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Can a Boolean ring be somehow extended to infinite binary sequences? If $2^{\Bbb N}$ is the Cantor space (the set of all infinite binary sequences), is it a ring if the operations are defined term by term?

For example, given the sequences $A:=\{a_n\}_{n\ge1}$ and $B:=\{b_n\}_{n\ge1}$, $\:A+B\:$ and $AB\:$ would be $\{a_n+b_n\}_{n\ge1}$ $\{a_nb_n\}_{n\ge1}$ respectively

One of the operations is $AND$/conjunction but I don´t know if the other should be $OR$/disjunction or $XOR/$exclusive disjunction. I´m also not sure about inverses: one sequence can have many different "inverses" under the operation $AND$. Does this mean that the conjunction is the multiplication and all those inverses are the "divisors of zero"?

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    In a boolean ring with identity, everything is a zero divisor except for the identity, which is the one and only invertible element of the entire ring. – 2017-02-20
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    @rschwieb you mean the additive identity? – 2017-02-20
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    no, the multiplicative identity. The additive identity is never invertible in a nonzero ring... – 2017-02-20

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You have indeed two options. If you equip the set $\{0, 1\}$ with $OR$ and $AND$ you have a semiring structure. If you make use of $XOR$ and $AND$, you have a Boolean ring. If $S$ is one of these two structures, the product $S^\mathbb{N}$ has the same structure, defined componentwise as you did.

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    $AND$ would be the multiplication? Is the "divisors of zero" part right? – 2017-02-20
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    To generalize, you can always lift a ring structure $R$ to a ring structure on $R^A$ (the set of functions from $A$ to $R$ for some set $A$) for any set $A$. In fact, this can be shown for all "essentially algebraic" (in a technical but broad sense) structures, e.g. groups and lattices. (This technical sense excludes fields though.) – 2017-02-20
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    $AND$ is indeed the multiplication. I am not sure to understand your remark on inverses because the neutral element for $AND$ is $1$. But the point is that in $2^\mathbb{N}$ every element $x$ is idempotent (that is, satisfies $x^2 = x$) and satisfies $x + x = 0$. Therefore $x(1+x) = x + x^2 = x + x = 0$, which is probably your remark on the divisors of $0$. – 2017-02-20
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    Derek Elkins, J.-E. Pin thank you so much I get it now – 2017-02-20