\eta$.

• Finally, we take $B=In_\mathfrak{c}$.

By construction, we have:

• $B\cap P=In_\mathfrak{c}\cap P_\eta\supseteq In_{\eta+1}\cap P_\eta\not=\emptyset$, and

• $B^c\cap P_\eta\supseteq Out_{\eta+1}\cap P_\eta\not=\emptyset$,

so $B$ is indeed Bernstein.

So what?

Well, my point is that to build a Bernstein set with additional properties, you'll want to repeat this construction but with more bells and whistles. Think of it like this - in the construction of a "vanilla" Bernstein set, you have continuum-many requirements to meet: "positive" requirements "the $\eta$th perfect set intersects the set I'm building," and "negative requirements "the $\eta$th perfect set intersects the complement of the set I'm building." Each of these requirements is atomic: you can satisfy any one requirement by putting some element into the set you're building, or promising to keep some element out of the set you're building.

Now, you want to build a set $B$ such that

• $(i)\quad$ $B$ is Bernstein, and

• $(ii)\quad$ $B+B$ is Bernstein.

We already know how to break $(i)$ down into a bunch of "atomic" requirements, so let's look at $(ii)$. $(ii)$ breaks down into continuum-many requirements: the "positive" requirements "$P_\eta$ intersects $B+B$," and the "negative" requirements "$P_\eta$ intersects the complement of $B+B$." The positive requirements are no harder to satisfy than the positive requirements in the vanilla case.

The negative requirements, on the other hand, are harder: if I want to decree "$x$ is an element of $P_\eta$ which is not in $B+B$," it's not enough to throw a single real into my $Out$-set - I have to promise that, whenever a real $a$ enters my $In$-set, the real $x-a$ enters my $Out$-set. So this is not an atomic requirement, and to accomodate this I'm going to need to make the construction slightly more complicated: at each stage, I'll have an $In$-set of size $<\mathfrak{c}$, an $Out$-set of size $<\mathfrak{c}$, and additionally a set of $<\mathfrak{c}$-many "rules" of the form "no two elements summing to $x$ wind up in my $In$-set."

Fortunately, this doesn't complicate things too much. Can you see how to proceed?