A single set can be said to be pairwise disjoint with itself, right? From the definition I've been given:

A collection of sets $\{A_\alpha : \alpha \in I\}$ is said to be pairwise disjoint if the following is satisfied: For all $\alpha$, $\beta$ $\in I$, if $A_\alpha \cap A_\beta \ne \emptyset$, then $A_\alpha = A_\beta$

In the case of one set, $A_1$, which is nonempty, we have $A_1 \cap A_1 = A_1$ and $A_1 = A_1$ so it satisfies the condition above. I just want to make sure this is a common understanding and not just a mistatement of the definition.
 A: The answer to the question in your title is "No" - a single set is not pairwise disjoint with itself. In fact, that "with itself" part is confusing, and not good mathematical language.
However, a collection of sets containing just a single set is pairwise disjoint, for the vacuous reason that you can't find any pairs of sets in it.
We're really strict about these uses of language in math, and neglecting to do so can lead one to some serious mistakes. As another example, the number of elements in the sets $\mathbb Z$ and $\{\mathbb Z\}$ are drastically different.
A: The correct definition is :

A collection of sets $\{A_\alpha \ : \ \alpha \in I\}$ is said to be pairwise disjoint if the following is satisfied: For all $\alpha, \beta \in I$, if $A\alpha \cap A \beta \neq \emptyset$, then $\boxed{\alpha= \beta}$

Yet, it is a clumsy way to formulate it. I hence prefer the nicer :

A collection of sets $\{A_\alpha \ : \ \alpha \in I\}$ is said to be pairwise disjoint if the following is satisfied: For all $\alpha, \beta \in I$, if $\alpha \neq \beta$, then $A_\alpha \cap A_\beta = \emptyset$.

And yes indeed the family $\{A_1\}$ is pairwise disjoint but the family $\{A_1, \ A_2\}$, where $A_1 = A_2$, is not pairwise disjoint.
Hence, the answer to the question as stated in the op title, namely

A single set can be said to be pairwise disjoint with itself, right?

is no.
A: The only set that is disjoint from itself is the empty set; a nonempty set is not disjoint from itself.
The definition you have for a pairwise disjoint collection of sets refers to an indexed collection that may contain duplicates (i.e. different indexes $\alpha,\beta$ with $A_\alpha=A_\beta$), and it considers the collection to be pairwise disjoint as long as any two unequal members are disjoint from each other.
An alternative definition for indexed collections would require that $A_\alpha\cap A_\beta=\emptyset$ whenever $\alpha\ne\beta$. The difference is that this disallows duplicate nonempty members.
A: As others have pointed out it's the COLLECTION of sets that is "pairwise disjoint".  That the sets within the collection.   So the collection $\{A\}$ is pairwise disjoint.  But that says nothing about whether $A$ itself is pairwise disjoint.  For $A$ to be pairwise disjoint the term "disjoint" must apply to the elements of $A$.  i.e. $A$ must be a collection of sets.  Now if $A$ is a collection of sets we have no idea how many set elements it has nor which of them are disjoint from each other.
So concl:  Yes, a collection $\{A\}$ of a single set is by definition pairwise disjoint. (Because $A\cap A \ne \emptyset \implies A=A$... [in fact "Kittens eat purple dinosaurs" $\implies A=A$ and $5$ is a prime number $\implies A=A$])
But, no, a set $A$ may or may not be pairwise disjoint. ( $A= \{\{1,2\},\{3,4\},\{5,6\}\}$ is.  $A = \{\{1,2\},\{2,3\}\}$ is not.  And for $A = \{apple, bannana, cantalope\}$ the question is meaningless.)
Once again, this comes down to a set (or a "collection") being a different thing with entirely different properties than its elements.  There are are just different animals altogether.
