1

Would someone please enlarge on Arturo Magidin's original answer ?

$1.$ Say the question didn't divulge $|S| = |X|$. Then how can $|S|$ be determined? Any intuition?

I recast it below with more details to try to apprehend it. Please feel free to edit errors/problems.
As defined therein, $X$ is an infinite set, and $ S := \{$ all finite subsets of $X \}$.

Since ● $X$ is given as infinite which means $|X| \ge |\mathbb{N}|$
and ● $S \quad \supseteq \quad \{\emptyset, \{1\}, \{2\},...,\{n\},...\} = \require{cancel} \cancel{\{\text{singletons}\}} = |\mathbb{N}|$ $\implies \emptyset, \{1\}, \{2\},...,\{n\},... \in S$
$\implies 1, 2, ..., n, ... \in X \iff \mathbb{N} \subseteq X$ thus $\color{limegreen}{|\mathbb{N}| \le |X| \le |S|}$.

For all $n \in \mathbb{N}$, define $S_n$ as did Prof Magidin: $S_n:= \color{violet}{\{}{\text{ all subsets of cardinality $n$ }} \color{violet}{\}}\subseteq S.$
This definition involves a subset of cardinality $n$ (ie with $n$ elements) so we now scrutinise it.

$2.$ How would you divine/previse to define $S$ wrt $S_n$ ?

$S_n$ has the form $\color{violet}{\{} \quad \{ \, (a_1, ..., a_n) \, \}, \{ \, (b_1, ..., b_n) \, \}, ...\quad \color{violet}{\}}$.
The $n$ elements in each $n$-tuple in each subset of cardinality $n$ can be chosen $n!$ ways,
ie: the $a_1, ..., a_n$ in $(a_1, ..., a_n)$ in $\{ \; (a_1, ..., a_n) \; \}$ can be chosen $n!$ ways.
Thus every such subset of cardinality $n$ determines $n!$ $n$-tuples of elements of $X$.
In general, an $n$-tuple included in any set will produce a subset (of that set) of cardinality $\le n$.
This paragraph implies: $ |S_n| \color{red}{\le} \, n!|X|^n \, = |X|$.

$3.$ How and why is $ |S_n| \color{red}{\le} \, n!|X|^n $?

In toto, $ \begin{align} |X| \color{lime}{\le} |S| & = \left|\biguplus_{n=0}^{\infty} S_n\right| = \sum_{n=0}^{\infty}|S_n| = |\underbrace{S_0}_{= \emptyset}| + \sum_{n=1}^{\infty}|S_n| \\ & = \sum_{n=1}^{\infty}|S_n| \\ & \color{red}{\le} \sum_{n=1}^{\infty}|X| = |\mathbb{N}||X| \; {\Large{\color{darkturquoise}{[}}} = {\Large{{\color{darkturquoise}{]}}}} \; |X| \end{align} $

$4.$ In the last line, how and why is $|\mathbb{N}||X| \; = \; |X|$ ?

