I'm trying to find a solution for exercise G1 of Chapter VII in Kunen's &quot;Set Theory - An Introduction to Independence Proofs&quot;. The question is in the context of Cohen forcing. We start with a c.t.m. $M$ for a sufficiently large segment of ZFC. In the ground model we fix three infinite cardinals $\kappa,\lambda,\vartheta$. The notion of forcing is $P := \mathrm{Fn}(\kappa\times\omega, 2)$. One now has to prove that $$(\lambda^\vartheta)^{M[G]} = ((\max\{\lambda, \kappa\})^\vartheta)^M$$ for any generic filter $G\subseteq P$. I'm guessing that the argument uses the fact that $P$ has c.c.c. and since $|P| = \kappa$ there are at most $\kappa^{\aleph_0}$ many antichains in $P$. However, I do not see how to start the argument. Would someone be so kind as to give me a hint?
123818
Let $\kappa, \lambda,\theta$ be infinite cardinals of $M$ where $M$ is a transitive model of ZFC. Let $\mathcal{P}=Fn(\kappa\times \omega,2)$ (finite functions from $\kappa\times \omega$ to 2) and G is a (M,P)-generic set. I want to show $(\lambda^\theta)^{M[G]}=(max(\kappa,\lambda)^\theta)^M$. One direction is easy $(\lambda^\theta)^M \leq (\lambda^\theta)^{M[G]}$ and $(\kappa^\theta)^M\leq (\kappa^\theta)^{M[G]}\leq ((2^\omega)^\theta)^{M[G]}=(2^\theta)^{M[G]}\leq (\lambda^\theta)^{M[G]}$ where $(\kappa \leq 2^\omega)^{M[G]}$ is by the forcing construction. But now I am not quite sure how to get the other direction. Any thoughts?
115074
I am reading through some course notes to prepare for an upcoming Real Analysis course I have to take in the fall, and found two definitions of the Axiom of Choice.    [AC] If $\Lambda \neq \emptyset$ and for each $\lambda \in \Lambda$, $X_{\lambda}$ is a non-empty set, then $\prod_{\lambda \in \Lambda} X_{\lambda} \neq \emptyset$. [ACD] Suppose that $\Lambda \neq \emptyset$ and that    $\forall \ \lambda \in \Lambda$, $X_{\lambda}$ is a non-empty set, and $X_{\lambda} \cap X_{\beta} = \emptyset$ if $\lambda \neq \beta \in \Lambda$   Then $\prod_{\lambda \in \Lambda}X_{\lambda} \neq \emptyset$.     It wasn't immediately apparent to me that they were equivalent, so I started trying to prove it.  I think that it is trivial to show that $1 \implies 2$, more or less just by saying that if every non-empty set has a choice function then every pairwise non-empty set will also have a choice function.   But I am stuck on trying to prove that $ 2 \implies 1 $ ... I can't seem to construct any meaningful way to use the disjoint version to prove the more general version.   Any tips?
117133
This follows  answer of .     Let $I$ an infinite set and $\{E_i\}_{i\in I}$ a family of finite   sets.      It is said that there exists an injection      $$\bigcup_{i\in I} \ E_i \to I.$$   In the comments, an argument with "$\aleph$ numbers" (which I haven't studied yet) is given.  I wondered how one could build this injection explicitely ? I have been trying for some time now, but it is quite hard to have intuition about that.
93710
Statement: There are uncountably many surjections from $\mathbb N$ to $\mathbb N$.  Is the following proof valid:  The real numbers are uncountable.   There exists a bijection between the set of real numbers and the set of all functions from $\mathbb N$ to $\{0,1\}$.   Given a function, $f$, from $\mathbb N$ to $\{0,1\}$, construct a surjection from $\mathbb N$ to $\mathbb N$, denoted $g$, as follows:  We map the $n$th composite number to $n$, and the nth prime, to $f(n)$.   Clearly this is a surjection since the composite numbers are countable. Clearly different functions $f$ lead to different surjections $g$. Since there are uncountably many $f$s, there are uncountably many surjections $g$.
