cardinality of infinite sets prove or disprove:
If two infinite sets $A$,$B$ have the same cardinality, then $A\cup B$  and $A$ have the same cardinality.
I even cannot make a judgement.
P.S: Can this be done without using cardinals? This concept has not been introduced in class yet.
 A: Since $A\cup B$ contains $A$, and we have an injection $A\cup B= A\sqcup(B\setminus A)\to A\sqcup A$, by the Cantor-Schroeder-Bernstein theorem it is enough to show $A\sqcup A$ has the same cardinality as $A$ whenever $A$ is infinite.
The idea behind the proof is that we know how to interleave two copies of $\mathbb{N}$, so write $A$ as a disjoint union of copies of $\mathbb{N}$ and interleaf each corresponding pair separately.  
In particular, form a bijection $\mathbb{N}\sqcup\mathbb{N}\to\mathbb{N}$ by surjecting each copy of $\mathbb{N}$ onto the sets of even and odd natural numbers.  For general $A$, consider all the ways of partitioning a subset of $A$ into countable subsets.  These are partially ordered by containment, i.e. partition $P$ is contained in partition $Q$ if, for any countable subset $S\subset A$ included in $P$, it is also included in $Q$.  Furthermore, if $\{P_\alpha\}$ is any ascending chain of such partitions, then $\bigcup_\alpha P_\alpha$ is also a partition of a subset of $A$ into countable subsets. So by Zorn's lemma, there is a maximal such partition $P$.
If $P$ is maximal, then $A\setminus\bigcup P$ must be finite, or else we could pull out another countable subset from $A$ and add it to $P$.  Then pick any element $S$ of $P$ and replace it with the countable set $S\cup (A\setminus\bigcup P)$; the modified partition $P'$ is therefore a partition of $A$ into disjoint countable subsets.
We can therefore write $A\cong \bigsqcup_{S\in P'}\mathbb{N}$.
Then $A\sqcup A\cong(\bigsqcup_{S\in P'}\mathbb{N})\sqcup(\bigsqcup_{S\in P'}\mathbb{N})\cong \bigsqcup_{S\in P'}(\mathbb{N}\sqcup\mathbb{N})\cong\bigsqcup_{S\in P'}\mathbb{N}\cong A$.
