Let $(A_\gamma)_{\gamma&lt;\alpha}$ and $(B_\gamma)_{\gamma&lt;\alpha}$ be two sequence of sets. They satisfy $|A_\gamma|=|B_\gamma|$ for all $\gamma&lt;\alpha$, where $|\cdot|$ denotes cardinality. Also, whenever $\beta&lt;\gamma$, we have $A_\beta\subseteq A_\gamma$ and $B_\beta\subseteq B_\gamma$.   My question is: is it always true that $$ \left|\bigcup_{\gamma&lt;\alpha}A_\gamma\right|=\left|\bigcup_{\gamma&lt;\alpha}B_\gamma\right|? $$ I am struggling quite a bit when trying to prove this, but I could not find any counterexamples either. This is not given in my book, and I come up with this myself, so I really do not know where to start.
121107
Let $(X_{\alpha})_{\alpha \in A}, (Y_{\alpha})_{\alpha \in A}$ be families of sets such that for $\{X_{\alpha} \mid \alpha \in A \}$ and $\{Y_{\alpha} \mid \alpha \in A \}$ are totally ordered by inclusion. We have $(X_{\alpha},Y_{\alpha}) \neq (X_{\beta},Y_{\beta})$ for $\alpha \neq \beta$. Suppose that $|X_{\alpha}| \leq |Y_{\alpha}|$ for any $\alpha \in A$. It is true that $|\bigcup_{\alpha} X_{\alpha}| \leq |\bigcup_{\alpha} Y_{\alpha}|$?
51648
Let $\alpha ,\beta ,\delta$ be ordinals such that $\alpha &lt; \beta$, prove that there is $\delta \leq \beta$ such that $\alpha+\delta = \beta$ ? I know that if $\beta$ is cardinal then we can take $\delta=\beta$.
6443
My motivation for asking this question is that a classmate of mine asked me some kind of question that made me think of this one. I can't recall his exact question because he is kind of messy (both when talking about math and when thinking about math).   I'm kind of stuck though. I feel like the set $A^{\mathbb{N}} = \{f: \mathbb{N} \rightarrow A, f \text{ is a function} \}$ should have the same cardinality as the power set of A, if A is infinite. On the other hand, in this , it is stated that the sequences with real coefficients have the same cardinality as the reals.   It's easy to see that $A^{\mathbb{N}} \subseteq P(A)$, but (obviously) I got stuck on the other inclusion. Is there any general result that says anything else? References would be appreciated.  EDIT To clarify the intetion of this question: I want to know if there are any general results on the cardinality of $A^{\mathbb{N}}$ other that it is strictly less than that of the power set of A.   Also, I was aware that the other inclusion isn't true in general (as the post on here I linked to gave a counterexample), but thanks for pointing out why too. :)
55029
Let $A,B$ be any two sets. I really think that the statement $|A|\leq|B|$ or $|B|\leq|A|$ is true. Formally:     $$\forall A\forall B[\,|A|\leq|B| \lor\ |B|\leq|A|\,]$$   If this statement is true, what is the proof ?
59425
Let $\Gamma$ be a set of formulas and $\phi$ be a formula. Show that $\Gamma \cup \{\neg \phi\}$ is satisfiable if and only if $\Gamma\not \models \phi$.   This seemed pretty obvious but I wanted to see if my proof made sense:  Proof: $(\Rightarrow)$  To derive for a contradiction, suppose that: $\Gamma \models \phi$. That means for all truth assignments $v$, for $\gamma \in \Gamma$, if $v(\gamma) = T$, then $v(\phi) = T$.  But this contradicts our assumption that $\Gamma \cup \{\neg \phi \}$ is satisfiable by the fact that $v(\phi) = T$ cannot happen so: $\Gamma \not \models \phi$ .  $(\Leftarrow)$   So by the definition of $\Gamma \not \models \phi$, we have that there is some truth assignment $v$ which satisfies $\Gamma$ but does not satisfy $\phi$. So  that means $v(\phi) = F$ since $v(\phi) \not = T$ which implies that $v(\neg \phi) = T$ which means $v$ satisfies $\Gamma \cup \{\neg \phi\}$.    I feel like I'm missing something in the forward direction, but at the same time... It looks pretty trivial as well. Am I missing anything crucial?  Thank you!
122768
Let $2\leq \kappa\leq \gamma$ where $\gamma$ is an infinite cardinal number and $\kappa$ is any cardinal. Prove that $2^\gamma = \kappa^\gamma$.  One direction is obvious but I'm stuck with the other. i.e. showing $\kappa^\gamma\leq 2^\gamma$.  Any hints or solutions will be appreciated.
122077
I want to proof that:  $A \subseteq B  \Leftrightarrow  A \cup B = B$  At the moment I have no idea on how to start. Please give me a hint.  Thx in advance!
116030
I know how to prove this using an element and subset proof but that's not what I'm being asked to do. I'm suppose to prove it using the distributive law and the associative law but I'm just not seeing how to do this. Can you give me some advice on how to start this. I figure the first thing would be $$((A \cup B) \cap A) \cup ((A \cup B) \cap B^c))$$ But I'm not sure where to go from there.
56817
I have alot of questions.   Do Infinite sets have the same cardinality when there is a bijection between them? Are $\mathbb{N}$ and $\mathbb{Z}$ infinite sets? I assume they are, but why? Why does there not exist a bijection between them? Why do $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$ have the same cardinality? I assume there exists a bijection between them, but how can I find this function?
