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It is known that sum-free
subsets of $mathbbN$ can have
natural density at most
$frac12$. This density is achieved by the odd numbers: the sum of two
odd numbers is even.

I ask now a similar question for XOR rather than addition.
For $a,b in mathbbN$, define $a oplus b$ as follows:
Represent $a ge b$ as binary numbers, pad the smaller $b$ with zeros,
and take the bit-wise XOR of the binary representation.
For example, $35 oplus 15 = 44$:
begineqnarray
35 = & ;100011\
15 = & ;001111\
oplus = & ;101100
endeqnarray
The condition for a set $S subset mathbbN$ to be XOR-free is that, for any $a,b in S$, $a oplus b notin S$.

Q. What is the largest density of an XOR-free subset $S$ of $mathbbN$?

Again the odd numbers with density $frac12$ are XOR-free.
I am not seeing an argument that $frac12$ is the maximum possible density.

Let $ain S$ be some fixed element. Note that $aoplus b le a + b$. Let $N$ be some big number. Put $M = [1, ldots , N]cap S$. We have $aoplus M cap M = varnothing$. We also have $aoplus M subset [1, ldots , N + a]$. Therefore $2|M| le N + a$ or $|M| le fracN2 + fraca2$. Since $a$ is fixed taking limit $Nto infty$ yields the desired result.

UPD Here is an answer for a perhaps more interesting question: what is limsup of the biggest density of the subset of $[0, ldots , N]$ which is XOR-free. Note that if $N = 2^k - 1$ for some $kin mathbbN$ then $|S| le 2^k-1$. Indeed, for any $a, bin [0, ldots, N]$ we have $aoplus b in [0, ldots , N]$ . Since $|Soplus S| ge |S|$ and $Scap (Soplus S) = varnothing$ we get $2|S| le (N+1)$ or $|S| le 2^k-1$.

For the general case assume that $2^k le N < 2^k+1 - 1$. Then on the one hand $|S| le 2^k$ since $Ssubset [0, ldots , 2^k+1-1]$ and on the other hand $|S| le 2^k-1 + (N - 2^k) = N - 2^k-1$ since $|Scap [0, ldots , 2^k - 1]| le 2^k-1$. Therefore $3|S| le 2N$ or $|S| le frac2N3$.

Here is an example (found partly via computer search) which shows that density $frac23$ is possible: put $N = 6*2^k$ and let $S$ be a set of all numbers consisting of $k + 3$ digits such that first three of them is from the following set $M = (0, 1, 0), (0, 1, 1), (1, 0, 0), (1, 0, 1)$. Note that $Soplus S cap S = varnothing$, all elements of $S$ are not greater than $N$ and $|S| = 4*2^k$. Therefore $fracSN = frac23$.