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In this post A = {c > 1 : there exists a natural number m, such that for every n > m, there is a prime between n and cn}.

Now I don't want to discuss on the problem itself, but only on how to write A = {c > 1 : there exists a natural number m, such that for every n > m, there is a prime between n and cn} using mathematics. Now here's what I wrote:

Let $P$ be a propriety of a number $c$; $P(c)=\exists m \in \mathbb{N} : \forall n>m, n \in \mathbb{N} \rightarrow \exists p: p\ \text{is prime}, cn>p>n$. Now let $A=\{c>1 \ | \ P(c)\}$

Is this correct? If it's wrong how can I write it?

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PunkZebra
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    In your formalization, it nowhere says that $p$ should be prime. It doesn’t say that $m$ and $n$ are assumed to be natural numbers either. – k.stm Jan 04 '15 at 14:34
  • Thanks a lot I didn't notice it – PunkZebra Jan 04 '15 at 14:36
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    To say that there exists a natural number m, simply write $\exists m\in\mathbb{N}$. There is no universal notation for prime numbers (that I know of, at least). However, it is usual to either write where p is a prime number, or $p\in\mathbb{P}$. – Demosthene Jan 04 '15 at 14:39
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    In most case, you should write it out in words, not symbols. Stylistically, probably put the text in Roman type, the mathematical symbols in italic. – GEdgar Jan 04 '15 at 14:42

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Using fairly standard notation, $P(c)$ could be described as follows.

$$\exists m \in \mathbb{N}\;\; \forall n>m\;\; \exists p \in \mathbb{N} \,:\, \mathrm{isPrime}(p) \,\&\, (cn > p > n)$$

You see a lot of variants on this basic pattern; for example, some people might write

$$\exists m \in \mathbb{N}, \forall n>m, \exists p \in \mathbb{N}, \,\mathrm{isPrime}(p) \wedge (cn > p > n)$$

However, people generally use words for such things, since experience has taught us that definitions are often easier to understand when they're phrased in English, rather than "Symbolish." So following quid's suggestion, one might write:

Let $A$ be the set of all $c∈ℝ_{≥1}$ such that there exists a natural number $m$, such that for every $n > m$, there is a prime $p \in \mathbb{N}$ satisfying $cn > p > n$.

This is the "usual" way of defining $A$.

However, I would actually go one step further.

Consider that most people avoid lambda abstraction in function definitions. So rather than

Let $f = \lambda x\in \mathbb{R} : x^2.$

we would probably say the following:

Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be given as follows.

$$f(x) = x^2$$

Sure, its a bit longer, but it is also a lot easier to understand.

For the same reasons, I think set-builder notation should be avoided in most circumstances, especially definitions, because it is basically lambda abstraction in disguise. For example, $\{x \in X \mid \varphi\}$ is analogous to $\lambda x \in X,\varphi$.

Therefore, I would probably define $A$ as follows.

Let $A \subseteq \mathbb{R}_{\geq 1}$ denote the unique set such that the following are equivalent.

  • $c \in A$
  • there exists a natural number $m$, such that for every $n > m$, there is a prime $p \in \mathbb{N}$ satisfying $cn > p > n$.

This has the advantage that its clear how to actually use the definition. Suppose we know that $\varepsilon \in A$, where $\varepsilon$ is some potentially complicated expression of type $\mathbb{R}_{\geq 1}$. Then we may conclude that that big long condition holds, except with the expression $\varepsilon$ substituted for $c$. Conversely, suppose we know that that big long condition holds, for some expression $\varepsilon$ is substituted for $c$. Then we may conclude that $\varepsilon \in A$.

goblin GONE
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  • Do you have references for showing that "Prime(p)" is fairly standard? The second part is nice; though personally I would say "Let $A$ be the set of all $c \in \mathbb{R}_{\ge 1}$ such that there exists [...]" – quid Jan 04 '15 at 14:46
  • @quid, I did not mean that $\mathrm{Prime}(p)$ is in any way standard, only that the logical aspects of the notation are. However I think that a safe general principle is that if you denote your predicates letter-for-letter the same as the corresponding adjectives, people will understand. And yes, "Let $A$ be..." is another good option. I'd like to add it to the body of the answer if that is okay with you? – goblin GONE Jan 04 '15 at 14:50
  • Please add it if you find it a good addition. I am glad that you like it. I think regarding Prime(p) the situation is a bit risky in my opinion. For example, Wolfram Alpha will reply 7, the 4th prime number, on input Prime(4). IsPrime or is_prime might be safer. Another option could be $p \in \mathbb{P}$, and possibly recalling $\mathbb{P}$ denotes the set of prime numbers before. – quid Jan 04 '15 at 14:55
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    @quid, thanks for the tip about "isPrime". I've edited in your suggestion, while adding some explanation as to why I prefer my own way of phrasing it. – goblin GONE Jan 04 '15 at 15:16