@BalarkaSen The inductive one doesn't seem too intuitive at the moment. I'm reading it, but it's not really springing any mental connections.

@robjohn And this method also works for that series I posted in the past, the one with $\zeta(2)$ and $\zeta(3)$ :-)

@Khalli First think of a different situation : Say you're given the n-object multiset $\{\{0, 1\}, \{0, 1\}, \cdots, \{0, 1\}\}$. Consider that your object is to pick up one element from each of the $\{0, 1\}$ set and form another set of $n$ elements.

@BalarkaSen Got it.

You can form 2 of those.

Right?

Since there are $2$ elements in each set and you have $n$ such 2-sets, you can form $2 \times 2 \times 2 \times \cdots \times 2 = 2^n$ such $n$-elt sets.

@BalarkaSen Woah, you've lost me.

@robjohn did I show you the application of Au-Yeung series on another series?

@Khallil In how many ways can you pick an element from $\{0, 1\}$?

@BalarkaSen there is a questoin about special linear groups interested?

@Alizter depends. what is le question?

@Chris'ssis i don't know A-Y by that name.

@BalarkaSen 2 ways. You can pick either $0$ or $1$.

@robjohn I mean $\sum (H_n/n)^2$

@Khallil And how many $\{0, 1\}$s are there?

@BalarkaSen Here

@BalarkaSen There are $n$ of the $\{0,1\}$ sets.

@Khallil So shouldn't there be $2^n$ ways to pick up elements from each set?

@robjohn See this one

@BalarkaSen It might help if understood the $n=3$ case.

How many ways can I pick an element from $\{0,1\}$, $\{0,1\}$, $\{0,1\}$? Isn't it only 6 since there are 6 elements?

You can pick 2 from each of the sets.

I might be lost when you say pick. What do you mean by pick?

Just pick one element from the first set, say $0$. Another from the second, say $1$ and another from the third, say $1$. Form a set $\{0, 1, 1\}$

That is what I mean by pick and you are asked to count such sets you can form.

I can see loads of such multisets, certainly more than 6. $\{0, 1, 1\}, \{1, 1, 1\}, \{0, 0, 0\}, \{1, 0, 1\}, \{1, 1, 0\}, \{0, 1, 0\}, \{1, 0, 0\}$ are 7 examples.

So they have to be distinct, right?

right, I am talking about multisets.

Oh, they are multi sets. The order does matter in multi sets, right?

are you convinced about $2^n$?

@Khallil yes.

Consider me convinced!

@Chris'ssis I worked on that using fourier series and converting to an integral on the circle.

@Khallil Now, back to the problem. Consider the nonempty set $A$ with $n$ elts.

Take any element $a \in A$. Now, given a subset of $A$, $a$ can be either inside it or outside it, right?

@robjohn You got a full proof?

@BalarkaSen Did you see the questoin

@BalarkaSen Yea. That makes sense.

@Chris'ssis I didn't finish, but it looked good so I left it for when I had some time.

@Alizter I did, and I am not familiar with such an isomorphism. It'd take some work.

@robjohn OK

@Chris'ssis Is another proof needed? I saw that there were some in an answer on main

@Khallil So for each element, there are two choices for different subsets, namely {inside, outside}. Since there are $n$ elements, the number of subset is then, as per out previous experiment with the binary sets, $2^n$. QED.

@robjohn Where on main?

@Alizter I believe both are false.

@BalarkaSen Nice! That's neat!

The inductive proof looked at that possibility as well.

@Khallil You have to have the power of counting stuffs whenever you are fiddling with sets. Combinatorics is a fun branch of mathematics.

"Consider a set $S$ with $k+1$ members. Mentally divide it into a set with $k$ members plus one new member $a$. Now consider all the subsets of $S$. We know that there are $2^k$ subsets of the $S – {a}$, so there are $2^k$ subsets of $S$ that do not include $a$, namely $T_i \cup \{a\}$ ..."

@Chris'ssis here

@Khallil Ah, interesting.

@robjohn Ah, OK.

@BalarkaSen Here's the full link. I didn't quite get the introduction of $T_i$. (people.umass.edu/partee/409/h15.pdf) Oh, I get it. It's the same thing you were talking about, kinda.

@Khallil If you have a subset without $a$, then subtracting (set theoretically) it out from $S$ will give you... ?

@BalarkaSen That's not clear enough. I don't know what you mean.

Let $A$ be a subset of $S$ not containing $a$. Then can you convince yourself that $a \in S - A$?

you are familiar with set subtraction, i believe?

@BalarkaSen It's referred to the set difference I believe, but if it's like normal subtraction, then we'll simply have a set $S-A$ that does contain $a$.

yes.

so for each set not containing a, you'll get a corresponding set containing a.

Yep!

since you have 2^k sets not containing a, there'd be 2^k sets containing a as well.

In other words, there'd be $2(2^k)$ i.e. $2^{k+1}$ subsets of $\mathcal{P}(A) = 2^{k+1}$ so long as $A$ contains $k+1$ elements where the extra element is what we've been referring to as $a$ the whole time.

yes

Yep, it's a really nice method. I better learn how to count. Counting is power!

I'm exhaused now. Back some hours later.

Later pal.

I've got a quick question, @BalarkaSen.

If I want to find all $X$ s.t. $\left| X \right| \leqslant 1$ that's an element of the power set, $\mathcal{P}(\{1,2,3\})$, is it as simple as listing the subsets of the original set with cardinality less than or equal to $1$?

If that's the case, would it be true that $\{ X \in \mathcal{P}(\{ 1, 2, 3 \}) : \left| X \right| \leqslant 1 \} = \{ \{1\}, \{2\}, \{3\}, \varnothing \}$?

Looks like so.

@BalarkaSen Cool. Would it also be the case that $\{ X \in \mathcal{P}(\{ 1, 2, 3 \}) : \left| X \right| \leqslant 1 \} = \{ X \subseteq \mathcal{P}(\{ 1, 2, 3 \}) : \left| X \right| \leqslant 1 \}$?

Looks like so.

Haha! Thanks!

I've got another strange looking question, @BalarkaSen. It's got to do with cardinalities of power sets under certain restrictions.

I mean, state your problem.

Sorry, got dozed off.

Haha, no problem!

Given that $|A| = m$, is is true that $| \{ X \in \mathcal{P}(A) : |X| \leq 1 \} | = m$?

The question was to find that cardinality and I ended up with $m$.

it is false.

it should be $m + 1$

you missed the null set

Surely that's not an element of the power set?

I thought it was a subset.

ah, fair point.

as I said I dozed off.

$m$ is fine then.

Is it late where you are?

4:14 AM

Yea, I thought that if they specified that $X \subseteq \mathcal{P}(A)$, then it'd be $m+1$.

WOAH!

That's awesome. I used to stay up late like that until a few weeks ago. Get some rest! It can't be good for you.

I will go to sleep when the hell freezes over.

slaps himself. slap slap slap

That's better.

@BalarkaSen You're crazy! I might watch the latest Hunter X Hunter episode and call it a night. It's almost midnight over here!