Showing that the empty set is subset of every set

[Mr Davis 97]

Homework Statement

Let $J = \emptyset$ and $A$ be any set. Then $J \subseteq A$.

Homework Equations

The Attempt at a Solution

Proof: Suppose that it is false that $J \subseteq A$. Then $\exists x (x \in J ~~\text{and}~~ x \not \in A)$ is true. But this is a contradiction, since $\exists x (x \in J ~~\text{and}~~ x \not \in A)$ is false also, since there exists no element in the empty set.

 

[fresh_42]

Homework Statement

Let $J = \emptyset$ and $A$ be any set. Then $J \subseteq A$.

Homework Equations

The Attempt at a Solution

Proof: Suppose that it is false that $J \subseteq A$. Then $\exists x (x \in J ~~\text{and}~~ x \not \in A)$ is true. But this is a contradiction, since $\exists x (x \in J ~~\text{and}~~ x \not \in A)$ is false also, since there exists no element in the empty set.

Correct. You can also express this positively. $J\subseteq A$ means all elements of $J$ are also elements in $A$. This is true, because all statements about the elements of the empty set are true, or as I like to say: all elements of the empty set have purple eyes.

 

[Mr Davis 97]

Correct. You can also express this positively. $J\subseteq A$ means all elements of $J$ are also elements in $A$. This is true, because all statements about the elements of the empty set are true, or as I like to say: all elements of the empty set have purple eyes.

One slightly tangential question. By definition, $A \subseteq B$ means that $\forall x (x \in A \implies x \in B)$. But for the latter statement, what exactly is the set that $x$ is quantified over?

 

[fresh_42]

One slightly tangential question. By definition, $A \subseteq B$ means that $\forall x (x \in A \implies x \in B)$. But for the latter statement, what exactly is the set that $x$ is quantified over?

What do you mean? "For all $x$" is the same as "given any $x$" with emphasis on any. But as soon as the quantifier is left, $x$ becomes a certain element, because $x\in A$ deals with only one $x$. However, it cannot be further quantified, since it might not exist.

 

[Mr Davis 97]

What do you mean? "For all $x$" is the same as "given any $x$" with emphasis on any. But as soon as the quantifier is left, $x$ becomes a certain element, because $x\in A$ deals with only one $x$. However, it cannot be further quantified, since it might not exist.

My question is what are the x's under consideration here? Where do they live?

 

[fresh_42]

My question is what are the x's under consideration here? Where do they live?

Nobody knows. They can be anywhere without having anything to do with $A$ or $B$. But as soon as one of them is in $A$, then it has to be also in $B$. There is no statement about the elements of $C$. However, all elements of $C$ which happen to be in $A$, too, have to be as well in $B$. Same for the sets $D,E,\ldots$ or whatever - even none.

 

[member 587159]

Given a set $A$, you have to prove $\forall x: (x \in \emptyset \implies x \in A)$

But this implication is always true, since the antecedent of the implication is always false, because the emptyset can not contain elements by definition.

Therefore, the statement is what they call "vacuously" true. There is nothing to check.