122333
Hi I have to proof that for $\Omega$ a nested family of odd subsets of $V$ with $V$ finite, we have that $|\Omega|\leq\frac{3}{2}|V|$. I can prove that for certain constructions of $\Omega$, it is satisfies $|\Omega|\leq \frac{3}{2} |V|$, the problem I encounter is that I have to proof that my construction of $\Omega$ is optimal. (Since I have to prove that this holds for every family of odd subsets) Can anyone give me a hint on how to proof that every $\Omega$ satisfies this equation.    $\Omega$ is nested if for all $A,B\in\Omega$ either $A\subseteq B$, $B\subseteq A$, or $A\cap B=\varnothing$; a set is odd if its cardinality is odd.
78946
I'm trying to show that club filter of $\kappa$ is $\kappa$-complete for uncountable regular cardinal $\kappa$:  Let $\kappa$ be uncountable regular cardinal, let $C(\kappa)$ be the club filter generated by $\kappa$.  To show that $C(\kappa)$ is $\kappa$-complete it is enough to show that for every sequence $\langle \alpha_i\mid i&lt;\gamma\rangle$ with $\gamma&lt;\kappa$ and $\alpha_i$ club of $\kappa$ the set $\alpha=\bigcap_{i&lt;\gamma}\alpha_i$ is a club of $\kappa$.  Showing that $\alpha$ is close is easy:     If $\beta_n$ is a sequence in $\alpha$ then for every $i$ we that $\beta_n$ is a sequence in $\alpha_i$, because $\alpha_i$ is a club it is close hence $\lim \beta_n$ is in $\alpha_i$ for every $i$ so it is also in $\alpha$   But I have a problem with unboundedness, my guess will be for each $\lambda&lt;\kappa$ to find a sequence for each $\alpha_i$ such that all of the sequences converge to some $\mu&gt;\lambda$, that way $\beta&lt;\mu\in\alpha$. I think I'll have to use the fact $\kappa$ is regular, but I don't know how to proceed and prove my idea.  So, is my idea correct, and if yes how can I proceed?
24304
I have a few questions about Godel's pairing function and proving that c = c * c for aleph cardinals.  Mostly, though, I'm concerned that most of the proofs I've seen are erroneous, and this concerns me. (And I've searched many times for a good proof.)  So we order the class On x On lexicographically, and this is a proper well-ordering, so that On x On is order-isomorphic to On by a unique mapping P: On x On -> On.  This makes perfect sense. But in every "proof" of c = c * c that I've seen, there's been some glaring mistake.   For example, one "proof" relies on the "proposition" that "If $\gamma$ = max($\alpha$,$\beta$) and at least one of $\alpha$,$\beta$ is infinite, then P($\alpha$,$\beta$) &lt;= card($\gamma$) x card($\gamma$)."  But it seems to me this is false, and here's why: so it seems clear that (0,$\omega$) is the first limit ordinal in On x On under the lexicographical ordering, and of course $\omega$ is the first limit ordinal in On--consequently, we must have P(0,$\omega$) = $\omega$; but then P(1,$\omega$) = $\omega$ + 1, which contradicts the "proposition" (since P(1,ω) = ω + 1 is not &lt;= $\omega$, which is card($\gamma$) x card($\gamma$), as $\gamma$ = $\omega$). Am I wrong here?  I guess what I'm asking for is some kind of real explanation as to how Godel's pairing function is used to prove that c = c * c for all aleph cardinals c. Even if this just involves pointing me toward a decent source, that would be awesome.
44151
I'd like some help making this argument complete and rigorous (if it's correct - if not, help with that would be nice).  Here $k$ is a field.     Let $A_1,\ldots,A_n \subseteq k$ be infinite subsets. Then any polynomial in $k[x_1,\ldots,x_n]$ that vanishes on $A_1\times\cdots\times A_n\subseteq k^n$ must be $0$ (as a polynomial).   This is what I have ...  For the case $n=1$, a non-constant polynomial can only have as many roots as its degree, and in particular, it must have a finite number of roots. The only polynomial in one variable that has an infinite number of roots is $0$, so if a polynomial in $k[x_1,\ldots,x_n]$ vanishes on an infinite subset then it must be $0$.  For the inductive step, suppose the proposition is true for less than $n$ subsets and variables. Let $p\in k[x_1,\ldots,x_n]$ vanish on $A_1\times\cdots\times A_n$. Fix $x_n$ as some $a\in A_n$, and we have a polynomial in $n-1$ variables that vanishes on the set $A_1\times\cdots\times A_{n-1}$, so by the inductive hypothesis it must be identically $0$.  (Now it gets sketchy). Since this is true for any of the infinite values in $A_n$, and , $p$ must be $0$.
