Let V be a finite set and let $\Omega \subseteq 2^{V}$ be a nested family of odd subsets of V, where subset $A\subseteq V$ is odd if |V(A)| is odd (if A has an odd amount of nodes).  Question: How do I show that $|\Omega| \leq \frac{3}{2}\cdot|V|$?  I have no clue on how I should tackle this problem, thanks in advance!
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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.
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Let $A$ be a non-empty set and $n$ be the number of elements in $A$, i.e. $n:=|A|$.  I know that the number of elements of the power set of $A$ is $2^n$, i.e.  $|\mathcal{P}(A)|=2^n$.  I came across the fact that exactly half of the elements of $\mathcal{P}(A)$ contain an odd number of elements, and half of them an even number of elements.  Can someone prove this? Or hint at a proof?
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Let $X$ be an infinite set of cardinality $|X|$, and let $S$ be the set of all finite subests of $X$. How can we show that Card($S$)$=|X|$? Can anyone help, please?
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A question from my homework I'm having trouble understanding. We are given:  $A(1) = \{\varnothing\}$, $A(n+1) = A(n)\cup (A(n)\times A(n))$  $A=A(1)\cup A(2)\cup A(3)\cup \cdots \cup A(n)\cup A(n+1) \cup \cdots$ to infinity  The questions are:  1) show that $A\times A \subseteq A$  2) Is $A \times A = A$?  Thank you for your help.  I've tried writing $A(2)$ but it gets really complicated and I'm having trouble understanding what the sets are. Let alone solve the question.
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I don't know if this question is appropriated for this site. Anyway, I'm searching for an isomorphism of order $f:K \longrightarrow \epsilon_o $, such that $(K, \leq)$ is a subset(proper or not) of $(\mathbb{Q}, \leq)$ and $\epsilon_o = \sup\{\omega, \omega^{\omega}, \omega^{\omega^{\omega}}, ... \}$. Actually,  I'm not finding neither an isomorphism between $K$ and $\omega^{\omega}$.  Thanks in advanced.
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Let $F,G,H$ be groups such that $F\trianglelefteq G \trianglelefteq H$.  I am asked whether we necessarily have $F\trianglelefteq H$. I think the answer is no but I cannot find any counterexample with usual groups. Is there a simple case where this property is not true?
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Let , $D$ be the set of tuples $(w_1,w_2,\cdots,w_{10})$ , where $w_i \in \{1,2,3\},1\le i\le10$ and $w_i+w_{i+1}$ is an even number for each $i$ with $1\le i\le 9$.   Then find the cardinality of $D$.   As , $w_i+w_{i+1}$ is an even number so either both even OR both odd. Again $w_i\in \{1,2,3\}$. So , both even possibility is exactly $1$ which is $(2,2,\cdots,2)$.  But I am unable to find the both odd possibility..
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Suppose that $f : \mathbb{C} \to \mathbb{C}$ is entire. Let $\Omega = \{ z : \frac{1}{2} &lt; |z| &lt; 2\}$. Furthermore, suppose that $ \mathcal{F} = \{f(kz) : k \in \mathbb{C}\}$ is a normal family of analytic functions defined on $\Omega$; the family $\mathcal{F}$ is normal on $\Omega$ iff each sequence from $\mathcal{F}$ has a subsequence converging uniformly on compact subsets of $\Omega$. Prove that $f$ is a polynomial.  I'm studying for my qualifying exam in complex analysis, and this question appears on an old exam. Usually, when I want to show that some entire $f$ is a polynomial, I just try to show that there exists some $N$ so that $f^{(k)}(0) = 0$, all $k \ge N$, but I'm having trouble getting that strategy to work here. I've been looking at the sequence $f(\frac{z}{n})$ trying to make something work, but have not had any success.   I would prefer to receive a hint or two, and then I'll try to post my own solution. Thanks in advance!
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Let n be an even integer. How many different ways can I choose subsets of the set {1,2...,n} if they all have to have odd size, but the intersection of any two of them has to have even size?  Please help me answer this question or lead me in the right direction. Thank you :)
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This is from Tao Exercise 19.1.3. Let $\Omega \subseteq \mathbb R^n$ measurable and $f: \Omega \rightarrow \mathbb R$ be measurable. Assume $f \geq 0$ on $\Omega$. Then there exists a sequence $\{f_n: \Omega \rightarrow \mathbb R\}_{n=1}^\infty$ s.t. all $f_n$ are simple, non-negative and increasing and $f_n \rightarrow f$ pointwise on $\Omega$. Tao gives the hint to define $$f_n(x) := \sup \left \{\frac j {2^n} : j \in \mathbb Z, \frac j {2^n} \leq \min(f(x),2^n) \right \}$$ Can someone help me to prove why all $f_n$ are measurable ?!
