Direct proof of empty set being subset of every set
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.
Solution 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!)
Solution 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.
Solution 3:
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 $\emptyset \subseteq A$.
Solution 4:
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}