How to interpret this definition of a subset?

[Robin04]

Hi,
I just had my first class of calculus where we mostly talked about sets. We defined the subset like this:
Let A and B be sets. We say that A is a subset of B if $x \in A \implies x \in B$
But if we look at the truth table of the implication (https://en.wikipedia.org/wiki/Truth_table#Logical_implication) we see that if the first statement if false, then the implication is true independently of the second statement. So if A and B are disjoint sets and as $x$ is a free variable we can choose it in a way that it is not in A, then the implication will be true, but this clearly contradicts with the concept of a subet relation. Don't we need a universal quantifier in this definition?

 

[fresh_42]

Hi,
I just had my first class of calculus where we mostly talked about sets. We defined the subset like this:
Let A and B be sets. We say that A is a subset of B if $x \in A \implies x \in B$
But if we look at the truth table of the implication (https://en.wikipedia.org/wiki/Truth_table#Logical_implication) we see that if the first statement if false, then the implication is true independently of the second statement. So if A and B are disjoint sets and as $x$ is a free variable we can choose it in a way that it is not in A, then the implication will be true, but this clearly contradicts with the concept of a subet relation. Don't we need a universal quantifier in this definition?

Yes, an all quantifier would be helpful. $A \subseteq B \Longleftrightarrow [\forall \,x\, : \,x\in A \Longrightarrow x\in B]$. Now your counterexample means $A=\emptyset$ and $\emptyset \subseteq B$. It is generally sloppy to introduce unquantified variables.

 

[Robin04]

Yes, an all quantifier would be helpful. $A \subseteq B \Longleftrightarrow [\forall \,x\, : \,x\in A \Longrightarrow x\in B]$.

Doesn't this have the same problem? I thought about writing the definition without implication like this: $\forall x \in A: x \in B$

Now your counterexample means $A=\emptyset$ and $\emptyset \subseteq B$. It is generally sloppy to introduce unquantified variables.

Do you mean that $A=\emptyset$ results from the original (wrong?) definition (if yes, how?) or it is the only example that satisfies my counterexample and still fits into the concept of a subset?

 

[fresh_42]

Doesn't this have the same problem? I thought about writing the definition without implication like this: $\forall x \in A: x \in B$

This depends on the axioms you set. E.g. you could write $A\subseteq B \Longleftrightarrow A \cup B =B \Longleftrightarrow A \cap B = A$

Do you mean that $A=\emptyset$ results from the original (wrong?) definition (if yes, how?) ...

You said:

if the first statement is false, then the implication is true independently of the second statement

Now to have $x\in A$ false all the time, i.e. for all $x$, such that the RHS is automatically true, we get $A=\emptyset$. If we quantify the RHS and say $x\notin A$ for one $x$, then the implication does not apply and thus makes no statement about the LHS, i.e. $x$ is in $B$ or not; except there is only one $x$ at all, and then $A$ is empty again by the same argument as before.

... or it is the only example that satisfies my counterexample and still fits into the concept of a subset?

It is quite difficult to argue on a sloppy worded statement. Fact is, the all quantifier belongs to the RHS implication, and with it, there is no contradiction. Without it, the definition is wrong and any deductions from it are necessarily true.

 

[Stephen Tashi]

Doesn't this have the same problem?

The definition: $A \subset B \iff \forall x ( x \in A \implies x \in B)$ doesn't have a problem when $A$ and $B$ are disjoint sets. In that situation, we have two possibilities - either there exists an element $x$ such that $x \in A$ or no such element exists.

If there exists an $x \in A$ then there exists an $x$ such that the implication $x \in A \implies x \in B$ is false in a non-vacuous way. Hence the universally quantified statement $\forall x (x \in A \implies x \in B)$ is false because there exists at least one $x$ that makes the implication false.

It there does not exist an $x \in A$ then $A = \emptyset$ and the definition says $A = \emptyset \subset B$.

( It's a tradition in mathematical writing to treat any variable variable that comes up as universally quantified unless specific restrictions are placed on it - a potentially confusing tradition. )

I thought about writing the definition without implication like this: $\forall x \in A: x \in B$.

That's notation, but what would you mean by that notation? The fact that notation doesn't contain the symbol "$\implies$" doesn't guarantee that a coherent interpretation of the notation can omit the concept of "if...then..".

 

[ad infinitum]

Hi,
I just had my first class of calculus where we mostly talked about sets. We defined the subset like this:
Let A and B be sets. We say that A is a subset of B if $x \in A \implies x \in B$
But if we look at the truth table of the implication (https://en.wikipedia.org/wiki/Truth_table#Logical_implication) we see that if the first statement if false, then the implication is true independently of the second statement. So if A and B are disjoint sets and as $x$ is a free variable we can choose it in a way that it is not in A, then the implication will be true, but this clearly contradicts with the concept of a subet relation. Don't we need a universal quantifier in this definition?

Often when a formula is written without a universal quantifier, like x in this case, it is intended that all variables (including A and B) be treated as universally quantified for the entire expression. So even though your teacher wrote $x \in A \implies x \in B$ what they really meant was ∀A∀B∀x($x \in A \implies x \in B$).