Theorem: If $\mathfrak a$ and $\mathfrak p$ are cardinals satisfying $2\mathfrak p = \mathfrak p$ and $\mathfrak a + \mathfrak p=2^\mathfrak p$, then $\mathfrak a \ge 2^\mathfrak p$.
Here's a proof:
Proof: Let $P$ and $P'$ be disjoint sets of cardinality $\mathfrak p$, and let $A$ be a set with cardinality $\mathfrak a$ disjoint from $P$. Then $|A\cup P|=\mathfrak a + \mathfrak p = 2^\mathfrak p=2^{\mathfrak p + \mathfrak p}=|2^{P \cup P'}| $. ($2^{P \cup P'}$ stands for/denotes the power set of the set $P \cup P'.$ )
Let $f$ be a one-to-one mapping of $A \cup P$ onto $2^{P \cup P'}.$ For any subset $E$ of $P',$ let $E'$ denote the set containing $E$ and all the elements $x$ of $P$ that don't belong to the set $f(x)$. Then $E'\subset P\cup P'$, and for all $x$ in $P$, $x \in E'$ iff $x \notin f(x)$ - in short, for all $x$ in $P$, $E'\not= f(x). $ It follows that $E'=f(y)$ for some $y$ in $A$. Now $E$ is any one of the $2^\mathfrak p$ subsets of $P'$ and the correspondence between $E$ and $E'$ is one-to-one therefore there are $2^\mathfrak p$ sets of the form $E'$, hence $2^\mathfrak p$ corresponding elements $y$ in $A$. Consequently, $A$ has at least $2^\mathfrak p$ elements and so $\mathfrak a \ge 2^\mathfrak p$.
My problems start in the second paragraph.
How can $E' \subset P\cup P'$ be true? $E'$ contains $E$ which isn't an element of $P\cup P'$, right? $E',$ therefore, can't be the image of any $y$ in $A$ because the images of all the elements of $A$ have to be subsets of $P\cup P'.$ Everything else later in the proof seems fine because for each set $E$, one can construct an $E'$ matching it and so if there are $2^\mathfrak p$ subsets $E$ of $P'$, there are the same number of sets $E'$ and the result would follow but all of this depends on the two ideas I pointed out that to me seem false. The proof isn't correct, yes? Can you help make it right?
