1. ## Combinatorics function

Q 28 from page 26 of Combinatorics A Guided Tour:

Define g: $\displaystyle 2^[$$\displaystyle ^2$$\displaystyle ^]$$\displaystyle \longrightarrow$$\displaystyle \mathbb Z$ by the rule g(S)=|S|, where S is any subset of [2]. Write g as a set of ordered pairs.

S could be {1} {2} and {1,2}.

So for S={1} would g(S) be 1? So would that be $\displaystyle 2^1$=2 and the ordered pair would be (2,1)?

But then for S={2} the cardinality of S would again be 1 giving the ordered pair (4,1).

I don't know how to do S={1,2}.

I am confused to the nth power.

Any clues towards unwarping my brain would be greatly appreciated.

2. You have used the notation $\displaystyle [2]$ before. But it is not standard.
Does it mean $\displaystyle \{1,2\}?$ If not please tell us how the textbook uses it.
Note that $\displaystyle \mathbb{P}(\{1\})=\{\emptyset,~\{1\}\}$
So $\displaystyle g=\{(\emptyset,0),(\{1\},1\}$ in this case .

3. Originally Posted by Plato
You have used the notation $\displaystyle [2]$ before. But it is not standard.
Does it mean $\displaystyle \{1,2\}?$ If not please tell us how the textbook uses it.
Note that $\displaystyle \mathbb{P}(\{1\})=\{\emptyset,~\{1\}\}$
So $\displaystyle g=\{(\emptyset,0),(\{1\},1\}$ in this case .
[n] as defined in the textbook is the set of all postive integers less than or equal to n.

I don't believe that the problem involved the Power set.

4. Originally Posted by oldguynewstudent
I don't believe that the problem involved the Power set.
Originally Posted by oldguynewstudent
Define g: by the rule g(S)=|S|, where S is any subset of [2].
Well you yourself said it does in the OP.
The function $\displaystyle g$ maps the subsets of $\displaystyle [2]$
to the integers. The power set of $\displaystyle [2]$ is the set of all subsets of $\displaystyle [2]$.

From the OP the function $\displaystyle g=\{(\emptyset,0),(\{1\},1),(\{2\},1),(\{1,2\},2)$.

So it has everything to do with power sets.

But once again you notation is confused.
$\displaystyle 2^{[2]}$ is the set of all functions from the set $\displaystyle [2]$ to the set $\displaystyle \{0,1\}$.
However, that is not the way $\displaystyle g$ was defined.

If I were you, I would get a textbook that uses standard notation.

5. Originally Posted by Plato
Well you yourself said it does in the OP.
The function $\displaystyle g$ maps the subsets of $\displaystyle [2]$
to the integers. The power set of $\displaystyle [2]$ is the set of all subsets of $\displaystyle [2]$.

From the OP the function $\displaystyle g=\{(\emptyset,0),(\{1\},1),(\{2\},1),(\{1,2\},2)$.

So it has everything to do with power sets.

But once again you notation is confused.
$\displaystyle 2^{[2]}$ is the set of all functions from the set $\displaystyle [2]$ to the set $\displaystyle \{0,1\}$.
However, that is not the way $\displaystyle g$ was defined.

If I were you, I would get a textbook that uses standard notation.
Yes, I see your point Plato. Thanks.

When it says any subset, then null would be one of those subsets.

I will have to ponder this and perhaps the professor will have time to expound on this. But since class hasn't started yet, perhaps he won't have time. Don't have a choice on the textbook however. Most of the text is very good, it's the unanswered problems that are a little fuzzy. I may try to pick up some extra material before the summer session starts.