Can we prove that there exists at least one chain $A$ in P(X), where X is a non-empty set (finite or infinite), s.t. $ |X|&lt;|A|\leq |P(X)|$?   If you can't solve it, ideas/possible directions are welcome :)  The two links mentioned above do not contain answers.
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$P(\mathbb N)$ = power set of $\mathbb N$. $A \subset P(\mathbb N)$ is a chain if $a,b \in A \implies$ either $a \subseteq b$ or $ b \subseteq a$ That is, we have something like this: $$\cdots a \subseteq b \subseteq c \subseteq\cdots$$ where $a,b,c \in A$ are distinct. We can show easy enough that there is an uncountable chain - this is done by noting $\mathbb N\sim\mathbb Q$ then using Dedekind cuts in $\mathbb Q$ to define $\mathbb R$ we see that a family of (nearly arbitrary) cuts satisfy the condition. For instance the family $L_r=\{q \in \mathbb Q : q &lt; r \}$ for $r &gt; 0$ gives us the sets we need and obviously we can pick others. I tried doing this for $ \mathbb R$ and don't seem to be getting anywhere.  To be more specific, does there exist a chain in $P(\mathbb R)$ with cardinality $2^ \mathbb R$? Given a set of cardinality $X$ (necessarily non-finite), is there a chain in $P(X)$ of the same cardinalty of $P(X)$?
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The problem stated as follow:     Suppose $X_1 \leq X_2 \leq \cdots$ and $X_n \xrightarrow[]{p} X$. Show that $X_n \to X$ a.s.   I'm think about may be use the continuity of probability measure, but I don't know if that's correct.
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I was wondering whether anyone would be so kind as to briefly check my proof? I am supposed to prove the statement without using any theorems which would render the proof trivial.     If $\displaystyle\int_a^b f$ exists then $f$ has infinitely many points of continuity in $[a,b]$   For the sake of contradiction suppose $f$ has finitely many. It suffices to show that there is an interval $[u,v]\subset [a,b]$ over which $f$ is not Riemann integrable. Pick any $[u,b]\subset [a,b]$ such that $f$ is discontinuous everywhere in $[u,v]$. For any finite partition $D=\{x_1\cdots x_n\}$ of $[u,v]$ let:  $$s(f,D)=\sum_i (x_{i+1}-x_i)\inf_{x\in[x_{i},x_{i+1}]}f\quad\text{and}\quad S(f,D)=\sum_i (x_{i+1}-x_i)\sup_{x\in[x_{i},x_{i+1}]}f$$ We prove $\sup_D s(f,D) &lt;\inf_D S(f,D)\;(1)$. To achieve this consider an arbitrary chain: $$D_0\subset D_1\subset \cdots$$ Pick any $I_0\subset I_1\subset \cdots$ where $I_k=[u_k,v_k]$ and $I_k$ is an interval of the partition $D_k$. Without loss of generality $\bigcap_i I_i=z$. Given that $f$ is discontinuous everywhere, there exists $\epsilon&gt;0$ such that for any $\delta&gt;0$ there exists $y$ such that $|z-y|&lt;\delta$ yet $|f(z)-f(y)|\geq \epsilon$.   This immediately implies $\displaystyle\sup_{I_n} f&gt;\inf_{I_n} f$ and hence $\displaystyle\sup_{D_n} s(f,D)&lt;\inf_{D_n} S(f,D)$.   This is valid for any sequence $D_n$ so claim $(1)$ holds, contradicting the integrability of $f$.
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I am curious about whether $|x|^p$ with $0&lt;p&lt;1$ satisfy $|x+y|^p\leq|x|^p+|y|^p$ for $x,y\in\mathbb{R}$.   So far my trials show that this seems to be right... So can anybody confirm whether this is right or wrong and give a proof or counterexample?
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Can someone help me? I can't even understand what the question is asking for.     Let $S = \{x_1,x_2,x_3,\ldots,x_n\}$. Show that ${\sim}[\forall x\in S,P(x)\lor Q(x)]\iff \exists x\in S,[{\sim}P(x)\land {\sim}Q(x)]$
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Problem statement:  Let A be an uncountable set, B a countable subset of A, and C the complement of B in A.  Prove that there exists a one-to-one correspondence between A and C.  My thoughts:  There's a bijection between A and A (the identity function).  There's a bijection between C and C (the identity function).  There's a bijection between B and $\mathbb{N}$.  That's all I know.
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I think that the question is clear from the title:     Are there sets $A$ and $B$ such that $A \in B$ and $B \in A$?   My feeling is that the definition of any two such sets will be circular (we can't define $B$ till we define $A$ and we can't define $A$ till we define $B$), and hence will be excluded by some axiom.  On the other hand, I can't see an inherent contradiction, though I wouldn't be surprised if some form of Russell's paradox pops up, pushed behind one more set of braces.
