I need to prove the following inequlity:$$ |P([0,1]) \backslash P((0,1))| &gt; |P(N)| $$ Now I have the intuition that in the left side of the inequlity there is all the subsets of [0,1] that contain the elements 0 or 1 and that is in the same cardinality is P[0,1] witch is greater then the right hand side (as Cantor proved) but I dont know how to write it down. Thank you
121387
Is it true or false that $\lvert P([0, 1]) \setminus P((0, 1)) \rvert &gt; \lvert P(\mathbb{N}) \rvert$ ?  I need a hint because I don't see a way to solve this
57507
I'm fixing an arbitrary $x$ in $a&lt;x&lt;b$ and I must prove that $x$ is the limit of a subsequence of $(x_n)$. I have a theorem that says that $x$ is a subsequential limit of a sequence $(x_n)$ iff every interval $(x-\epsilon, x+\epsilon)$ containts infinitely many terms of $(x_n)$. I also know that $a$ and $b$ are the smallest and largest subsequential limits of $(x_n)$.  I don't know where to go from here.
22623
I'd like to have my proof verified and if possible, to see other solutions that are interesting.  Proof: Suppose $n(n+1)$ is a square. Then we write $$n(n+1) = \prod_{p} p^{c(p)}$$ where $c(p) = a(p) + b(p)$ are such that \begin{align*} n &amp;= \prod_p p^{a(p)} \\ n+1 &amp;= \prod_p p^{b(p)} \end{align*} Now, by our hypothesis, $c(p)$ is even for all primes $p$. As $(n,n+1)=1$ for all $n$, it must be that $a(p)$ and $b(p)$ are even for all primes $p$ and moreover, $a(p) = 0$ whenever $b(p)&gt;0$ and reversely.  This indicates that both $n$ and $n+1$ are squares. This is impossible as there are no consecutive squares in the natural numbers.
59723
Let $x_n \in[0,1)$ s.t $x_n - \sqrt{n}\in \mathbb{Z}$. Prove that $X=\{x_n | n\in \mathbb{N}\}$ is dense in $[0,1]$.  It it suffices to prove that $\overline{X}=[0,1].$ I was trying to use the fact that $p+q\sqrt{2}$ $(p,q \in \mathbb{N})$ is dense in $\mathbb{R}$. So I can choose $n=2q^2$ and $x_n=p+q\sqrt{2}$ close enought of $1$. Hence, $x_n=p+q\sqrt{2} - q\sqrt{2} = p \in \mathbb{Z}$. I am pretty sure that this argument is invalid but I feel that the idea can be useful somehow.
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I was given the following question: Let $q$ be a real number such that $0 &lt; q &lt; 1$, and let ${a_n}$be a sequence that follows the rule : $|a_{n+1}-a_n| &lt; q|a_n-a_{n-1}|$ for any $n&gt;1$. Prove that ${a_n}$ converges. I already know this can be proven using Cauchy's criterion, but I had a different proof in mind (using Cantor's lemma), and wanted to know if it's valid, or if I might be mistaken. So here it is: First - I prove that the sequence is bounded. Since the sequence is bounded, the set of its members (and the set of any sub sequence of ${a_n}$) has a Supremum and an Infimum. Then, for any natural number $k$, I define the closed interval $A_{k}$, such that $A_{n}=[\inf(a_k), \sup(a_k)]$, where inf and sup of $k$ are the Infimum and Supremum of the set of the members of $a_n$, starting from $a_k$ (meaning, any member with an index greater than or equal to $k$). Naturally, $A_1$ is the &quot;biggest&quot; interval, and because of how the sequence is defined, $A_{n+1}$ is always a proper subset of $A_n$ for any $n$. Now I use Cantor's lemma - because the interval gets infinitely smaller and smaller (after n iterations, it gets at least $q^n$ times smaller), the intersection of all intervals $A$ (meaning, $A$ of $n=\infty$) contains one single point, let's call it $L$. Now, I state that $L$ must be the limit of $a_n$ - for any epsilon, there exists a natural number $k$, such that for any $n&gt;k$, $A_k$ is contained in the epsilon neighborhood of$ L$, and therefore all members of $a_n$ starting from $a_k$ are in the epsilon neighborhood of $L$, which means $L$ must be the limit. So do you think this is a valid proof? Or am I making some false assumptions? I know this proof is a bit longer than using Cauchy's criterion, but I like it better because I feel like it gives a better explanation as to 'why' the series converges. Thanks in advance!
