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I was wondering if one could assign random attributes to the elements empty sets, even contradictions. Because there are no elements, I can say something like:

$\forall x \in \emptyset.P(x) \land \neg P(x)$

Hence a set containing contradictions for all elements must be empty, I can assign any attribute (even its opposite).

Is my reasoning wrong, or is this just the way it is and it shouldn't be mind-boggling?

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    I don't have a background in logic (at all), but I'm pretty sure that for all $x\in\varnothing$, it is true that $Q(x)$ regardless of the details of $Q$. We certainly act like this is true in less formal contexts.2012-12-27

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"Hence a set containing contradictions for all elements must be empty"

The empty set doesn't contain any element, so it doesn't contain propositions that are contradictions.

But it is true that $\forall x \in \varnothing : P(x) \land Q(x)$. That is, $\forall x: x\in \varnothing \rightarrow (P(x) \land \lnot P(x)).$ Since there is no $x \in \varnothing$, $x\in \varnothing \rightarrow (P(x) \land \lnot P(x))$ is vacuously true for any (every) such (nonexisting) $x$.

Perhaps what you mean to be saying is:

  • $\varnothing = \{x \mid P(x) \land \neg P(x)\}:\;\;$ "The set of all $x$ such that $P(x) \land \lnot P(x)$."

    Since there is no $x$ such that $P(x) \land \lnot P(x)$, the empty set remains empty.

The empty set can be defined by any condition(s) that no element can satisfy.

Examples:

  • $\varnothing = \{x\in \mathbb{R} \mid x^2 = -1\}$

  • $\varnothing = \{x \mid x + 3 = x\}$

  • See this post, too.


Note: every universal statement about the elements of the empty set is true; this is known as vacuous truth. One might say that universal statements are "true until proven false." Since there are NO elements in the empty set, such statements cannot be proven false (as there is no element to serve as a counterexample).

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    @amWhy: They had a bad attack on me. I didn't have any support there. Enclosed to anything.....2013-02-09
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It is worth pointing out that in most formalization of logic, "$\forall x\in\varnothing:P(x)$" is considered an abbreviation of $ \forall x: (x\in\varnothing \to P(x)) $

Since $x\in\varnothing$ is always false for every $x$, the implication $x\in\varnothing\to P(x)$ is always true (because the only way $Q\to P$ can be false is when $Q$ is true and $P$ is false), and therefore the entire quantified formula is true.