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This is a follow-up question to the discussion about the finite axiom of choice .  Suppose we have a countable collection of non-empty sets $\{A_1, A_2, A_3,\cdots\}$ Reasoning as indicated in that discussion, we can prove the existence of a choice function $c_n$ such that $c_n(i)$ belongs to $A_i$ for $i = 1,\cdots,n$  Again reasoning as suggested and using induction, we can prove the existence of a function $C$ defined over $\{1, 2, \cdots\}$ such that $C(n)$ is a choice function defined for $i=1,\cdots,n$ with $C(n)(i)$ an element of A_i. We can require in addition that $C(n)(i)$ and $C(n-1)(i)$ are equal for $i=1,\cdots,n-1$.  Now it would it seem that we can prove the countable axiom of choice by taking the union of $C(1)$, $C(2)$, $\cdots$  The question is as follows: why is that proof wrong? I suspect the answer may have something to do with the axiom of union.
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I am having trouble getting started with this particular problem.  Let $A$ be a nonempty set, and let $\mathcal{B}$ be the set of all functions $f:A\to\{0,1\}$. Show that $\mathcal{B}\sim\mathcal{P}(A)$; that is, construct a bijection $\phi: \mathcal{B} \rightarrow \mathcal{P}(A)$.   Faced with this problem, I know that the power set contains all subsets of $A$. Now, does that mean all possible combinations of the functions which map to $\{0,1\}$? Also, does the fact that the characteristic function is associated with the power set have anything to do with the problem? Thanks.
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Does there exist a set $x$ such that $\mathcal P(x)\in x$, where $\mathcal P(x)$ denotes the power set of $x$?  In Zermelo-Fraenkel set theory, the axiom of regularity implies that for all sets $x$ and $y$, we cannot have $x\in y\in x$. Since $x\in\mathcal P(x)$, we cannot have $\mathcal P(x)\in x$, which gives an elementary proof.  However, many authors do not include the axiom of regularity in Zermelo-Fraenkly theory, mainly because almost all mathematical results can be proved without it, and the point of this axiom is basically a matter of convenience.  Therefore, I would like a proof which does not use the axiom of regularity either that we cannot have $\mathcal P(x)\in x$, or that this question is undecidable without this axiom.  In order to prove that this question is undecidable, I tried to show that the existence of a set $x$ such that $\mathcal P(x)\in x$ amounts to the existence of a set $a$ such that $a\in a$, or a set $b$ such that $b=\{b\}$, which (correct me if I'm wrong) are known to be two undecidable questions without the axiom of regularity.
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Claim: If $(X,\tau)$ is not $T_1$ then $A'$ is not necessarily closed for any $A$.  Proof: Here is an example. Take $X=\mathbb{N}$ and let $\tau=\lbrace u\subset\mathbb{N}\mid 1\in u\iff 2\in u\rbrace$. Let $A=\lbrace 2,3 \rbrace$. Since any open set containing $1$ also contains $2$ it follows that $1\in A'$. Hence $\overline{A}=\lbrace1,2,3\rbrace$, as it is the smallest closed set containing $A$. But  $X\setminus A'=\mathbb{N}\setminus\lbrace 1\rbrace$ is not open because it contains $2$ and not $1$. The set of accumulation points $A'$ is not closed.  So, in general, trying to prove that $A'$ is closed in a topological space is an effort in futility?
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Is there any way to solve this problem using the diagonalization method? I know there's the proof that uses $\mathcal{P}(\mathbb{N})$ = set of all finite subsets $\cup$ set of all infinite subsets, but I'm trying to solve it using the diagonalization method specifically. I was thinking that we can assume the two have the same cardinality and arrive at a contradiction, but I don't know how to get there. How can we show there has to be some infinite set in the set of all infinite subsets that cannot match 1-1 with a natural number?
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I cannot prove this statement so need help. This problem is one of exercises right after the chapter about Hausdorff's maximal principle and Zorn's Lemma. Thus, you cannot use the concept of cardinal number. I found the exact same question on this website, but the proof is done by using the concept of cardinal number. Please make sure that this question is at right after introducing the axiom of choice and Zorn's lemma.  Let $A$ be any set with more than one element. Prove that there exists a bijective function $f:A \to A$ such that $f(x)\neq x$ for every element $x$ in $A$.
