# Prove that the empty set is a subset of every set

Does this proof work?

By definition:

$$[A \cap B = A] \wedge [A \cup B = B] \implies [A \subseteq B]$$

Therefore:

$$[\emptyset \cap B = \emptyset] \wedge [\emptyset \cup B = B] \implies [\emptyset \subseteq B]$$

Thanks in advance.

• meta.math.stackexchange.com/questions/5020/… – Timbuc Feb 7 '15 at 13:44
• What about: for any set $\;A\;$ we have that for all $\;x\in\emptyset\implies x\in A\;$ and thus $\;\emptyset\subset A\;$ ? – Timbuc Feb 7 '15 at 13:45
• understanding $\emptyset \subseteq B$ is easier than understanding $[A \cap B = A] \wedge [A \cup B = B] \implies [A \subseteq B]$ – user 1 Feb 7 '15 at 13:50
• If you take that relation as a definition, then this is valid, as long as you can show that $\emptyset\cap B=\emptyset,\emptyset\cup B=B$. – Wojowu Feb 7 '15 at 13:56
• @G Ch Wojowu is right. So the question is how do you say $\varnothing \cap B = \varnothing$ and $\varnothing \cup B = B$ – R_D Feb 7 '15 at 14:11

## 1 Answer

If you use this as your definition of subset, you need to justify that $\varnothing\cap B=\varnothing$ and $\varnothing\cup B=B$. I will do these here.

Note that $\varnothing\cap B=\left\{x:x\in \varnothing\land x\in B\right\}$. Since the condition $x\in\varnothing$ is never satisfied, the set is defined by an always-false condition, and so we have that $\varnothing\cap B=\varnothing$.

Now, note that $\varnothing\cup B=\left\{x:x\in\varnothing\lor x\in B\right\}$. Since, again, the condition $x\in\varnothing$ is never satisfied, the defining condition for this set is $x\in B$, and so we have that $\varnothing\cup B=B$.

Therefore, since $\varnothing\cap B=\varnothing$ and $\varnothing\cup B=B$, we have that $\varnothing\subseteq B$. Since $B$ was no set in particular, this makes $\varnothing$ a subset of every set, as required.