Union of sets without axiom of choice Are there models of ZF (without choice) with one of the following properties?
Question 1
The cardinality $|A|+|B|$ of two sets is defined as the cardinality of the union of two disjount sets $A'$ and $B'$ with $|A|=|A'|$ and $B=|B'|$. Is it possible, that the cardinality of the union of $A$ and $B$ is (strictly) smaller than that of the union of $A'$ and $B'$?
Question 2
Can there be a set $A$, so that the union of $A$ with a singleton has greater cardinality than $A$ itself? This is true for a cardinal which has a countable infinite subset, but I did't find an general answer.
 A: As Arturo points out, both Q1 and Q2 have an easy answer of "yes"; just take $A=B$ to be any finite set. If you mean to restrict to infinite sets, then the answer is still yes, but it requires a little more thought. For question 1, let $A,B$ be any Dedekind-finite sets such that $A\cap B$ has size one, say $A\cap B=\{c\}$. Then $A\cup B$ is also Dedekind-finite. (Why?) Now, let $A'$ and $B'$ be disjoint sets as in your question, with bijections $g:A\to A'$ and $h: B\to B'$. Then the map $f:A\cup B\to A'\cup B'$ given by $f(a)=g(a)$ for $a\in A$ and $f(b)=h(b)$ for $b\in B\setminus\{c\}$ is an injection, so $|A\cup B|\leqslant |A'\cup B'|$. But $\operatorname{im} f$ is a proper subset of
$A'\cup B'$. Hence, if $f':A'\cup B'\to A\cup B$ were any bijection, then $f'\circ f$ would be an injection from $A\cup B$ to a proper subset of itself, a contradiction. So $|A\cup B|\neq|A'\cup B'|$, as desired.
For question 2, let $A$ be any Dedekind-finite set. Let $c$ be any set not an element of $A$; then the inclusion $i:A\hookrightarrow A\cup\{c\}$ is an injection, so $|A|\leqslant |A\cup\{c\}|$. But, if $g:A\cup\{c\}\to A$ were any bijection, then $g\circ i$ would be an injection from $A$ onto a proper subset of itself, a contradiction. So $|A|\neq |A\cup\{c\}|$ and the desired result follows.
