[Math] Semi-colon in set notation

elementary-set-theorynotationprobability theory

In a math text, what does something like
$$
\{ (1,2,3,\dots, n); n\in \mathbb{N}\}
$$
mean?

More specifically, would it be $\{(1,2,3,\dots, n)\}$ for a specific $n \in \mathbb{N}$, or would it be $\{ (1), (1,2), (1,2,3) , \dots \}$. That is, would it apply to all $n\in \mathbb{N}$?

I ask because I guess I am confused on the use of $;$ in set notation versus $:$. Semi-colon is such that, so if the set was $\{ (1,2,3,\dots, n); n\in \mathbb{N}\}$ I believe it would apply for all $n\in\mathbb{N}$. Semicolon is for a specific $n$ perhaps though? I can't find much reference to this, although the wiki says the semicolon serves to add an additional rule. That is, the semicolon (or comma) is like an "and".

Where I have encountered this notation? Rosenthal's introduction to probability theory.

Best Answer

$$ '\{ (1,2,3,\dots, n); n\in \mathbb{N}\}' $$

is the same thing as

$$ \{(1), (1,2),\ldots \} $$

Check out:

I don't think we have $\Omega = \{(0,0,...) \cup (1,1,...)\}$

Related Solutions

Combinatorics – Direct Formula for Elements of Power Set

The easy thing to do is to notice that the $2^n$ subsets of an $n$-element set correspond in a natural way to the binary expansions of the numbers $0\ldots 2^n-1$:

000 {     }
001 {    a}
010 {  b  }
011 {  b,a}
100 {c    }
101 {c,  a}
110 {c,b  }
111 {c,b,a}

If you're willing to order the sets in this way, then the $n$th set contains element $m$ exactly when the binary expansion of the number $n$ has its $m$th-least significant bit set.

Let's write $f_m(n)$ for the function that has value 1 if the $m$th bit of the binary numeral $n$ is a 1, and 0 if it is a 0; it has value 1 if the $n$th subset contains the $m$th element, and 0 if not. Then $f_0(n) = n\bmod 2$, and

$$ \begin{align} f_m(n) & = f_{m-1}\left(\left\lfloor \frac n2\right\rfloor\right) \\ & = f_{0}\left(\left\lfloor \frac n{2^m}\right\rfloor\right) \\ & = \begin{cases}1,&\text{if $\left\lfloor \frac n{2^m}\right\rfloor$ is odd}\\0,&\text{if it is even}\end{cases} \end{align} $$

[Math] What if there are multiple conditions in set-builder notation

The set $\{x| x\in \mathbb N\land x<1000\}$ is finite and it can easily be presented in the set builder notation...

All integers greater than $0$ but less than $6$:

$$\{n| n\in\mathbb N\land n>0 \land n<6\}$$

All positive even integers less than $11$:

$$\{n| n\in\mathbb N\land n<11\land (\exists k\in\mathbb N: n=2k)\}$$ or, alternatively,

$$\{2k| k\in\mathbb N\land 2k<11\}$$

All perfect squares less than $101$:

$$\{n| n\in\mathbb N\land n<101\land (\exists k\in\mathbb N: n=k^2)\}$$

or, again, alternatively as

$$\{k^2| k\in\mathbb N\land k^2<101\}$$

Best Answer

Related Solutions

Combinatorics – Direct Formula for Elements of Power Set

[Math] What if there are multiple conditions in set-builder notation

Related Question