Definitions of sets

Question

How would I define the set $\Omega_2(x)\ =\ \{4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 33, 34, 35, 38, 39\dots\}$, where $\Omega_2$ is the set of semi primes (ie - numbers with $2$ not necessarily distinct prime factors)? I would like to define $\Omega_3$, etc. in a similar way.

I was thinking something along the lines of $\Omega_2(x)\ :=\{\mathbb{P}_p\cdot \mathbb{P}_q\ \rm{s.t}\ p=q\ \vee p \neq q \} $

Update

$\Omega_2 \in \mathbb{N}\ \text{s.t.}\ \Omega_2:=\{\mathbb{P}_p\cdot \mathbb{P}_q\ \rm{s.t.}\ p=q\ \vee p \neq q \}\text{ where } p \wedge q \in \mathbb{N}$

I assume $\mathbb{P}$ is the set of primes. What does the $_p$ do? — Karolis Juodelė, Dec 27 '13 at 18:58
Good point! Would it be ok to omit it, or would I need to say that $\Omega_2 \in \mathbb{N}$ ... — martin, Dec 27 '13 at 18:58
You seem lost. $p$ is an integer, so what does $p \wedge q$ mean? You wanted $p, q \in \mathbb{N}$. Also, $\mathbb{THAT}$ font is usually used for special sets, so using it for numbers is odd. You can write $\Omega_2$ in therms of $n$th prime, ${ p_i \cdot p_j \mid i, j \in \mathbb{N} }$. — Karolis Juodelė, Dec 27 '13 at 19:10

score 5 · Accepted Answer · edited Dec 27 '13 at 19:23

5

$$\Omega_k = \underbrace {\mathbb{P}\cdot\mathbb{P}\cdots\mathbb{P}}_{\text{k times}}$$

Here $\mathbb{P}$ is a set of primes and $A \cdot B = \{a \cdot b \mid a \in A, b \in B \}$.

edited Dec 27 '13 at 19:23

Cameron Buie

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answered Dec 27 '13 at 19:05

Karolis Juodelė

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I don't quite follow, but it looks impressive! would this suffice as a definition of all sets of k-almost primes then? – martin Dec 27 '13 at 19:22
+1: I have altered the formatting slightly. Hopefully, it is to your liking. – Cameron Buie Dec 27 '13 at 19:24
I don't really understand where $A,B,a$ & $b$ come in. Not that I am questioning it, just that I am curious as to how they complete the definition. – martin Dec 27 '13 at 19:25
What are $A$ & $B$ here? – martin Dec 27 '13 at 19:26
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@martin: $A$ and $B$ are intended to be arbitrary sets of numbers. Karolis is simply defining the notation $A\cdot B$ in this context. – Cameron Buie Dec 27 '13 at 19:59

score 3 · Answer 2 · answered Dec 27 '13 at 19:09

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What's wrong with the definition, "$\Omega_k$ is the set of all products of exactly $k$ not-necessarily-distinct primes"?

answered Dec 27 '13 at 19:09

Gerry Myerson

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score 1 · Answer 3 · edited Apr 13 '17 at 12:21

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Personally, I like Gerry Myerson's suggestion, which is brief, clear, unambiguous, and to the point. But if you really want a pile of notation, you might say:

$$\begin{align}\Omega_0 & = \{1\} \\ \Omega_{i+1} & = \{ n \mid \exists a\in\Omega_i: \exists p\in\Bbb P: n = ap\} \end{align}$$

(I have marked this suggestion community wiki because I don't think it is a good idea.)

edited Apr 13 '17 at 12:21

Community

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answered Dec 27 '13 at 19:29

MJD

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${ n \mid \exists a\in\Omega_i: \exists p\in\Bbb P: n = ap}$ is a longer way to write ${ap | a \in \Omega_i, p \in \Bbb P}$. – Karolis Juodelė Dec 27 '13 at 20:03
I would have put it the other way around: ${ap\mid p\in\Omega_i, p\in \Bbb P}$ is an abbreviation for the set that is normally written ${ n \mid \exists a\in\Omega_i: \exists p\in\Bbb P: n = ap}$. See my discussion of this abbreviation here. – MJD Dec 27 '13 at 21:06

Definitions of sets

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