106511
Is there any example that for cardinal numbers $\kappa &lt; \lambda$, we have $2^\kappa = 2^\lambda$?  My guess is that it only depends on whether GCH holds. Is it true?
16420
My attempt to prove the following is below:     Let X be an infinite set. Show that $\mathscr{T}_1=\{U \subseteq X : U = \emptyset $ or $ X\setminus U $ is finite $ \}$   My book calls this set the ""  I just want to know if this proof is correct. (I intuitively understand why its a topology, but I'm not sure if I'm showing it properly)  Using the definition of topology:     $\mathscr{T}$ is a topology on X if and only if the following are true:   (i) X and Ø are elements of $\mathscr{T}$.   (ii) $\mathscr{T}$ is closed under finite intersections.   (iii) $\mathscr{T}$ is closed under arbitrary unions.   Let:$V_k \subseteq X$ be any finite set. (ie: $V_1, V_2...$ are all finite sets)  So Then, $U_k = X \setminus V_k$, would be the subsets of $\mathscr{T}$.   Now to verify the definition.     (i) X and Ø are elements of $\mathscr{T}$.   Ø is given in the definition of $\mathscr{T}_1$. Also, the empty set is considered finite, with cardinality zero. Thus $X$ is in $\mathscr{T}_1$ because its complement is finite.      (ii) $\mathscr{T}$ is closed under finite intersections.   $U_k \cap U_j = X \setminus (V_k \cup V_j) $, due to De Morgan's law. Since the Union of two finite sets is also finite, this set is still in $\mathscr{T}_1$. Or Symbolically (using bars || for cardinality): If $|V_k \cup V_j| &lt; |\mathbb{N}|  $, then $(U_k \cap U_j) \in \mathscr{T}_1$. This reasoning can be extended to any finite amount of intersections.       (iii) $\mathscr{T}$ is closed under arbitrary unions.   $U_k \cup U_j = X \setminus (V_k \cap V_j) $, due to De Morgan's law.Since the Intersection of two finite sets is also finite, this set is still in $\mathscr{T}$. Or Symbolically: If $|V_k \cap V_j| &lt; |\mathbb{N}|  $, then $(U_k \cup U_j) \in \mathscr{T}_1$. This reasoning can be extended to any finite amount of Unions.  Now let's consider the infinite case: $V_k \cap ... \cap V_j$ will only contain points that are in EVERY set being intersected. So if $V_k \cap ... $ was infinite, that would imply the every $V_k$ being intersected is also infinite, but $V_k$ is specifically defined as being finite. Thus even with infinite Unions, $\mathscr{T}_1$ is still closed.
32315
I'm reading the proof of lemma 15.43 of Jech's Set Theory:     Let $G$ be generic on a complete Boolean algebra $B$. If $M$ is a model of $\mathsf{ZFC}$ such that $V\subset M\subset V[G]$, then there exists a complete subalgebra $D\subset B$ such that $M=V[D\cap G]$.   In the proof of this theorem the author states the following fact: "First we note that since $M$ satisfies the Axiom of Choice, there is for every $X\in M$ a set of ordinals $A_X\in M$ such that $X\in V[A_X].$"  My question is, why is this fact true?  Thanks
43745
The following is a question I'm interested in solving:    The idea is to solve it without using AC. As @Asaf Karagila pointed out in the comment section:  "Even if $\kappa$ is an infinite cardinal, you can't assume that $\kappa \geq \omega$ without using AC".  I am still trying to think this through: My reasoning being, that as $\kappa &gt; n \forall n \in \omega$, then $\kappa \geq \sup\{n:n\in \omega\}=\omega$ as $\omega$ is a limit ordinal.  Where does my reasoning fail? And also, a detailed solution would be very appreciated.
15842
If $\lambda\leq\kappa$ are infinite cardinals, how many subsets of $\kappa$ of size $\lambda$ are there? And of size $\leq\lambda$? Is there some sort of explicite formula for this? The internet isn't as helpful on this as one would expect (unless one knows what to look for, I guess).
