Direct proof of empty set being subset of every set

Question

Recently I finished my first pure mathematics course but with some intrigue about some proofs of definitions by contradiction and contrapositive but not direct proofs (the existence of infinite primes for example), I think most of them because the direct proof extends away from a first mathematics course or the proofs by contradiction/contrapositive are more didactic. The one that most bothers me in particular is the demonstration that the empty set is a subset of every set, and it is unique. I understand the uniqueness and understand the proof by contradiction:

"Suppose $\emptyset \subsetneq A$ where $A$ is a set. So it exists an element $x \in \emptyset$ such that $x \notin A$ wich is absurd because $\emptyset$ does not have any elements by definition."

but I would like to know if there exists a direct proof of this and if indeed extends from a first course. Thanks beforehand.

Answer 1

For any $B\subseteq A$, we have $A\setminus B\subseteq A$. So $A\setminus A=\varnothing\subseteq A$.

(This proof operates at a slightly higher level of abstraction than verifying the definition of $\varnothing\subseteq A$. Since the definition is so easy to verify, you might think that it's silly to take a different strategy. If so, fair enough!)

Answer 2

There is a direct proof, if you know what a vacuous truth is. But the problem is that when one sees this statement $\varnothing\subseteq A$, it's usually before fully understanding vacuous arguments. So it's slightly more instructive to first give a proof by contradiction, and then discuss vacuous arguments. At least from my experience teaching this argument.

The proof is simple. We verify that $\forall x\in\varnothing$ it holds that $x\in A$. However, since $\forall x(x\notin\varnothing)$, the argument holds vacuously. And we are done.

Answer 3

Here is another direct proof, more calculational, where we first use the definitions and basic properties of $\;\emptyset,\subseteq\;$ and then simplify using predicate logic: \begin{align} & \emptyset \subseteq A \\ \equiv & \qquad\text{"definition of $\;\subseteq\;$"} \\ & \langle \forall x :: x \in \emptyset \Rightarrow x \in A \rangle \\ \equiv & \qquad\text{"basic property of $\;\emptyset\;$"} \\ & \langle \forall x :: \text{false} \Rightarrow x \in A \rangle \\ (*) \quad \equiv & \qquad\text{"logic: false implies anything"} \\ & \langle \forall x :: \text{true} \rangle \\ \equiv & \qquad\text{"logic: leave out unused quantified variable"} \\ & \text{true} \\ \end{align} Note how the last steps are really just a more detailed proof of the principle of 'vacuous truth', as used by earlier answers.

If you want more detail on the third step $(*)$: \begin{align} & \text{false} \Rightarrow P \\ \equiv & \qquad\text{"rewrite"} \\ & \lnot \text{false} \lor P \\ \equiv & \qquad\text{"simplify"} \\ & \text{true} \lor P \\ \equiv & \qquad\text{"simplify"} \\ & \text{true} \\ \end{align}

Answer 4

Yes, there's a direct proof:

The way that we show that a set $A$ is a subset of a set $B$, i.e. $A \subseteq B$, is that we show that all of the elements of $A$ are also in $B$, i.e. $\forall a \in A, a\in B$.

So we want to show that $\emptyset \subseteq A$. So consider all the elements of the empty set. There are none. Therefore, the statement that they are in $A$ is vacuously true: $\forall x \in \emptyset, x \in A$. So $\emptyset \subseteq A$.

Answer 5

For any set A, union of set A and nullset, gives set A. this proves that null set is subset of every set A. Using union operation for subset definition is the trick.

Answer 6

Let say we have two sets

A = {1,2,3,4} and B = {2,3,4}

A $\cup$ B = {1,2,3,4} = A

Applying Union operation does not changes anything to set A. Which means we are performing Union operation between set and its subset.

It implies B $\subset$ A

In same manner, if we perform Union operation between A and empty set.

A $\cup$ $\emptyset$ = A

it implies $\emptyset$ $\subset$ A.

Answer 7

Another way to look at vacuous proofs. Make use of the logical principle that anything follows from a falsehood (the arbitrary consequent rule): $$P\implies[\neg P \implies Q]$$

Your proof:

1. $\forall a: \neg a\in \emptyset$ (by defintion, better to use $\neg$ than $\notin$ in this case)

2. Let $S$ be any set.

3. Suppose $x\in \emptyset$

4. $\neg x\in \emptyset$ (from 1)

5. $\neg\neg x\in \emptyset\implies x\in S$ (arbitrary consequent rule applied to 4)

6. $x\in \emptyset\implies x\in S$ (from 5)

7. $x\in S$ (from 3 and 6)

8. $x\in \emptyset \implies x\in S$ (conclusion from 3 and 7)

9. $\forall a:[a\in\emptyset\implies a\in S]$ (generalizing from 8)

Yes, lines 6 and 8 look the same, but they play different roles in the proof. We can't immediately generalize on line 6.

Answer 8

A set $A$ is a subset of a set $B$ if $A$ has no elements that are not also in $B:$

$¬∃x∈A:x∉B$

Since the $empty$ $set$ has $no$ $elements$, it is clearly a subset of every set.

Answer 9

Here's a definition of "subset" that works:

"If set A contains an element that is not also in set B, then A is not a subset of B; otherwise, it is."

So, it is not true that the empty set contains an element that is also not in some (other) set or another. Therefore, the empty set is a subset of every set.

The problem is that the definition of a "subset" is sometimes (or even usually) stated like this:

"If all of the elements of set A are also in B, then A is a subset of B; otherwise, it is not."

But according to general English usage, this definition pre-supposes at least one element in set A and therefore can't be applied to the empty set -- at least, not according to general English usage.

An interesting follow-on question is, I think: "Why do mathematicians even conceive of the empty set, given that it constitutes 'nothingness'?"