Combinatorics function

• Apr 25th 2010, 01:06 PM
oldguynewstudent
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.
• Apr 25th 2010, 01:21 PM
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 .
• Apr 25th 2010, 02:22 PM
oldguynewstudent
Quote:

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.
• Apr 25th 2010, 02:48 PM
Plato
Quote:

Originally Posted by oldguynewstudent
I don't believe that the problem involved the Power set.

Quote:

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.
• Apr 25th 2010, 03:06 PM
oldguynewstudent
Quote:

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.