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I got this problem from Shiryaev's book "Probability-1". I tried to use technique involving $\pi - \lambda$ system but I failed.  Let $\mathcal{F}_1, \mathcal{F}_2$ be two $\sigma$-algebras of subsets of the set $\Omega$. I want to prove  $$d(\mathcal{F}_1,\mathcal{F}_2)=4 \sup_{A_1 \in \mathcal{F}_1,A_2\in \mathcal{F}_2} |P(A_1 \cap A_2)-P(A_1)P(A_2)|$$  satisfies inequality $0\leq d(\mathcal{F}_1,\mathcal{F}_2) \leq 1$.
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Why is the cardinality of set S is 3^n, where S is the collection of all ordered pairs of disjoint subset of {1,2,3,...,n} where n is non negative. if n = 2, then subset A and B are disjoint and can only contain elements of 1,2, or the empty set, is this correct? I can't get 9 ordered pair. Can someone explain how do I get 9 ordered pair?
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This is from exercise 3.4 of Thomas Jech's &quot;Set Theory&quot;, stated:  &quot;Show that $\Gamma(\alpha\times{\alpha})\leq{\omega^{\alpha}}$. Thus $\Gamma(\alpha\times{\alpha})=\alpha$ for all indecomposable ordinals $\alpha$&quot;.  $\Gamma$ is the Gödel pairing function on ordinals. An ordinal $\alpha$ is indecomposable if there are no $\beta&lt;\alpha$ and $\gamma&lt;\alpha$ such that $\beta+\gamma=\alpha$. The first part is trivial, since it holds for $\alpha=0$, and the function $\omega^{\alpha}$ is normal in the variable $\alpha$. I'm having trouble to show the second part, the only connection I see is the fact that $\gamma$ is indecomposable if and only if $\gamma=\omega^{\alpha}$ for some $\alpha$, which is easily proven from Cantor's normal form theorem. Any help will be appreciated.
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Could anyone provide me with a nice proof that the  $\leq$ on partitions of an integer $n$ satisfies the following: if $\lambda, \tau$ are 2 partitions of $n$, then $\lambda \leq \tau \Longleftrightarrow \tau &#39; \leq \lambda &#39;$, where $\lambda&#39;$ is the conjugate partition of $\lambda$ (i.e. the transpose of the set of 'dots' which $\lambda$ represents).   I feel like there must be a nice clever and concise/intuitive proof of this, but all I can come up with is an ugly brute force approach based on the definition of sums of the components $\lambda_i$. Could anyone suggest a particularly nice way to obtain this result? Many thanks.
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Question: Take any $37$ integers from the set {${1,2,…,112}$}. Show that there will always exist two integers $x,y$ out of those $37$ integers such that $ \mid x-y \mid \in \{9,10,19\}$. Approach: What have I done so far: Let $S=$ {${1,2,…,112}$} We partition the set into subsets $S_i$ of the form $\{x, x+9, x+19\}$. Then the difference of any two elements is in $\{9, 10, 19 \}$, which is easy to see. The following partitions are made (for $x=1, 2, \cdots , 9$): $$\{1, 10, 20 \}, \{2, 11, 21 \}, \cdots, \{9, 18, 28\}$$ Then next we choose $x=29$ to maintain the disjoint sets essential for our proof. This gives the disjoint sets $$ \{29, 38, 48\}, \cdots, \{37, 46, 56 \} $$ and similarly $$\{85, 94, 104 \}, \cdots, \{93, 102, 112 \}$$ However as the quick reader may see that the numbers $\{19, 47, 75, 103\}$ do not occur in any set. However, I decide to do a case by case analysis of the situations. Name this set as $X$. Suppose, initially, that out of the $37$ integers in our selection, none belong to $X$. Then it follows from Dirichlet's box principle that at least two elements are from the same set (i.e. from one of the subsets that we partitioned $S$ into) then the claim is valid. Now suppose that one element from $X$ is in our selection, say $19$ then if at least one of $9, 28, 29, 38$ is in our selection, then the claim is valid. Suppose none of these are present, and we are deficit in four integers. We  can now select any four integers from the remaining numbers and the box principle guarantees that at least two are from the same set. For, even if all $3$ other elements of $X$ are now present, there is one empty slot which can be filled using the other subsets from where we already have at least one element in our choice of $37$ numbers. I hope this is comprehensible because my combinatorics proof writing abilities are horrible. I am preparing for Olympiads and also for being a good Mathematician in general. Thank you for the help :) Edit: It seems I didn't enter my main question, apologies. Is my solution correct?
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I am assuming already that a) the union of countably many countable sets is countable and  b) $\omega_1$ is the least uncountable ordinal, so $x &lt; \omega_1$ if and only if $x$ is a countable ordinal.   I'm not sure if it is relevant but this question also allows The Axiom of Choice.   Thanks
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It seems to be known that the minimum size of a generating set for a finite group of order $n$ is at most $\log_2 n$. Can someone explain why this is true?  Edit: noted that the logarithm is base 2, and $n$ represents the order of the group.