Possible to construct a disjoint sequence of infinite subsets of $\mathbb{N}$ whose union is not $\mathbb{N}$?   I am thinking this must be possible.  I've considered trying to define each sequence as the range of some function. I have to find a way to hop over natural numbers in some way to leave infinitely many left for each subsequent subset of $\mathbb{N}$. Thinking about just leaving out $1$ so then the union isn't $\mathbb{N}$. I haven't tried to do anything like this before. Any possible hints on how this can be done and I can think a little more to see if I can come up with something? If not I may just ask for an answer heh.  Thanks much for your time and input I appreciate it very much.
122045
The title says it. I thought of the following: we want $$\Bbb N = \dot {\bigcup_{n \geq 1} }A_n$$ We pick multiples of primes. I'll add $1$ in the first subset. For each set, we take multiples of some prime, that hasn't appeared in any other set before. Then $$\begin{align} A_1 &amp;= \{1, 2, 4, 6, 8, \cdots \} \\ A_2 &amp;= \{3, 9, 15, 21, 27, \cdots \} \\ A_3 &amp;= \{5, 25, 35, 55, \cdots \} \\ A_4 &amp;= \{7, 49, 77, \cdots \} \\ &amp;\vdots \end{align} $$  I'm heavily using the fact that there are infinite primes. I think these sets will do the job. Can someone check if this is really ok? Also, it would be nice to know how I could express my idea better, instead of that hand-waving. Alternate solutions are also welcome. Thank you!  Edit: the subsets must be also infinite.
53757
Define a sequence $(a_n)_{n = 1}^{\infty}$ of real numbers by $a_n = \sin(n)$. This sequence is bounded (by $\pm1$), and so by the , there exists a convergent subsequence.  My question is this: Is there any way to construct an explicit convergent subsequence?  Naïvely, I tried considering the subsequence $$\sin(3), \,\sin(31), \,\sin(314), \,\sin(3141),\, \sin(31415),\, \dots$$ hoping it would converge to $0$, but I was unable to prove this (and I think it's probably not even true).
44326
I am trying to prove that $\mathbb{R}$ with the lower limit topology is not second-countable.  To do this, I'm trying to form an uncountable union $A$ of disjoint, half-open intervals of the form $[a, b)$, $a &lt; b$. Is this possible? I think this would imply the $A$ is open but no countable union of basis elements could coincide with $A$ therefore making the real numbers with the lower limit topology not second-countable.  I think there must exist something like $A$ described above but I am having trouble visualizing it and coming up with a formula to represent it.  Maybe there is some other way to show it is not second-countable.
79181
I would like to find a way to construct uncountably many non-isomorphic linear (total) orderings of natural numbers (as stated in the title).  I've already constructed two non-isomorphic total orderings. They are $$ 1 &lt; 2 &lt; 3 &lt;\ldots\\ 1 &gt; 2 &gt; 3 &gt;\ldots $$ where "$&lt;$" and "$&gt;$" are  strict total ordering relations.  Thanks in advance for your help.
99650
$(\to)$ Assume that $S \subseteq R$ is bounded. If $(x_n)$ is a sequence in $S$ , then $(x_n)$ is bounded and thus it has a convergent subsequence.  $(\leftarrow)$ Assume that $S$ is not bounded. Then for any integer $n$ there is an $x_n$ in $S$ such that $|x_n| &gt; n$. Since any convergent sequence is bounded, the sequence $(x_n)$ cannot have a convergent subsequence.  Can someone check to see whether my proof is missing something.  thank you.
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.
32976
The set of all strictly increasing sequences  ($a_n$) of natural numbers has cardinality $\mathbb{N}$, $\mathcal{P}(\mathbb{N})$ or $\mathcal{P}( \mathcal{P}(\mathbb{N}))$?  I answered $\mathcal{P}(\mathcal{P}(\mathbb{N}))$ because as the set can be described by $X=\{\{\ldots, a_n, \ldots \},\{\ldots,b_n,\ldots\}, \ldots\}$ as it is not known if there are any limited sequences (so, assuming all have infinite elements), the cardinality of $X$ could not be the first option $\mathbb{N}$ because considering any function $\phi : \mathbb{N} \rightarrow X$ there would always be elements not listed from $X$. So I thought it would possible to say that $|X|\ge | \mathcal{P}(\mathbb{N})|$ and thinking that trying to build any bijection $\varphi: \mathcal{P}(\mathbb{N}) \rightarrow X$ would also not be possible as there would be elements from $X$ not listed too. So than I got $|X| = |\mathcal{P}( \mathcal{P}(\mathbb{N}))|$, but I would like to know if I answered correct and, if not, why?
30985
I have to prove that $\ell^{\infty}(\mathbb N)$ is not separable.  My attempt  Consider a SUBSET $V$ of $\ell^{\infty}(\mathbb N)$ consisting of bounded sequences that have only $0$, $1$ entries, e.g. $(0,1,1,0,0,0,0,1,0,0,\dots)$  Now assume that this SUBSET is uncountable.  If we take a radius $r=1/4$ then balls with origins that are elements of $V$ would be disjoint and because any base of  $\ell^{\infty}(\mathbb N)$ must contain a subset of each element of the set of these balls, base of  $\ell^{\infty}(\mathbb N)$ can't be countable so it doesnt satisfy the second axiom fo countability and thus is not separable (since  $\ell^{\infty}(\mathbb N)$ is a metric space).  Could someone check this? And I still need to prove that $V$ is uncountable...
