If $(a \in A) \land (b \in B) \implies (a \in B) \land (b \in C)$ is true, then is $a \in A \implies a \in B$ also true?

Question

Let A,B and C be sets.

If $(a \in A) \land (b \in B) \implies (a \in B) \land (b \in C)$ is true, then is $a \in A \implies a \in B$ also true?

Thanks!

Answer 1

Let $X = A \cup B \cup C$.
Assume that $\forall a, b, c \in X : (a \in A) \land (b \in B) \implies (a \in B) \land (b \in C)$ is true and that $A, B, C$ are not empty.

Choose any $a \in A$ and $b \in B$. By the implication, $a \in B$, thus $A \subseteq B$, which is equivalent to $\forall a \in X : a \in A \implies a \in B$.

If $B = \emptyset$, then $a \in A \implies a \in B$ is only true if $A = \emptyset$ as well.

So in conclusion, it is true if $A$ is empty or $B$ is not empty.

Answer 2

Think of it this way: $a\in A$ can be any value completely independent from any value $b\in B$. But somehow simply $B$ having any element at all will "force" $a\in B$.

So that means if $B$ is not empty then $a\in A\implies a\in B$.

But what if $B$ is empty. Then $a\in A\land b\in B$ is false and as $F\implies anything$ is true there is no reason that $a\in A$ would mean $a \in B$.

So the result is true if $B$ is non-empty, but false otherwise.

....

Oh, I guess if we know $A$ to be empty then $a\in A\implies a\in B$ is vacuously true as well. But I was going on the idea of what we can conclude with no preknowledge of the sets.

If we know $A,B, C$ are sets and we know

$(a\in A)\land (b\in B) \implies (a\in B)\land (b\in C)$ then one of the things we can conclude is:

$B\ne \emptyset \implies (a\in A\implies a\in B)$.

Or in other words $A\subset B$.

Actually, we can conclude.

$B\ne \emptyset \implies A\subset B$ and $A\ne \emptyset \land B\ne \emptyset \implies A\subset B \subset C$.