24278
Let $A,B$ and $U$ be sets so that $A\subseteq U$ and $B\subseteq U$. Prove that $A\subseteq B$ iff $(U\setminus B)\subseteq(U\setminus A)$.  The forward implication is easy to prove but i got stuck at the backward implication, i.e. $(U\setminus B)\subseteq(U\setminus A)\Rightarrow A\subseteq B$. Can someone help me out with it? Thank you.
106818
Full Question: Show that if $\Delta(\Gamma)$ is the deductive closure of a consistent $\Gamma$ then $\Delta(\Gamma)$ is the intersection of all maximal consistent sets of formulas $\Lambda$ with $\Gamma \subseteq \Lambda$.  I'm really stuck with this question, I honestly don't know how to even start. Can someone please provide some assistance?
16166
I'm having trouble to prove the following question:  Supose $\{A_n\}_{n\in\mathbb{N}}$ is a family of sets such that $A_1\subset A_2\subset A_3\subset\ldots$ (it's possible to have $A_n=A_{n+1}$). I need to prove that $$\lim_{n\to\infty}A_n = A_1\cup\bigcup_{n=2}^\infty(A_{n}\backslash A_{n-1}).$$  I'm not sure if this is useful, but I proved that $A_{n}\backslash A_{n-1}$ and $A_{n'}\backslash A_{n'-1}$ are disjoint if $n\neq n'$.  Thank you.
33208
Let $C$ be a chain of countable sets, i.e. $\forall S, T \in C: S \subseteq T \lor T \subseteq S$, and that every $S \in C$ is countable. Then, is $\bigcup C := \{ t \mid \exists S \in C: t \in S\}$ countable?  The answer is no, and a counter-example is $C = \omega_1$, the first uncountable cardinal/ordinal. Is there a more elementary example that doesn't involve, say, cardinals and ordinals?
15390
So first let me state my homework problem:  Let $X$ be a set, let $\{A_k\}$ be a sequence of subsets of $X$, let $B = \bigcup_{n=1}^{+\infty} \bigcap_{k=n}^{+\infty} A_k$, and let $C = \bigcap_{n=1}^{+\infty} \bigcup_{k=n}^{+\infty} A_k$. Show that (a) $\liminf_k\; {\xi_A}_{_k} = \xi_B$, and $(b)$ $\limsup_k \;{\chi_A}_{_k} = \chi_C.$  I know that, in the context I am familiar with, that $$\liminf_{k\rightarrow +\infty}\; X_k = \bigcup_{k=1}^{+\infty} \bigcap_{n=k}^{+\infty} X_n$$ and $$\limsup_{k\rightarrow +\infty}\;X_k = \bigcap_{k=1}^{+\infty} \bigcup_{n=k}^{+\infty} X_n.$$   I also know that the characteristics (indicator) function is defined as $\chi_A(x) = \begin{cases} 1, &amp; x \in A \\ 0, &amp; x \notin A .\end{cases} $  So I wrote out $B$ in some of its `glory': $B= (A_1 \cap A_2 \cap A_3 \cap \cdots) \cup (A_2 \cap A_3 \cap \cdots) \cup (A_3 \cap A_4 \cap \cdots) \cup \cdots$, and as the first argument is the smallest, with increasing size to the right, the last term in the expression for $B$ would be $B$, which would be the largest.  So if I replace the $X_k$'s above with $\chi$'s I still don't see how I can get the correct answer - though it looks pretty clear from the definition of $B$ and that of $\liminf$ being basically the same, except in this case for the $\chi$.  Any direction would be greatly appreciated. By the way, I have checked out  but I was somewhat confused by the topology, the meets/joins, etc.  Thanks much, Nate
94919
Given two sets$$A=\{1,3,5\}\quad,\quad B=\{1,3,8\}.$$ Then I compute the $A\setminus B=\{5\}.$   But my book$^\dagger$said: ... . Also, observe that $x\notin A\setminus B$ does not mean that $x\notin A\lor x\in B$. Why?  I don't know how to explain the Why, please give me some example.    $^\dagger$, second edition, page $4$.  EDIT: My intuition tells me that it does mean that. I guess there is some special case about null-set or about the restriction of the universal set.
2479
Im doing a course on measure theory and I'm stuck on one of the exercises.  Take $\{Y_{\gamma}:\gamma \in C\}$ as an arbitrary collection of random variables and $\{X_{n}: n \in N\}$ to be a countable collection of random variables  I now want to show that $$\sigma \{Y_{\gamma} : \gamma \in C\}=\sigma\{Y^{-1}_{\gamma}(B): \gamma \in C, B \in Borel\}$$  So i need to show that both $\sigma \{Y_{\gamma} : \gamma \in C\}\subset\sigma\{Y^{-1}_{\gamma}(B): \gamma \in C, B \in Borel\}$ and $\sigma\{Y^{-1}_{\gamma}(B): \gamma \in C, B \in Borel\}\subset\sigma \{Y_{\gamma} : \gamma \in C\}$ hold.  I know that for a single random variable X you have $\sigma(X)=\{X^{-1}(B): B \in Borel\}$. How do I expand this for the collection of random variables?  I then also have to show that if $Fn=\sigma\{X_{1},...,X_{n}\}$ and $A=\cup_{n=1}^{\infty}F_{n}$ then $$\sigma(A)=\sigma\{X_{n}: n \in N\}$$  Can anyone help me with this?
61050
Show that $| \mathbb N | = | \mathbb N \times \mathbb N |$, i.e., the cardinality  of $\mathbb N$ is the same as the cardinality of $\mathbb N \times \mathbb N$.   How do I show it using the Fundamental Theorem of Arithmetic and the Schröder–Bernstein Theorem?