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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?

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This is a theorem of Specker, I believe. For a wonderful proof see Andres Caicedo answer here. –  Asaf Karagila Aug 18 '12 at 20:36
    
Also, can you try and give better linebreaks? I find it very hard to follow what is written in this post. Both your given proof as well the problem you are having with it. –  Asaf Karagila Aug 18 '12 at 21:33
    
Is this supposed to be in ZF, without choice? If so, you should say so. If not, this is much easier than that proof suggests. –  Chris Eagle Aug 18 '12 at 22:12
    
@Chris: Indeed. –  Asaf Karagila Aug 18 '12 at 22:15
    
@Chris Eagle: Yes, not assuming the AC. This was in fact a lemma which was needed in the proof that the GCH implies the AC. –  Mark Aug 19 '12 at 13:21

1 Answer 1

up vote 2 down vote accepted

Let $F=\{x\in P:x\not\in f(x)\}$. Then $E'=E\cup F$. $E$ is assumed to be a subset of $P'$. The set $F$ of all $x\in P$ such that $x\not\in f(x)$ is a subset of $P$. Hence $E'\subseteq P\cup P'$.

The rest of the proof seems clear to you, so I won't elaborate on that.
I believe you only misunderstood the definition of $E'$: "Let $E'$ be the set containing $E$ and ..." is supposed to mean that $E'$ is the union of $E$ and the set $F$ that I defined above. Funnily enough, this kind of ambiguity was just discussed here: How do you pronounce the inverse of the $\in$ relation? How do you say $G\ni x$?

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I don't know why I didn't think of it that way. Thanks. –  Mark Aug 19 '12 at 13:29

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