Is "The empty set is a subset of any set" a convention?

Question

Recently I learned that for any set A, we have $\varnothing\subset A$.

I found some explanation of why it holds.

$\varnothing\subset A$ means "for every object $x$, if $x$ belongs to the empty set, then $x$ also belongs to the set A". This is a vacuous truth, because the antecedent ($x$ belongs to the empty set) could never be true, so the conclusion always holds ($x$ also belongs to the set A). So $\varnothing\subset A$ holds.

What confused me was that, the following expression was also a vacuous truth.

For every object $x$, if $x$ belongs to the empty set, then $x$ doesn't belong to the set A.

According to the definition of the vacuous truth, the conclusion ($x$ doesn't belong to the set A) holds, so $\varnothing\not\subset A$ would be true, too.

Which one is correct? Or is it just a convention to let $\varnothing\subset A$?

Answer 1

There’s no conflict: you’ve misinterpreted the second highlighted statement. What it actually says is that $\varnothing$ and $A$ have no element in common, i.e., that $\varnothing\cap A=\varnothing$. This is not the same as saying that $\varnothing$ is not a subset of $A$, so it does not conflict with the fact that $\varnothing\subseteq A$.

To expand on that a little, the statement $B\nsubseteq A$ does not say that if $x\in B$, then $x\notin A$; it says that there is at least one $x\in B$ that is not in $A$. This is certainly not true if $B=\varnothing$.

Answer 2

From Halmos's Naive Set Theory:

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A transcription:

The empty set is a subset of every set, or, in other words, $\emptyset \subset A$ for every $A$. To establish this, we might argue as follows. It is to be proved that every element in $\emptyset$ belongs to $A$; since there are no elements in $\emptyset$, the condition is automatically fulfilled. The reasoning is correct but perhaps unsatisfying. Since it is a typical example of a frequent phenomenon, a condition holding in the "vacuous" sense, a word of advice to the inexperienced reader might be in order. To prove that something is true about the empty set, prove that it cannot be false. How, for instance, could it be false that $\emptyset \subset A$? It could be false only if $\emptyset$ had an element that did not belong to $A$. Since $\emptyset$ has no elements at all, this is absurd. Conclusion: $\emptyset \subset A$ is not false, and therefore $\emptyset \subset A$ for every $A$.

Answer 3

What confused me was that, the following expression was also a vacuous truth.

For every object of $x$, if $x$ belongs to the empty set, then $x$ doesn't belong to the set $A$.

As a complement (heh) to Brian Scott's (+1) answer, your argument shows that $\varnothing \subset A^{c}$, the complement of $A$. This statement is also (vacuously) true.

Answer 4

Very subtle point:

"All x are not something" does not imply "Not all x are something".

The first may be vacuously true. The second one can not. If the x are vacuous then the second one has to be "vacuously false" as all x of nothing are any property so it is impossible for them not to be any property.

So "all elements of the empty set are not in $S$" does not imply "Not all elements of the empty set are in $S$" $\iff$ "It is not true that all elements of the empty set are in $S$" $\iff$ "There are some elements of the empty set that are not in $S$".

The first is vacuously true (and is equivalent to $\emptyset \subset S^c$ which is true) and the second set of equivalent statements are all equivalent to $\emptyset \not \subset S$ which is not true.

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The thing is what you say is absolutely correct for non empty sets.

More formally:

All elements $x$ in $S$ are not in $A$ $\implies$

$S \subset A^c$ $\implies$

$\color{red}{\text{There is an } x\in S \text{ where } x \not \in A}\implies$

It is not true that all $x \in S$ are also in $A$ $\implies$

$S \not \subset A$.

However the red line can only be concluded if $A$ is non-empty. If $A$ is empty the red line is simply false.

And without the red line there is simply no logic or means to jump from the line before to the line after:

All elements $x$ in $S$ are not in $A$ $\not\implies$

It is not true that all $x \in S$ are also in $A$.

That simply is not true for an empty $S$.

Answer 5

Every theory has axioms, which are some propositions held to be true without being proven from anything else, and are not provable from each other. Subsequent truths of the theory derived from the axioms are theorems.

The properly termed question is whether the empty set being a subset of every other set is axiom of set theory, or a theorem.

It depends on how "subset" is defined. If $A\subset B$ means that every element of $A$ is in $B$, it is not necessarily true that $\emptyset$ is a subset of anything, since it has no elements. In this case, $\emptyset \subset A$ can be added as an axiom. It doesn't conflict with anything, and simplifies all reasoning about subsets. Alternatively, if $A\subset B$ is defined as "$A$ has no elements that are not also in $B$", then we do not require the extra axiom for the $\emptyset$ case. If $A$ has no elements at all, it has no elements that are not in $B$.

Suppose that we use the first, positively termed definition of subset, and then adopt as an axiom not $\forall A:\emptyset \subset A$, but rather its negation: $\exists A:\emptyset \not\subset A$, or the outright proposition $ \forall A:\emptyset \not\subset A$.

This is just going to cause problems. We can "do" set theory as before, but all the theorems will be uglified by having to avoid the special cases involving the empty set. In any derivation step in which we rely on a subset relation being true, or assert one, we will have to add the verbiage of an additional statement which asserts that the variable in question doesn't denote the empty set. This proposition then has to be carried in all the remaining derivations, unless something else makes it superfluous (some unrelated assurance from elsewhere that the set in question isn't empty).

Working with this clumsy subset definition that doesn't work with the empty set very well, someone is eventually going to have an epiphany and introduce a new subset-like relation which doesn't have these ugly problems: a new $A\ \mathbf{subset*}\ B$ binary relation which reduces exactly to $A\subset B$ when neither $A$ nor $B$ are $\emptyset$, and which, simply by definition, reduces to a truth whenever $A = \emptyset$, regardless of $B$. That person will then realize that all the existing work is simpler if this $\mathbf{subset*}$ operation is used in place of $\subset$.

At the end of the day it boils down to criteria like: is the system consistent (doesn't contradict itself), is it complete (does it capture the truths we want) and also is it convenient: are the rules configured so that we do not trip over unnecessary cases and superfluous logic.

Answer 6

This is a very mundane explanation (with finite sets). A subset is made of any combination of elements from the set. Suppose a set$ S$ is made of three elements: $\{a,b,c\}$. Each element $a$, $b$ or $c$ can be, or not, in the combination. If all of them are not in the combination, they STILL form a combination of "absent elements", or $\emptyset$, a subset of $S$. They are the dual of the subset of "all elements", $\{a,b,c\}$.

In the same way that $\{a,b\}$ and $\{c\}$ are (complementary) subsets, $\emptyset$ and $\{a,b,c\}$ belong to the set of subsets.

Answer 7

What confused me was that, the following expression was also a vacuous truth.

For every object of $x$, if $x$ belongs to the empty set, then  $x$ doesn't belong to the set $A$.

The contrapositive of the above conditional is:

"For every $x$, if $x$ belongs to set $A$, then $x$ doesn't belong to the empty set" which is easy to understand as the empty set has no elements at all.

NOTE- $p\rightarrow q$ has same meaning as $\lnot q\rightarrow \lnot p$