relations between equinumerosity and infinite sets Given: $A\cap B$ has a one-to-one correspondence to $A$.
If $A\cup B \neq A,$ then is it true then that $B$ must then be an infinite set?
I managed to prove that A is an infinite set in the case of A∪B≠B but could not apply the same logic to the case above.
Thanks in advance
 A: Suppose we are given two sets: $A$ and $B$, such that $A\cap B$ is in one-to-one correspondence with A. 
Let $A= \{1, 2\},$ $\;B=\{1,2,3, 4\}.\;$  Then  $A\cap B = \{1, 2\}=A$ and we see that  $A$ is in one-to-one correspondence $A\cap B$.  
Then, we have $A\cup B = \{1, 2, 3, 4\} =  B \neq A$.  And $B$ is finite. 
So the implication you write at the top of your post is incorrect, since we have found $A, B$ such that the hypotheses are all true, but the conclusion (that B must therefore be infinite) is false.
A: This is not true in general: consider $A = \{ 1,2,3 \}$ and $B = \{ 1,2,3,4 \}$. 
We have that $A \cap B=A$ is in one-to-one correspondence to $A$ but $B$ is not infinite. 
What we need is that $A \cap B$ must be a proper subset of $A$ that is in one-to-one correspondence to $A$.
In this case, by definition of finite set, $A$ is not finite.
Of course, the same apply with $B$: if $A \cap B$ is a proper subset of $B$ that is equipotent with $B$, then $B$ is infinite.

We may consider the case: $A= \mathbb N$ and $B= \{n \mid n \in \mathbb N \text { and } n \text { is even } \}$.
In this case $A \cap B = B \subsetneq A$; $A \cap B$ is equipotent with both $A$ and $B$ and thus both are infinite.
But in this case: $A \cup B = A \ne B$.