115713
What are the consistency strengths of $$ZF+``\text{The club filter on $\omega_1$ is an ultrafilter}"$$ and $$ZF + DC + ``\text{The club filter on $\omega_1$ is an ultrafilter}"?$$  I know that the latter (stronger) statement follows from AD, and has consistency strength much weaker than AD (I think a measurable is enough), but I don't know what the precise strength of either statement is. In particular, I suspect that the latter has strictly greater consistency strength than the former, but I don't know if this is the case.    EDIT: of course, if the club filter is an ultrafilter, then it is countably complete (since it is countably complete as a filter), so $\omega_1$ is measurable; so it may appear that the answer to this question is trivially "a measurable." However, this isn't quite right: it is not obvious (in fact, I suspect it's not true) that $ZF+$"there is a countably complete ultrafilter on some cardinal $\kappa$" and $ZFC+$"there is a countably complete ultrafilter on some cardinal $\kappa$" are equiconsistent.  FURTHER EDIT: As Asaf shows below, the club filter need not be countably closed in $ZF$, even though the intersection of countably many clubs is always club!
24105
Let $ A $ be a ring.  Let $ I $ be a preordered set, filtering.  Let $ \Sigma $ a multiplicative subset of $ A $.  Suppose for any given $ i \in I $ a multiplicative subset $ S_i $ of $ A $ contained in $ \Sigma $.  Make the following assumptions:   $ \Sigma = \displaystyle \bigcup_i S_i $ For $ (i, j) \in I^2 $ with $ i \leq j $, elements $ S_i $ become invertible in $ S_ {j}^{-1} A $.   $ \mathcal{D} = ((S_{i}^{-1} A)_{i \in I} \ , \ (S_{i}^{-1} A \to S_{j}^{-1} A)_{i \leq j}) $ is a commutative diagram in the category of filter $ A $ - algebras.  Question :  Show that $ \displaystyle \lim_{\longrightarrow} \mathcal{D} \simeq \Sigma^{-1} A $.  Thank you in advance for your help.
21836
Let $M_n(\mathbb{F})$ be the set of all $n\times n$ with entries in $\mathbb{F}$ and let $\exp:M_n(\mathbb{C})\to M_n(\mathbb{C})$ be defined by  $$ \exp(A)=\sum_{k=0}^{\infty}\frac{A^k}{k!},$$ for all $A\in M_n(\mathbb{C}).$  I want to prove that $\exp$ is a surjective map from $M_n(\mathbb{C})$ to $GL(n,\mathbb{C})=\left\{A\in M_n(\mathbb{C})\,\middle| \det(A)\neq0\right\}$, how do I go about that?   I mean saying that $\exp:M_n(\mathbb{R})\to GL(n,\mathbb{R})$ is an analogous to saying $\exp:\mathbb{R}\to \mathbb{R}_{&gt;0}$ and this is also pretty intuitive, since, in analogy with the case of numbers, $A^0=I\;\forall A$, so $\exp(0)=I+0+\frac{0^2}{2!}+\dots=I$, so even for $A=0$ we get $\det\left(\exp(A)\right)\neq0$ and so because of the first term we can never get a zero determinant. But I have no idea how to prove the subjectiveness.  Thanks in advance.
95639
I understand how we can show the existence of a choice function for any (finite or infinite) collection of (finite or infinite) subsets of, say, $\mathbb{N}$ or $\mathbb{Z}$ without using the axiom of choice, by showing that a well-ordering over the union of sets in the collection exists.  What I don't see though is how to do the same for what appears to be a much simpler case: a (possibly infinite) collection of finite sets, where nothing else is said about the sets other than that each of them is finite.  I would need to show a well-ordering exists over the union of those sets, but since they were not explicitly defined to be subsets of $\mathbb{N}$ or $\mathbb{Z}$, I first need to show the following, for example:  (i) there exists a surjection from $\mathbb{N}$ to $\bigcup X_i$ for an arbitrary collection of finite sets $X_i$  Correct so far?   Alternatively, I'm on the wrong track, and this is precisely a case that (unintuitively for me then) requires application of the axiom.  If so, I suppose I don't see why (i) wouldn't hold, when it seems that any definition within ZF of a finite number of elements is equivalent (in some sense to be made precise) to a finite subset of $\mathbb{N}$, so, given that the union of a collection of finite subsets of $\mathbb{N}$ is a subset of $\mathbb{N}$, the required well-ordering exists.