How can I show there's exist Bijection between Q x Q and natural numbers (Q × Q ~ N)
112443
I am new to functions and relations, and with some concepts I am not so familiar. I have a question in an homework:     Show that $\mathbb{Q} \times\mathbb{Q}$ is denumerable.   From what I understood, denumerable means that it is infinitively countable. There are some  on the web about this topic, but I am still not understanding their explanation, maybe because they are not explained so easily (for me).  From what I understood, a set is denumerable if there's a relation with the natural numbers (but I am still not understanding what is this relation).  I have heard also about equinumerous sets (contain a bijective function = onto + one-to-one function), but I cannot relate this information with the problem to try to solve it.
68581
I have to show that set of all $x$ such that $x \in P$ and $x \notin Q$ exists. So, I think I have to assume that sets $P$ and $Q$ already exist. So, with that, I can use  . This axiom is $$ \forall w_1,\cdots, w_n \, \forall A \,\exists B \,\forall x \left( x \in B \Longleftrightarrow [ x \in A \wedge \varphi(x,w_1, \cdots , w_n, A)] \right) $$ So, I will let $\varphi(x, Q, P) = x \in P \text{ and } x \notin Q $. Also, I will let $A = P$ and  $w_1 = Q $. So using universal instantiation, I get the following $$ \exists B \,\forall x \left( x \in B \Longleftrightarrow [ x \in P \wedge \varphi(x, Q, P)] \right) $$ And using, existential instantiation, there exists a set $B$ such that $$ \forall x \left( x \in B \Longleftrightarrow [ x \in P \wedge \varphi(x, Q, P)] \right) $$ Using the definition of $\varphi(x, Q, P)$, this can be simplified as following $$ \forall x \left( x \in B \Longleftrightarrow [ x \in P \wedge x \in P \wedge x \notin Q] \right) $$ $$ \forall x \left( x \in B \Longleftrightarrow [ x \in P \wedge x \notin Q] \right)$$ $$ \forall x \left( x \in B \Longleftrightarrow x \in \{ x \,| x \in P \text{ and } x \notin Q \} \right)$$ Now, using , it will follow that $$ B = \{ x \,| x \in P \text{ and } x \notin Q \}  $$ And, since $B$ exists, this means that $\{ x \,| x \in P \text{ and } x \notin Q \} $ also exists. Is the proof good ?
82640
I was wondering if it is possible to produce an explicit bijection $h\colon \mathbb{R} \rightarrow \mathbb{R}/\mathbb{Q}$. If we can produce an explicit injection $i\colon \mathbb{R} \rightarrow \mathbb{R}/\mathbb{Q}$, can the  be used constructively?  It is clear that the two sets have the same cardinality, so the existence of such a bijection is trivial. What I am really looking for is a nice-looking bijection, or a proof that no such nice-looking bijection exists, for a definition of "nice-looking" which I cannot quite figure out.  One problem which I think makes finding such a bijection difficult is that any natural injection of $\mathbb{R}/\mathbb{Q}$ (ie, those injections in which one representative is chosen from each coset) produces a non-measurable set, specifically a .  I hate to ask such a vague question, but I'm really not sure about whether the correct answer is constructive, or whether it is a proof that any such bijection is in some sense "very complicated."  As a final note, the motivation for this question came from , in which I was somewhat astonished to see such a clear, constructive bijection given between $\mathbb{R}$ and $\mathbb{R} \setminus S$, where $S$ is countable.
23040
I'm looking for a 'simple' bijection $$ \pi \colon \mathbb N_0 \times \mathbb N_0 \to \mathbb N_0, $$  where 'simple' in this context means that it should be as easy as possible to define and it should be self-evident that said function is in fact a bijection. It's a quest for convenience - not minimal complexity in any rigorous sense.  Here are a couple of examples I came up with - none of which satisfy both requirements:   Let $\pi(m,n) = c(2^m3^{n+1})$, where $c \colon \{2^m3^{n+1} \mid m,n \in \mathbb N_0 \}\to \mathbb N_0$ is the order isomorphism (or in fact Mostowski collapse) of its domain under the natural ordering, Let $\pi(m,n) = \langle n,m \rangle$ - Gödel's pairing function, Let $\pi(m,n) = m \oplus n$ - the number resulting from '' $m,n$ (where we imagine both $m$ and $n$ as in infinite decks of cards whose $i$th element is labeled with its respective $i$th digit). ...
50097
I have $A$ as an infinite set and $S$ as a countably infinite set, (so that means there exists a one-to-one correspondence between $S$ and $\mathbb{N}$).  How do I show that there always exists a bijection between $A$ and $A\cup S$? Can it be done by showing that there is a bijection from $A$ to $S$ or from $A$ to itself? I am lost on this one.   Oh, and can it be possible that there is no bijection if it is between $S$ and $A\cup S$? What about a map that maps $\mathbb{N}\to \mathbb{N}\cup\mathbb{R}$?
12771
Recall that a countable set $S$ implies that there exists a bijection $\mathbb{N}\to S$. Now, I consider (0,1). I want to prove by contradiction that $(0,1)$ is not countable.  First, I assume the contrary that there exists a bijection $f$, and I can find an element in $S$, but not in the range of $f$. But I can't find such element. How can you construct such $f$?
118550
What is the number of injective maps from a set of cardinality $m$ into a set of cardinality $n$ $(m \leq n)$?