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Here the problem 1.4.26 from Sohrab, Basic Real Analysis  Duplicate  This are not duplicate of this question:    the answer gives no proof  despite the title is not really connected  only finite set   Problem  Show that, if $A$ is an infinite set, then $|A\times\mathbb{N}|=|A|$. Hint: Let $\mathcal{F}$ denote the set of all bijective maps $f:S\times\mathbb{N}\to S$, where $S\subset A$. Since $|\mathbb{N}\times\mathbb{N}|=|\mathbb{N}|$, we have $\mathcal{F}\neq \varnothing$ (Why?) Show that Zorn's Lemma can be applied in $\mathcal{F}$ to produce a maximal bijection $h:B\times\mathbb{N}\to B$, with $B\subset A$, and that we must have $B=A$, by examining the cases where $S\setminus B$ is finite or infinite.  Proof  Let $\mathcal{F}$ denote the set of all bijective maps $f:S\times\mathbb{N}\to S$, where $S\subset A$. Since $|\mathbb{N}\times\mathbb{N}|=|\mathbb{N}|$ (see exercise 1.4.10(i)), we have $\mathcal{F}\neq \varnothing$ (From Exercise 1.4.22(d) we know that for a infinite set $A$, we have $|A|\geq\aleph_0$, and $|\mathbb{N}|=\aleph_0$, thus there is an injection from $g:\mathbb{N}\to A$, and, by defining $S:=g(\mathbb{N})\subset A$, we get a bijective map (composition of two bijective maps) $S\times\mathbb{N}\to\mathbb{N}\times\mathbb{N}\to\mathbb{N}\in\mathcal{F}$).  If we consider the inclusion as partial order, then every chain $S_i\subset A$, $i\in I$, has $A$ as upper bound and applying Zorn's Lemma, we get the existence of a maximal $\bar{S}$ and hence a bijection $h:\bar{S}\times\mathbb{N}\to\bar{S}$, with $\bar{S}\subset A$.  Now we prove that $\bar{S}=A$. $R:=A\setminus \bar{S}$. $\bar{S}$ is infinite, since the example of element of $\mathcal{F}$, that we gave above is bijective to $\mathbb{N}$. We have to do this. The second is trying to follow the hint and is incomplete:   Consider that $A\neq\bar{S}$, then there is $r\in R:=A\setminus\bar{S}$. We chose an element $s\in\bar{S}$ and define the bijection $h':\bar{S}\cup\{r\}\times\mathbb{N}\to \bar{S}\cup\{r\}$ as follow  $h'(a,n)=\begin{cases}h(s,2n-1)&amp;\text{for }a=s\\ r&amp;\text{for }a=r,n=1\\ h(s,2n-2)&amp;\text{for }a=r,n\neq 1\\ h(a,n)&amp;\text{otherwise}\end{cases} $  Since $\bar{S}\subsetneq \bar{S}\cup\{r\}$ we get a contradiction to the maximality of $\bar{S}$. Now either $R$ is finite or infinite. 1) $R$ finite, let say $R=\{r_1,\cdots,r_n\}$. Then chose $n$ elements $s_1,\cdots,s_n\in\bar{S}$ and we define the bijection $h':A\times\mathbb{N}\to A$ as follow $h'(a,n)=\begin{cases}h(a,2n-1)&amp;\text{for }a\in\{s_1,\cdots,s_n\}\\ a&amp;\text{for }n=1,a\in\{r_1,\cdots,r_n\}\\ h(s_i,2n-2)&amp;\text{for }a=r_i\\ h(a,n)&amp;\text{otherwise}\end{cases}$  In case $n&gt;0$ we get a contradiction with $\bar{S}$ being the maximal element.   Thus we have only to exclude the case 2) $R$ infinite. HOW TO DO THAT?   Question  Above I tried two proofs. The first part is the same. The two second parts are given by the two bullets.  The second proof tries to follow the hint.  My questions are:   - Is my proof correct?   - In which direction should go the proof from the hint?
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Possible Duplicate:          Let $X$ and $Y$ be sets, and suppose that $|\mathscr{P}(X)| = |\mathscr{P}(Y)|$ (where $\mathscr{P}$ denotes the power set).  Does it follow that $|X|=|Y|$?  Remark: It's obviously true for finite sets, as $2^m=2^n$ implies $m=n$.
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I'm doing exercises. In the related book, there is a claim. Is this right? I'm not sure.     For a sequence $\{a_n\}$, there exists a limit $a$ such that $\lim_{n\to\infty} a_n=a$ if and only if for any $p\in \Bbb N$, $\lim_{n\to\infty} |a_{n+p}-a_n|=0$   If not, could you kindly give some counterexamples?  Thanks.
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Is the finite Cartesian Product of Countable sets countable?  If possible, could you give a bijection that would work for any example?  Any help would be appreciated.