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If the 0(x)=0 for all x in $\emptyset$ it is also true that $0(x)=1$ for all x in $\emptyset$ It seems there are infinite possibilities for functions acting on nothing.

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Maybe you meant "infinitely many possibilities"? The phrase "infinite possibilities" would mean "possibilities, each of which, by itself, is infinite". – Michael Hardy May 22 '12 at 17:34
up vote 2 down vote accepted

No, there is exactly one function from the empty set to any set, so the ring in question is the "trivial ring" with exactly one element $0 = 1$. (This is a Boolean ring!)

Let's check this carefully: a function $f: X \rightarrow Y$ is a subset $R$ of $X \times Y$ such that for every $x \in X$, there is exactly one element of $R$ with first coordinate $x$. Now if $X = \varnothing$ then $X \times Y = \varnothing \times Y = \varnothing$, so the only subset of $X \times Y$ is $\varnothing$. This subset does satisfy the defining property of a function, since for every $x \in \varnothing$..., well never mind.

You might want to check for yourself the similar case of functions $f: X \rightarrow \varnothing$: there are none except in the case $X = \varnothing$ (in which, as a special case of what we checked above, there is exactly one).

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Thanks that's great. Only if ordered pairs define a function ...ok all ordered pairs in the "special case" are again no sets at all – user31953 May 22 '12 at 17:46

If $f:\varnothing\to\varnothing$ is the only function from $\varnothing$ to itself, then it is certainly true that $\forall x\in\varnothing,\ f(x)=1$.

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