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In Algebra by Serge Lang, the author asserts, to prove the existence of a field extension where an irreducible polynom has a root, that if you take one set $A$ and a cardinal $\mathcal{C}$, that you can find a set $B$ such that $\text{card}(B)=\mathcal{C}$ and such that $A \cap B=\emptyset$.

How can one build $B$ ?

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5 Answers 5

up vote 3 down vote accepted

Another approach, that avoids having to deal with ordinals or to invoke the axiom of foundation, is to note that there is a set $t$ not in $\bigcup\bigcup A$ (just on cardinality grounds: for any set $X$ there is a set $z$ not in $X$). Therefore if $B$ has size $\mathcal C$, letting $B'=\{t\}\times B$, we see that $B'$ is both of size $\mathcal C$, and disjoint from $A$. The point is that the usual set theoretic formalization of ordered pairs gives us that $(t,a)=\{\{t\},\{t,a\}\}$, so if such a pair $(t,a)$ is in $A$, then $t$ is in $\bigcup\bigcup A$.

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For any set $X$, there is a set $z$ not in $X$ because otherwise $X$ would be the set of all sets, which leads to the Russel paradox. –  Florian Apr 9 at 17:29
    
@Florian: Or Cantor's paradox (which is what I used to guarantee the existence of $t$, or $X$ in my answer). –  Asaf Karagila Apr 9 at 18:10

If you use the axiom of foundation, you can set $$ B := \left\{\{A,c\} \,\big|\, c \in \mathcal{C} \right\} \text{.} $$ $B$ obviously has cardinality $\mathcal{C}$, and if $x \in B \cap A$, then the axiom of foundation is violated because you then have an $\in$-circle $$ x \in A \in \{A,c\} = x \text{.} $$

Note that this assumes that cardinalities are sets - in particular, that $\mathcal{C}$ is a set. If your definition of cardinal doesn't make them sets, e.g. if your idea of a cardinal is the class of all sets with a given cardinality, then instead of $\mathcal{C}$, pick some set $C$ with $card(C) = \mathcal{C}$.

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There's an ordinal $\alpha$ such that all ordinals $\beta\ge\alpha$ are not in $A$. Take $\mathcal C$ ordinals starting at $\alpha$ as your set $B$.

Why does such ordinal $\alpha$ exist? Assume it doesn't, then the set $A$ contains arbitrarily big ordinals. By a transfinite recursion, we can construct a set of increasing ordinals from $A$ indexed by all ordinals. Just take an ordinal bigger than the supremum of all ordinals so far. This is a contradiction since ordinals are too numerous to form a set.

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Why is there an ordinal $\alpha$ such that all $\beta \ge \alpha$ are not in $A$ ? –  Florian Apr 9 at 16:23
    
@Florian I added details. I assume you know the ordinals aren't a set. Just take supremum of all ordinals you found so far to construct a new one. –  user2345215 Apr 9 at 16:27
    
Yes, one can build a sequence $(A_n)_{n\in\mathbb{N}}$, by recursion (why bother with limit ordinals ?) of strictly increasing ordinals. $\cup Im((A_n)_{n\in\mathbb{N}})$ is a set which is exactly the set of all ordinals (take an ordinal $\gamma$, then $\exists n\in\mathbb{N},\gamma<A_n$). And the set of all ordinals does not exist. We can then take the $\alpha$ described in the answer from user2345215, and have $B=(\mathcal{C}+A)\textbackslash A$. –  Florian Apr 9 at 16:57
    
@Florian Uhm no. Your sequence is too short. No matter how much a sequence of ordinals indexed by natural numbers (or any ordinal) increases, it's bounded. You need to do the full thing here. –  user2345215 Apr 9 at 17:06
    
@Florian Also you can add ordinals, so just set $B=\{\alpha+\gamma\mid\gamma\in\mathcal C\}$, that's what I meant in my answer. –  user2345215 Apr 9 at 17:09

Given any two sets, $A$ and $B$ we can find $A'$ and $B'$ such that $A'\cap B'=\varnothing$ and $|A|=|A'|$ and $|B|=|B'|$.

Simply take $A'=\{0\}\times A$ and $B'=\{1\}\times B$.


Since $A$ is fixed, using Cantor's theorem we know that that there exists some $X\subseteq A$ such that there are no ordered pairs of the form $\langle X,a\rangle\in A$ (for any $a$, that is). So now we can take $B'=\{X\}\times B$.

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2  
But $A$ is fixed, you cannot replace it with an equipotent $A'$. –  Andres Caicedo Apr 9 at 16:32
    
True. But now it works. –  Asaf Karagila Apr 9 at 16:35

Take the Cartesian product of $A$ with the set $\{1\}$.

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You miss-read, $B$ is supposed to have cardinality $\mathcal{C}$, which isn't necessarily the same as the cardinality of $A$. –  fgp Apr 9 at 15:16
    
There is also the requirement of a prescribed cardinality. –  Thomas Apr 9 at 15:16
    
Oops, you are right. What I ought to have typed was "of $\cal{C}$ with the set $\{a\}$ for some $a \in A$" which is what fgp got right first. –  Lee Mosher Apr 9 at 15:28

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