Suppose we have three sets $A$, $B$, and $C$ that we know are infinite sets, but we do not know anything else about the cardinality of $A$, $B$, and $C$.
Is $|A|=|B|$ and $|A|=|C|$ $\Longleftrightarrow |A|=|B\cup{}C|$?
In other words, if we can create a one-to-one correspondence between set $A$ and set $B$ and also can create a different one-to-one correspondence between set $A$ and set $C$, then do we also know that we can create a one-to-one correspondence between $A$ and the union of $B$ and $C$ (and the converse)?
I'm trying to prove that a certain set can be shown to have the same cardinality as the union of two others. It is easy to show the mapping directly for $A$ to $B$ and $A$ to $C$ but difficult to show it directly for $A$ to $B\cup{}C$. That's why I want to know.
The left to right direction certainly works assuming AC. Without AC it seems likely to have issues though I'm not an expert. It's worth noting that depending on what the sets look like if you are able to in some way actually construct the two bijections between $A$ and $B$ and $C$, you might be able to prove $|A|=|B\cup C|$ even without AC. For example if $A=\mathbb{N}$ then even just knowing that the two bijections exist allows you to prove a bijection exists between $A$ and $B\cup C$. That's mainly because instantiation doesn't need choice and $\mathbb{N}$ has enough structure too allow you to define the new function on odds and evens. This would even work for larger ordinals in place of $A$. Being even and odd still makes sense there.
But the right to left direction fails even under AC.
For an easy counterexample just take $A=\mathbb{R}$, $B=\mathbb{R}$ and $C=\mathbb{N}$, then you get $|A|=|B\cup C|$ but obviously $|\mathbb{R}|\neq |\mathbb{N}|$.