I'm trying to find all the possible premises for $A \setminus B = \emptyset$ and so far I have $A \subseteq B$ or $A = \emptyset$ or $A=B$ but I'm not sure if these are ALL the possible premises.

  • 2
    $\begingroup$ Isn't $A \setminus B = \emptyset$ simply equivalent to $A \subseteq B$, which includes the other two? This is because $A \setminus B = \{ x \in A : x \notin B \}$, so if this is empty, it means that for all $x \in A$ one has $x \in B$. $\endgroup$ – Andreas Caranti Sep 23 '18 at 17:32
  • $\begingroup$ Nvm it is. My bad. $\endgroup$ – Hai Sep 23 '18 at 17:35

$A\subseteq B$ is equivalent to $A\setminus B = \emptyset$, let's try to prove it.

Being more formal, $A\subseteq B$ means $\forall x \in A$, $x\in B$.

You can deduce from this that the empty set is a subset of any set. The sentence $\forall x \in \emptyset$, $x\in B$ is true since there are no elements $x\in\emptyset$ to begin with. This means that $A=\emptyset$ is just a case of $A\subseteq B$. It's also true to see that $A=B$ is just a case of $A\subseteq B$. All in all this means that

$$ A\subseteq B\qquad \text{OR} \qquad A=\emptyset\quad \text{OR} \quad A = B$$ is equivalent to

$$A \subseteq B.$$

Now, you should be able to prove that $$A\subseteq B\quad \Leftrightarrow \quad A\setminus B=\emptyset.$$

As for the premises, well all that imply $A\subseteq B$ are true. So $A=\emptyset$ and $A=B$ are correct.


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