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I understand the concept of vacuous truth and its use in showing that the empty set is a subset of every set. Based on my understanding of vacuous truth (for example https://en.wikipedia.org/wiki/Vacuous_truth) we have

$\forall x \in \varnothing: P(x)$

or

$\forall x(x \in \varnothing \implies P(x))$

Now $P(x)$ can be $x \in A$ so $\forall x (x \in \varnothing \implies x \in A)$. This establishes that $\varnothing \subset A$ (so the empty set is a subset of any set $A$).

However, my understanding is that it doesn't really matter what $P(x)$ is---the vacuous truth holds regardless (I find this quite obvious from what the vacuous truth is and how it works; wikipedia illustrates that if there are no phones in a room then we can say either 1) all phones are on or 2) all phones are off or even 3) all phones are both on and off). Therefore $P(x)$ can instead be $x \notin A$ and we have $\forall x (x \in \varnothing \implies x \notin A)$. It seems to me this is just as vacuously true. But then this would mean that the empty set is not a subset of any set $A$ ($\varnothing \not \subset A$) (along with being a subset of any $A$). If this is so then:

  1. Why is only the $\varnothing \subset A$ part mentioned (I've never seen $\varnothing \not \subset A$ stated)?
  2. If both $\varnothing \subset A$ and $\varnothing \not \subset A$ are (vacuously) true then how can any coherent statement or proof be based on any of them?
  3. Further, if $\varnothing \subset A$ is true "above" $\varnothing \not \subset A$ then doesn't that require a proof beyond merely vacuous truth?
  4. Lastly, can we extrapolate this further and say that both $\forall x(x \in \varnothing \implies x = x)$ and $\forall x(x\in \varnothing \implies x \neq x)$ are vacuously true?

This should be simple and clear so I'm wondering if I'm missing something. Thanks!

(PS I did search for for this before posting and the closest I came to was Is this statement "Every element in the empty set is greater than itself." true or its negation is true? which is really a different case where $\neg \forall x \in \varnothing : P(x)$ is negated to become $\exists x : P(x)$. Our case here instead is: $\forall x \in \varnothing:P(x)$ becoming $\forall x \in \varnothing:\neg P(x)$.)

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    $\begingroup$ The negation of $A\subset B$ is $\exists x\in A : x\notin B$. $\endgroup$ – Qiyu Wen Jun 25 '16 at 4:34
  • $\begingroup$ That seems entirely valid but how does this not violate the law of the excluded middle? @QiyuWen $\endgroup$ – Prince M Jun 25 '16 at 4:36
  • $\begingroup$ For all x in A then x not in B does NOT imply A is not a subset of B. The only way to show that A is not a subset of B is to show that there is an element of A that is not in B. Vaccuously claiming all elements are not is not the same as declaring there is an element that is not. $\endgroup$ – fleablood Jun 25 '16 at 5:53
  • $\begingroup$ $\forall x \implies P$ and $\forall x\implies -P$ are not contradictory statements. ~$(\forall x \implies P) \ne (\forall x \implies -P)$. Instead ~$(\forall x \implies P)= (\exists x \not \implies P)$. Those are entirely different statement. $\endgroup$ – fleablood Jun 25 '16 at 6:08
  • $\begingroup$ "we have ∀x(x∈∅⟹x∉A)." Yes, that is true. "It seems to me this is just as vacuously true." Yes, it is vacuously true. "But then this would mean that the empty set is not a subset of any set A (∅⊄A)" !NOPE! It absolutely does not mean that at all! It means that $\emptyset \subset A^c$. Which it is. $X \subset A^c \not \implies X \not \subset A$. (Think about it.) $\endgroup$ – fleablood Jun 25 '16 at 6:15
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"$X$ is a subset of $A$", in symbols $X\subset A,$ means $\forall x(x\in X\implies x\in A).$ If $X$ has no elements, this is true vacuously; hence $\emptyset\subset A,$ i.e., the empty set is a subset of $A.$

"$X$ is disjoint from $A$", in symbols $X\cap A=\emptyset,$ means $\forall x(x\in X\implies x\notin A).$ If $X$ has no elements, this is true vacuously; hence $\emptyset\cap A=\emptyset,$ i.e., the empty set is disjoint from $A.$

The empty set is a subset of every set; the empty set is disjoint from every set; the empty set is the one and only set which is both a subset of $A$ and disjoint from $A.$

"$X$ is not a subset of $A$", in symbols $X\not\subset A,$ means $\exists x(x\in X\text{ and }x\notin A).$ If $X$ has no elements, this existential statement is not true vacuously, it is simply false.

To answer your subquestions:

  1. The reason you've never seen $\emptyset\not\subset A$ stated, only $\emptyset\subset A,$ is that $\emptyset\not\subset A$ is false and $\emptyset\subset A$ is true.

  2. See answer to #1.

  3. See answer to #1.

  4. Yes, $\forall x(x\in\emptyset\implies\text{ anything})$ is true.

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  • $\begingroup$ Oops. Sorry. It's late and ... I missed it. $\endgroup$ – fleablood Jun 25 '16 at 7:16
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"$S \subseteq T$" is short-form for "$\forall x \in S\ ( x \in T )$".

"$S \nsubseteq T$" is short-form for "$\neg( S \subseteq T )$", which would be equivalent to "$\neg \forall x \in S\ ( x \in T )$", which is equivalent to "$\exists x \in S\ ( \neg( x \in T ) )$", which is not equivalent to "$\forall x \in S\ ( \neg( x \in T ) )$". Notice that the latter implies the former in all cases except when $S$ is empty, which is the case here.

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  • $\begingroup$ So your question (1) is answered, question (2) is irrelevant since (1) is false, question (3) is meaningless because there's no 'higher' or 'lower' truth, and question (4) is answered with "yes" because they are both vacuous truths, but (4) has nothing to do with (1,2,3). $\endgroup$ – user21820 Jun 25 '16 at 4:44
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Vacuous truth "all elements of A are not elements of B" is NOT the same as saying "there is an element of A that is not an element of B".

$\forall x \in A; x \not \in B$ is NOT enough to show $A \not \subset B$.

To show $A \not \subset B$ you must show $\exists x \in A; x \not \in B$ and you can not do that if $A = \emptyset$.

To pound it home. $A \subset B$ is defined as $\forall x \in A; x \in B$. The negation is "it is not true that all x in A are in B" which means "there exists an x in A that is not in B". Indeed "it is not true that all x in A are in B" is impossible to be true if A is the empty set, because it is true that all x in A are in B.

To really flog a dead horse "all elements are" and "all elements are not" are not contradictory statements at all. The contradictory statements are actually "all elements are" and "there is an element that is not". (Or "all elements are not" and "there is an element that is".

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    $\begingroup$ But if A is NOT empty then "not all elements are" IS equivalent to "there is an element that is not". But this is only true for non-empty sets. Hence your misunderstanding. $\endgroup$ – fleablood Jun 25 '16 at 6:18

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