112407
Let $A,B\subseteq\mathbb{R}$ two non-empty sets such that $a&lt;b$ for all $a\in A$ and $b\in B$.   Which of the following properties hold:  (1) $\sup A&lt;\infty$  (2) $\sup B&lt;\infty$  (3) $\sup A&lt;\inf B$  (4) $\sup A=\inf B$ iff there exist sequences $(a_n)_{\mathbb{N}}$ in $A$ and $(b_n)_{\mathbb{N}}$ in $B$ such that $\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}y_n$  (5) $\sup A&lt;\inf B$ iff there exist sequences $(a_n)_{\mathbb{N}}$ in $A$ and $(b_n)_{\mathbb{N}}$ in $B$ such that $\lim\limits_{n\to\infty}x_n&lt;\lim\limits_{n\to\infty}y_n$  My ideas:  (2) is false:  Consider $B=\{x+y: x,y\in \mathbb{R}_{&gt;0}\}$ and $A=\{0\}.$ Then $A$ and $B$ satisfy $a&lt;b$ for all $a\in A$ and $b\in B$, but $\sup B$ does not exist in $\mathbb{R}$.  (3) My attempt: Fix $b\in B$, by assumption it is $a&lt;b$ for all $a\in A$. Thus, $b$ is an upper bound for $A$, therefore $\sup A&lt;b$. Or follows $\sup A\le b$ (I'm not sure at this point)? Since $\sup A$ is a lower bound for $B$, $\sup A&lt;\inf B$...or is $\sup A=\inf B$ as well possible?  (1) is correct: $\sup A\le \inf B$ and $\inf B$ exists in $\mathbb{R}$. (Suppose $\inf B$ doesn't exist in $\mathbb{R}$. $\inf B=-\infty$ only can happen if $B$ is not bounded below which is a contradiction to $B$ is bounded below by a non-empty set $A$.)   (4) Suppose $\sup A=\sup B$. Take $(a_n)_{\mathbb{N}}$ in $A$ converging to $\sup A$ and $(b_n)_{\mathbb{N}}$ in $B$ converging to $\sup B$, for the existence see here  . Then: $\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}y_n$. However, I don't know about the other direction.  (5) I don't know..  Is everything what I have tried correct? How does it work for (3) and the remaining directions in (4) and (5)? Thank you
74643
I have the following question:  $\mathbb R^3\setminus \mathbb Q^3$ is a union of disjoint lines. This exercise is in the book: Set theory for the working mathematician of Krzysztof Ciesielski.  someone can give me a help on how to proceed?
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 $g$.
89132
Is it (and if yes how?) possible to define an an binary operation $*$ for an arbitrary set $M$ such that $(M,*)$ is a group?  If $M$ is finite or countable infinite this is trivial, but is it also possible for any other sets? And if yes how can you construct $*$? If no can you provide a counterexample?
12018
Let $U$ be an open set in $\mathbb R$. Then $U$ is a countable union of disjoint intervals.    This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.
73143
Assuming the axiom of choice, show that there exists a sequence of function $\langle f_\alpha\mid\alpha&lt;\omega_1\rangle$ such that: (1) each $f_\alpha:\alpha\rightarrow\omega$ is injective and (2) for all $\alpha&lt;\beta$, the set $\{\gamma&lt;\alpha\mid f_\alpha(\gamma)\not=f_\beta(\gamma)\}$ is finite.   My knee-jerk reaction in trying to solve this problem is to define the sequence by recursion. But I don't see how to actually carry out the construction (in particular, I don't see how I could ensure that (2) remains to be the case once I pass $\omega$). Any help is appreciated.
76540
Given a sequence of sets, is there some well-defined notion of a limit of a set?   In other words, given some universe set $U$, I am wondering if there is a topology on $2^U$ (the powerset of $U$) such that the usual intersection and the union limits converge in that topology.  As an explicit example, let $U=\mathbb{N}$,  $S_n = \{x\in \mathbb{N} | n&lt; x \le 2n \}$,   $T_n = \{n\}$.  The limit of both sequences above should be the empty set by the following argument:  \begin{align} S_n &amp;\subset (n,\infty) \\\\ \lim_{n\to\infty} S_n &amp;\subset \lim_{n\to\infty} (n,\infty) = \cap_{n\in\mathbb{N}} (n,\infty) = \emptyset  \end{align}  (I'm not sure how to justify passing a set inclusion to the limit.)
23431
Is the following true? Let $x_n$ be a sequence with the following property: Every subsequence of $x_n$ has a further subsequence which converges to $x$. Then the sequence $x_n$ converges to $x$.  I guess that it is true but I am not sure how to prove this.
4127
Can someone give me an example if a not recursively enumerable set $A \subseteq \mathbb{N}$ ?   I came up with this question, when trying to show, that there exist partial functions $f: \mathbb{N} \rightsquigarrow \mathbb{N}$ that are constant for some value (when they halt), but not recursive:  Since I know that a set is recursively enumerable iff it is the domain of a partial recursive function, to find an example of such a set, I would have to show that no matter which partial function whose domain is $A$ I would take, it wouldn't be recursive. I couldn't show the above, although it seems intuitively to be true to me. But since I could not show it and the existance of such a counterexample is even stronger (because I would have to show that a whole class of those partial functions aren't recursive) it made me wonder, wether such a set can exist.