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If $A, B, C$ are sets prove that $(A\setminus B)\setminus C\subseteq A\setminus(B\setminus C)$. Find a description of when it is that we have equality, and give an example where the inclusion is strict.   I'm not sure how to do this one I know that to prove that two sets $X$ and $Y$ are equal, we need to show that if $x$ is any element of $X$, then $x\in Y$, and that if $x$ is any element of $Y$, then $x\in X$: the first of these means that $X\subseteq Y$ and the second that $Y\subseteq X$, so the two together establish that $X=Y$. I'm not sure how to use this for a question and how to finish it.
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Here is a proof of the sequence lemma I lifted from this . It seems to me naively that the axiom of choice is being used to selected the $x_n$'s. Is the axiom of choice being invoked in this argument? Is it necessary?  The top answer in this  says that we can choose some element from each set in a finite collection of non-empty subsets without choice, though I'm not sure how to do that rigorously. Could some inductive argument be used in the case above?
101815
Prove that:   $$p \,\,\left|\, {p \choose k} \right., \quad 0&lt; k \lt p$$   if $p$ is prime.   how to prove that with direct proof?
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I tried to write a formal proof for the theorem: $A$ subset of $\mathbb R$ well ordered by the normal order $\implies A$ is at most of cardinality $\aleph_0$.  Any suggestions?  Thanks.
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Proving that $[0,1)$ and $(0,1)$ have the same cardinality (without assuming any previous knowledge) can be done easily using Cantor-Bernstein theorem.  However I'm wondering if someone can build an explicit bijection between these sets.  It's easy to build a bijection between $(0,1)$ and $\mathbb R$, so a bijection from $[0,1)$ to $\mathbb R$ will also fit the bill.
4210
Given any positive integer $n$, is there a way to quickly construct a statement $S$, such that the shortest proof of $S$, if it exists, must have length at least $n$?  And such that one out of $S$ or not $S$, is provable.    And question 2: If S, then a proof of S must have length atleast n. If not S, then a proof of not S must have length atleast n.  Edit:  I had in mind that the number of symbols in S be of the same order as the number of symbols required to specify n.
15892
I have to prove that  $$\forall x \in \mathbb{R},\exists !\,n \in \mathbb{Z} \text{ s.t. }n \leq x &lt; n+1\;.$$  where $\exists !\,n $ means there exists a unique (exactly one) $n$. I'm done with proving that there are at least one integers for the solution. I couldn't prove the &quot;uniqueness&quot; of the solution, and so I looked up the internet, and here's what I found:  Let $\hspace{2mm}n,m \in \mathbb{Z} \text{ s.t. }n \leq x &lt; n+1$ and $m \leq x &lt; m+1$. Since $n \leq x \text{ and } -(m+1) &lt; -x$, by adding both, $n + (-m-1) &lt; (x-x) = 0$. And (some steps here) likewise, $(x-x) &lt; n+m+1$.  Now, can I add up inequalities like that, even when the book is about real analysis (and in which assumptions are supposed to be really minimal)? Or should I also prove those addition of inequalities? Thank you :D
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The question says: $S$ is a set of cardinality $c$. Given two distinct elements $x, y ∈ S$, prove that there exist two disjoint subsets $S_1$ and $S_2$ of $S$ and each of cardinality $c$ such that $x ∈ S_1$ and $y ∈ S_2$.  Here is what I am done so far:  Since we know that $|[0,1]|=|\mathbb{R}|=c$. Let $S=[0,1]$, then pick $x=1/2, y=7/10$. Define $S_1=[0,1/2]$ and $S_2=[3/5,1]$, where $x∈S_1$, $ S_1 \subset S$ and  $y∈S_2$, $ S_2 \subset S$.  Since $|[0,1/2]|=c$ and $|[3/5,1]|=c$; therefore, we can conclude that there exist two disjoint subsets $S_1$ and $S_2$ of $S$ and each of cardinality $c$ such that $x ∈ S_1$ and $y ∈ S_2$.  Can someone give me some suggestions on my proof? Thanks!
22520
I came to point where I suppose for case 1 that $A\subseteq B$ and conclusion is trivial. For case 2 I suppose that $A\not\subseteq B$ and try to prove $B\subseteq A$, but that gets me nowhere. Any pointers here are most welcome.
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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.
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There are several proofs I have seen of this, but they all seem to use choice subtely at some point. Is there any way to prove this without choice, or is it possibly unproveable?
61762
Can anyone please help me in this question:     $(X,\leq )$ is a well-ordered set. $\forall x\in X$, either $x$ is the largest element of $X$ or there exists $y\in X$ such that   (i) $x&lt;y$ and   (ii) if $x_0\in X$ such that  $x\leq x_0\leq y$, then either $x = x_0$ or $x_0 = y$.   Will this be sufficient to start the proof:  If $x$ is the largest element then there is nothing to prove. Suppose $x$ is not the largest element. $$A=\{ z\in X \mid x&lt;z\}\neq \emptyset$$ How do I proceed?