Can't make sense out of this definition of the $xor$ function Can someone clear up the following definition for me?
Let $U \subseteq \mathbb{N}$.
We say that a function $x: \{0,1\}^U\to \{0,1\}$ is a $xor$ on the set $U$ if the following condition is met:
$x(a)=1-x(a')$, if $a$ and $a'$ are function that differ for exactly one argument $a \in U$. That is, there is such $n \in U$, that $a(n)\neq a'(n)$ and for all $m \in U-\{n\}$ $a(m)=a'(m)$.
Example (not a very helpful one):
For $U=\{0,...,10\}$ function $x:\{0,1\}^U\to \{0,1\}$ defined as:
$$ x(a)=(\sum_{i=0}^{10}a(i))(\mod 2)$$ is a xor on the set U.
That's all I got, and here are my thoughts: in the above example, does $x(\lambda a.1)=11 \mod 2=1$? What does that have to do with $xor$? And how does the definition work? It looks like an infinite recursion to me...
Thanks!
 A: I will try to rephrase the definition with fewer formulas.
Given a fixed index set $U$, a (an?) xor as defined here is a function that assigns to each sequence of $0$s and $1$s indexed by $U$ a value of $0$ or $1$ in such a way that if you change the input at exactly one place (from $0$ to $1$ or vice versa) then the result also changes. 
For a finite set $U$ such a function can be obtained by $x(a)=0$ if the sequence $a$ contains an even number of $1$s and $x(a)=1$ otherwise (this is “a bitwise xor”). This is the example that you have given. And indeed, if $U$ is finite, then it is easy to show that each xor for $U$ is either this one or the opposite one ($0$ for an odd number of $1$s). To see this, we only have to remark that we can get from any finite sequence to the all-$0$-sequence (let's call it $\mathbf 0$) by changing each $1$ to $0$, one after the other. So if $a$ has an even number of $1$s then $x(a)=x(\mathbf 0)$, and if $a$ has an odd number of $1$s then $x(a)=1-x(\mathbf0)$.
A: It seems like the following:
Your definition of $x$ being an xor function is actually more like that it is in $( U \to \{0,1\} ) \to \{0,1\}$; $x$ takes a function $f$ as an input, where $f$ is a function from $U$ to $\{0,1\}$, and returns either $0$ or $1$.
In your example, $x(a)$ is defined (for a finite $U$) as the remainder when the sum of $a$ on all elements of $U$ is divided by $2$. In the case of $U=(0,1,...,10)$, this is the odd-ness of the number of $1$s in the sequence $(a(0),a(1),...,a(10)$, also called the parity of the sequence. It clearly satisfies the condition for being an xor function, because changing one element in the sequence will always change odd to even and even to odd, and equivalently change the parity from 1 to 0 or 0 to 1. So yes $x$ applied to the function that always returns $1$ will result in $1$.
As a note, typically in programming contexts we use "%" for a binary modulo operation, rather than "(mod ...)".
