2
$\begingroup$

Let $2^{\Bbb N}$ be the set of infinite binary sequences $\{x_n\}$ where $x_n\in \{0,1\}$ for every $n\in \Bbb N$. I want it to fit the axioms of a known algebraic structure such that the following subset $I$ has "interesting" properties:

A sequence $X:=\{x_n\}_{n\ge1}$ has the "poperty $S$" if and only if for every positive integer $r\in \Bbb N$ there is at least a pair $s,t\in \Bbb N$ such that $s+t-1=r$ and $x_s=x_t=1$

I´ve been able to classify $2^{\Bbb N}$ as the ring $(2^{\Bbb N},\oplus,\cdot)$ with $\oplus$ defined as the exclusive difference between two sequences term by term (working as addition) and $\cdot$ defined as the conjunction again term by term (working as multiplication). The resulting structure is a conmutative boolean ring with unity

The problem is that $I$, the set of every sequence with the property $S$, isn´t any kind of "interesting" subset of a ring: it fails to be an ideal because although it absorbs products it isn´t an additive subgroup

That makes me think it could be better described with other type of structure like a boolean algebra or a lattice (join being disjunction and meet being conjunction) wich would be distributive (and satisfy DeMorgan´s laws)

$\endgroup$
1
  • $\begingroup$ The Cantor set also can be given the structure of a torsion free abelian group. $\endgroup$ Mar 2, 2017 at 4:34

1 Answer 1

1
$\begingroup$

By the canonical identification of $\scr P(\Bbb N)$ with $2^{\Bbb N}$, you can say that $$I = \{N\subset\Bbb N : N + N = \Bbb N + 1\}$$ Your additive ring operation is the symmetric difference $A \triangledown B = (A \cup B) \setminus (A \cap B)$. The multiplicative operator is just intersection.

So yes, it is expressible as a boolean algebra.

$\endgroup$
2
  • $\begingroup$ and what type of subset would $I$ be? $\endgroup$ Mar 2, 2017 at 6:38
  • $\begingroup$ Since all I ve done here is translate your definitions into a slightly different construction, you have already given the answer in your question: I is not "interesting" from your point of view. The point of this post was to translate your work into a simpler (IMO) expression that may be easier to work with, and suggests other variations, such as replacing $A \triangledown B$ with $A \cup B$. $\endgroup$ Mar 2, 2017 at 17:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .