# Is the null/empty set both a subset of and disjoint from every non-empty set?

For any set (nonempty or otherwise) $A, \varnothing \subseteq A$.

Two sets are said to be disjoint if they contain no elements in common, so for any nonempty set $A, \varnothing$ is disjoint from $A$.

Can the null set therefore be considered BOTH a subset of and disjoint from $A$? Doesn't that contradict the definitions of these terms?

• Where is the contradiction? Suppose, by contradiction, that $A$ and $\varnothing$ are not disjoint. Then, there exists $x\in A \cap \varnothing$. In particular, $x\in \varnothing$, which is a contradiction. Then, $A$ and $\varnothing$ are disjoint. – Laura Feb 17 '18 at 2:58
• It is both a subset of and disjoint from every set, empty or nonempty. – bof Feb 17 '18 at 3:03
• It's not a contradiction at all; rather, it's one example of many of the weirdness of the empty set from a "natural language" perspective. An easier-to-digest example might be: "every element of $\emptyset$ is positive" and "every element of $\emptyset$ is negative" are each true, and don't contradict each other. – Noah Schweber Feb 17 '18 at 4:46

Going straight to the definition: $A\subseteq B$ if whenever $a\in A$ we must have that $a\in B$. Well if $A=\emptyset$ this is true for every $B$ since there is no $a\in A$. Hence $\emptyset$ is a subset of every set.

$A$ and $B$ are disjoint if $A\cap B=\emptyset$. Clearly if $B=\emptyset$ then this is true for all $A$.

It may conflict with the English definition of the words, but math is not English. Words mean different things.

Let $A$ be an arbitrary set.
Suppose that $A\cap \varnothing\neq \varnothing$ (suppose they are not disjoint).
Then there exists $x\in \varnothing$ such that $x\notin A$, which is a contradiction since the empty set contains no elements by definition.
Now, suppose that $\varnothing \nsubseteq A$. Again, we can see that this cannot be true since this would imply that there exists some element in $\varnothing$ which is not an element of $A$.