statement 1 : If $x \in \emptyset$ then $x \in A$.

statement 2 : If $x \in \emptyset$ then $x \notin A$.

I know that both statements are true since the hypothesis is false.

But first statement says that $\emptyset \subset A$ while second statement says that $\emptyset$ is not a subset of $A$.

My question is why we prefer $\emptyset \subset A$ over the other implication that $\emptyset$ is not a subset of $A$? Thanks.

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    $\begingroup$ "$x \in \emptyset$ then $x \notin A$ (..) says that $\emptyset$ is not a subset of $A$". .. if such an $x$ exists, which does not. $\endgroup$ – mvw Jul 27 '17 at 11:30
  • $\begingroup$ @mvw Thanks. This comment helped me too. +1 $\endgroup$ – Error 404 Jul 30 '17 at 5:17

Second statement shows that $$\emptyset \subseteq A^c$$ which is also true. We are not preferring one over other. Both are true.

Edit 1: The second statement does not imply that $\emptyset$ is not a subset of $A$. For that you would need that there exists $x \in \emptyset$ such that $x \in A$. This is indeed false.

Edit 2: What second statement implies is $\emptyset \cap A=\emptyset.$

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  • $\begingroup$ So the statement "empty set is not a subset of every set" is also true? $\endgroup$ – Error 404 Jul 27 '17 at 11:11
  • $\begingroup$ That is not what the second statement implies. $\endgroup$ – Sahiba Arora Jul 27 '17 at 11:12
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    $\begingroup$ @Mathmore yes this wrong. But it is only wrong for the empty set. $\endgroup$ – quid Jul 27 '17 at 11:18
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    $\begingroup$ @Mathmore, no. $X \subseteq Y, X \subseteq Y^c \implies X \subseteq Y \cap Y^c=\emptyset.$ Hence, $X=\emptyset$. $\endgroup$ – Sahiba Arora Jul 27 '17 at 11:40
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    $\begingroup$ @SahibaArora Awesome! +1 $\endgroup$ – Error 404 Jul 27 '17 at 11:43

The statement

(For all $x$,) if $x\in B$ then $x\notin A$.

does not say that $B$ is not a subset of $A$. In order to say that $B$ is not a subset of A, you would need to say

There is an $x\in B$ such that $x\notin A$.

This is vacuously false then $B$ is $\varnothing$.

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  • $\begingroup$ What is vacuously false here when $B$ is $\emptyset$? $\endgroup$ – Error 404 Jul 27 '17 at 11:34
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    $\begingroup$ @Mathmore: "There is an $x\in\varnothing$ such that ..." is vacuously false. $\endgroup$ – hmakholm left over Monica Jul 27 '17 at 11:44
  • $\begingroup$ Gotcha! ${}{}{}{}$ $\endgroup$ – Error 404 Jul 27 '17 at 11:58

In elementary set theory, each set can be defined as $A=\{x\mid P(x)\}$ where $P(x)$ is a predicate with free variable $x$, and each subset as $B=\{x\mid x\in A\wedge Q(x)\}$ with predicate $Q$.

Define $\emptyset_A=\{x\mid x\in A\wedge x\not\in A\}$. This is a subset of $A$, the empty subset of $A$. For any other set $B$, define similarly $\emptyset_B=\{x\mid x\in B\wedge x\not\in B\}$ as the empty subset of $B$.

But $\emptyset_A=\emptyset_B$, since $x\in\emptyset_A\Leftrightarrow x\in A\wedge x\not\in A\Leftrightarrow x\in B\wedge x\not\in B\Leftrightarrow x\in\emptyset_B$. In the 2nd equivalence, a false assertion is replaced by another false assertion.

Hence, the empty subset of a set $A$ is independent of the set $A$ itself and so we can safely speak of the empty set by putting $\emptyset=\emptyset_A$.

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    $\begingroup$ What is this "elementary set theory"? Also, $P(x)$ given by $x\notin x$ or $x=x$ will probably cause you some issues with that. $\endgroup$ – Asaf Karagila Jul 27 '17 at 11:20
  • $\begingroup$ Axiomatic set theory aside. Russel's paradox is glooming. $\endgroup$ – Wuestenfux Jul 27 '17 at 13:35

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