5828
If $N$ is the set of all natural numbers, $R$ is a relation on $N \times N$, defined by $(a,b) \simeq (c,d)$ iff $ad=bc$, how can I prove that $R$ is an equivalence relation ?
118176
Looking at the top answer on this page,    A user said,      There are models of ZF such that $A,B$ are sets for which exists   surjections from $A$ onto $B$ and vice versa, however there is no   bijection between the sets.   Can anyone give an example of this?  If A is the set of real numbers and B is the set of natural numbers , then there will be a surjection from A to B, but no surjection from B to A.  Is this what surjection even means.  Here is a surjection from real to naturals  1.5 → 1  1.7 → 1  2.3 → 2  2.6 → 2  f(A) = int(A) → B  But there is no surjection from the naturals to the reals (Cantor's diagonal proof)  So, how is it even possible to have "There exists surjections from $A$ onto $B$ and vice versa, and there is no bijection between the sets"? Did he, maybe, mean there is no way to find the formula for the bijection?
14753
I'm examining a proof I have read that claims to show that the Cartesian product $\mathbb{N} \times \mathbb{N}$ is countable, and as part of this proof, I am looking to show that the given map is surjective (indeed bijective), but I'm afraid that I can't see why this is the case. I wonder whether you might be able to point me in the right direction?  Indeed, the proof begins like this:  "For each $n \in \mathbb{N}$, let $k_n, l_n$ be such that $n = 2^{k_n - 1} \left(2l_n - 1 \right)$; that is, $k_n - 1$ is the power of $2$ in the prime factorisation of $n$, and $2 l_n - 1$ is the (necessarily odd) number $\frac{n}{2^{k_n - 1}}$."  It then states that $n \mapsto \left(k_n , l_n \right)$ is a surjection from $\mathbb{N}$ to $\mathbb{N} \times \mathbb{N}$, and so ends the proof.  I can intuitively see why this should be a bijection, I think, but I'm not sure how to make these feelings rigorous?   I suppose I'd say that the map is surjective since given any $\left(k_n , l_n \right) \in \mathbb{N} \times \mathbb{N}$ we can simply take $n$ indeed to be equal to $2^{k_n - 1} \left(2l_n - 1 \right)$ and note that $k_n - 1 \geq 0$ and thus $2^{k_n - 1}$  is both greater or equal to one so is a natural number (making the obvious inductive argument, noting that multiplication on $\mathbb{N}$ is closed), and similarly that $2 l_n - 1 \geq 2\cdot 1 - 1 = 1$ and is also a natural number, and thus the product of these two, $n$ must also be a natural number. Is it just as simple as this?  I suppose my gut feeling in the proving that the map is injective would be to assume that $2^{k_n - 1} \left(2 l_n - 1 \right) = 2^{k_m - 1} \left(2 l_m - 1 \right)$ and then use the Fundamental Theorem of Arithmetic to conclude that $n = m$. Is this going along the right lines? The 'implicit' definition of the mapping has me a little confused about the approach.    On a related, but separate note, I am indeed aware that if $K$ and $L$ are any countable sets, then so is $K \times L$, so trivially, taking the identity mapping we see trivially that this map is bijective and therefore that $\mathbb{N}$ is certainly countable (!), and thus so is $\mathbb{N} \times \mathbb{N}$. Hence, it's not really the statement that I'm interested in, but rather the exciting excursion into number theory that the above alternative proof provides.
26846
Is there a proof for the logical equivalence of $a \rightarrow b$ and $\lnot a \vee b$?
80640
$Z$ is the set of non-negative integers including $0$. Show that $Z \times Z \times Z$ is countable by constructing the actual bijection $f: Z\times Z\times Z \to \mathbb{N}$ ($\mathbb{N}$ is the set of all natural numbers). There is no need to prove that it is a bijection.   After searching for clues on how to solve this, I found $(x+y-1)(x+y-z)/z+y$ but that is only two dimensional and does not include $0$.  Any help on how to solve this?
112702
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.
5771
I need to construct a bijection from an arbitrary interval $(a,b) \to\Bbb R$. I was thinking of somehow using the tangent function because it's asymptotic, but i'm not sure how to get started.
45786
Suppose that $\;f_1,\;f_2:\;A\to B$ such that $f_1$ is injective and $f_2$ is surjective.   I was trying to find out whether there exists $f_3:\;A\to B$ such that $f_3$ is bijective. Is it possible to construct it from $f_1$ and $f_2$?  I tried proving the statement non-constructively via a cardinality argument, but ended up hand-waving a lot (i.e., since the cardinality is the same, there exists a bijection, a bit of circular reasoning). But, I'd be more interested in a construction based argument, as I can't think of one.
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Here $n$ belongs to natural numbers. Firstly, I proved the relation by putting $n = 1$ . Then, taking $$|\sin(mx)| \leq m|\sin x|$$ true, I had to prove $$|\sin(m + 1)x| \leq (m + 1)|\sin x|$$  Now, here I got stuck. How to prove it?? Please help.
103593
Assume that a,b,n are all natural numbers.  I was going to set it up as:  na = q(1)*n(b) + r(1)     where a>b  and go down the chain:  nb = q2 * r(1) + (r2)  but something seems off. Someone told me that in the first part, it should be n*r(1), but I don't understand why. May I please have some direction?