Community
  • 1
  • The third lime is nonsense as written. What is $X$ and what does it have to do with $S$? Also, please write the inequalities in the right order. This gives you $|X|\ge|\mathbb N|$ and $|\mathbb N|\le |S|$, but no relation so far between $X$ and $S$. – Andrés E. Caicedo Jan 06 '14 at 19:25
  • @AndresCaicedo: Thank you! I hope I've emended the errors. Please feel free to edit my post. Is it correct now? –  Jan 06 '14 at 19:34
  • The fourth line is incorrect. That $|X|\ge|\mathbb N|$ is very different from $\mathbb N\subseteq X$. The latter is needed for line four to make sense. Are you assuming this? – Andrés E. Caicedo Jan 06 '14 at 19:43
  • Once you fix line 4, it is not clear what line 6 means. $S_n$ is the collection of all subsets of size $n$, not just a single one. Line 7 is incorrect. What you wrote is a singleton set, whose only element is an ordered $n$-tuple. There are no $n$-tuples anywhere in $S$ or $S_n$. The whole talk of $n!$ etc is very confused: Sets do not have an order. – Andrés E. Caicedo Jan 06 '14 at 19:46
  • @AndresCaicedo: Thank you very much again. About line $4$, I was trying to write: Because $\mathbb{N}$ is countably infinite and $X$ is infinite, thus $|\mathbb{N}| \le |X|$? Is this right? Since ${\text{ singletons }}$ can $\not\subseteq$ any infinite set, thus I never assumed $\mathbb{N} \subseteq X$. Is this correct? –  Jan 06 '14 at 19:59
  • Line 4 says that ${1},{2},\dots$ are elements of $S$, that is, $1,2,\dots$ are elements of $X$, that is, $\mathbb N\subseteq X$. – Andrés E. Caicedo Jan 06 '14 at 20:01
  • @AndresCaicedo: For line 5, I did mean to write: $S_n$ is the $\color{#FF4F00}{\text{ collection }}$ of all subsets of size $n$. I just rewrote it with brackets: $S_n:= \color{#FF4F00}{{}{\text{ all subsets of cardinality }} n \color{#FF4F00}{}}$. Is this correct? –  Jan 06 '14 at 20:02
  • 1
    To your first question, I don't see a red bracket. To the second one, I don't see a red bracket. – Asaf Karagila Jan 06 '14 at 20:13
  • @AndresCaicedo: Thank you deeply. I edited those lines. Are they correct now? I might've misapprehended the original answer; it does show $n!$ in the last line of the 3rd paragraph? –  Jan 06 '14 at 20:25
  • @AsafKaragila: Are you alluding to http://meta.math.stackexchange.com/questions/4195/on-the-use-of-color-in-equations/4196#4196? Could you please elucidate? –  Jan 06 '14 at 20:27
  • 3
    Allow me to elaborate, and my thoughts to emancipate. My eyes, while capable of discerning colors and shades, do not do so in the greatest manner. It is unfortunate, and for my fortune unpropitious, that you use so many colors. Not just that, it is that brackets and separators are made of thin and hardly visible lines, whose inherent contrast is dull. And thus, you knowingly and willingly (and quite frankly, quite disrespectfully) exclude me from reading your questions properly, and antagonize me as to not to post an answer. Even when I can, often after grave effort, read your posts through. – Asaf Karagila Jan 06 '14 at 20:33
  • 5
    @AsafKaragila: I regret to learn about your condition. I only use colours simply because it helps me. Nonetheless, I realise there are those who would favour no colour and had thought that the answers at http://meta.math.stackexchange.com/questions/4195/on-the-use-of-color-in-equations/4196#4196 would've resolved this? As concerns the brackets and separators, please feel free to let me know how my OP can be improved. How can I lessen them? –  Jan 06 '14 at 20:57
  • 5
    @AsafKaragila: As concerns your last sentence, I had never any intent "to exclude [anyone] from reading your questions properly, and antagonize me as to not to post an answer," still less ""knowingly and willingly (and quite frankly, quite disrespectfully)" doing so. In light of my previous comment, this sentence sounds unwarranted. I'm very grateful for your contributions and expertise here so no disrespect has ever been intended. Thus, could we please resolve any antagonism? –  Jan 06 '14 at 21:11
  • 1
    In line 4, what you wrote is not the set of singletons. Lines 4,5,6 are terribly incorrect. Line 9 is incorrect, no matter how many brackets you add. It will remain incorrect as long as you keep mentioning $n$-tuples. The next 5 lines make no sense. We then get to the displayed equation. $S_0$ is not empty. Its size is certainly not the empty set. Even if you replace the empty set with $0$, this is still incorrect. Do you understand what the infinite sum means here? Do you understand what the notation $|A|\le |B|$ means? Nothing in the question makes this apparent; in fact, quite the opposite. – Andrés E. Caicedo Jan 06 '14 at 21:16
  • 1
    @AndresCaicedo: Thank you again. How does $n$-tuple feature in the original answer? Where would it here, if at all? I was trying to write this explicitly here but may have misapprehended it. About the infinite sum, I now realise that I don't; I was trying to determine how $\sum\limits{n \ge 0}|S_n| = \sum\limits{n le 1}|S_n|$. $|A| \le |B|$ means that the cardinality of $A \le$ that of $B$? –  Jan 07 '14 at 06:06

0 Answers0