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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?

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    $\begingroup$ 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. $\endgroup$ – Laura Feb 17 '18 at 2:58
  • $\begingroup$ It is both a subset of and disjoint from every set, empty or nonempty. $\endgroup$ – bof Feb 17 '18 at 3:03
  • $\begingroup$ 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. $\endgroup$ – Noah Schweber Feb 17 '18 at 4:46
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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.

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I think that this may answer your question.

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 $A$.

So in particular, the empty set is a subset of every set, and also that the empty set is disjoint with every set. Almost every time you prove something regarding the empty set, the best route to go is usually to use a proof by contradiction, since working with the empty set is almost always counter intuitive.

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