MHB What is the definition of 2^M and how can it be determined?

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Set
AI Thread Summary
The discussion centers on the definition and properties of the powerset, denoted as 2^M. Participants confirm that 2^M represents the set of all subsets of M, with examples provided for specific sets like {7, 4, 0, 3} and {a, b}. There is a debate about whether the expression 2^{A×B} can be equated to the set of all subsets formed by the Cartesian product of subsets of A and B, leading to the conclusion that they are not equivalent. Additionally, it is established that no set M can exist such that 2^M equals the empty set, as the empty set is always included in any powerset. The conversation concludes with inquiries about formal disproving methods and the implications of set order in Cartesian products.
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

  1. Let $M:=\{7,4,0,3\}$. Determine $2^M$.
  2. Prove or disprove $2^{A\times B}=\{A'\times B'\mid A'\subseteq A, B'\subseteq B\}$.
  3. Let $a\neq b\in \mathbb{R}$ and $M:=2^{\{a,b\}}$. Determine $2^M$.
  4. Is there a set $M$, such that $2^M=\emptyset$ ?

First of all how is $2^M$ defined? Is this the powerset? (Wondering)
 
Physics news on Phys.org
mathmari said:
First of all how is $2^M$ defined? Is this the powerset?

Hey mathmari!

Yep.
So $2^{\{a,b\}}=\{\varnothing,\{a\},\{b\},\{a,b\}\}$. (Thinking)
 
Klaas van Aarsen said:
Yep.
So $2^{\{a,b\}}=\{\varnothing,\{a\},\{b\},\{a,b\}\}$. (Thinking)

So we have the following:
  1. $2^M=\{\emptyset, \{7\}, \{4\}, \{0\}, \{3\}, \{7,4\}, \{7,0\}, \{7,3\}, \{4,0\}, \{4,3\}, \{0,3\},\{7,4,0\}, \{7,4,3\}, \{7,0,3\}, \{4,0,3\}, \{7,4,0,3\}$
  2. $2^{A\times B}=\text{ set of every subset of } A\times B=\{(a,b)\mid a\in A, b\in B\}$.

    So it is of the form $2^{A\times B}=\{\emptyset, \{(a_i, b_i)\}, \{(a_i, b_i), (a_j, b_j)\}, \cdots \}$, or not?

    This is not of the form $\{A'\times B'\mid A'\subseteq A, B'\subseteq B\}$, is it? (Wondering)
  3. We have that $M=\{\emptyset, \{a\}, \{b\}, \{a,b\}\}$.

    The $2^M=\{\emptyset, \{\emptyset\}, \{\{a\}\},\{\{b\}\}, \{\{a,b\}\}, \{\emptyset, \{a\}\}, \{\emptyset, \{b\}\}, \{\emptyset, \{a,b\}\}, \{\{a\}, \{b\}\}, \{\{a\}, \{a,b\}\}, \{\{a,b\}, \{b\}\}, \{\emptyset, \{a\}, \{b\}\}, \{\emptyset, \{a\}, \{a,b\}\}, \{\emptyset, \{a,b\}, \{b\}\}, \{\emptyset, \{a\}, \{b\}, \{a,b\}\}$

    Is this correct? Or do I miss something? (Wondering)
  4. There isn't such a set, since at least the emptyset is contained in the powerset. Is this correct? (Wondering)
 
mathmari said:
So we have the following:

1. $2^M=\{\emptyset, \{7\}, \{4\}, \{0\}, \{3\}, \{7,4\}, \{7,0\}, \{7,3\}, \{4,0\}, \{4,3\}, \{0,3\},\{7,4,0\}, \{7,4,3\}, \{7,0,3\}, \{4,0,3\}, \{7,4,0,3\}$

2. $2^{A\times B}=\text{ set of every subset of } A\times B=\{(a,b)\mid a\in A, b\in B\}$.

So it is of the form $2^{A\times B}=\{\emptyset, \{(a_i, b_i)\}, \{(a_i, b_i), (a_j, b_j)\}, \cdots \}$, or not?

This is not of the form $\{A'\times B'\mid A'\subseteq A, B'\subseteq B\}$, is it? (Wondering)

3. We have that $M=\{\emptyset, \{a\}, \{b\}, \{a,b\}\}$.

The $2^M=\{\emptyset, \{\emptyset\}, \{\{a\}\},\{\{b\}\}, \{\{a,b\}\}, \{\emptyset, \{a\}\}, \{\emptyset, \{b\}\}, \{\emptyset, \{a,b\}\}, \{\{a\}, \{b\}\}, \{\{a\}, \{a,b\}\}, \{\{a,b\}, \{b\}\}, \{\emptyset, \{a\}, \{b\}\}, \{\emptyset, \{a\}, \{a,b\}\}, \{\emptyset, \{a,b\}, \{b\}\}, \{\emptyset, \{a\}, \{b\}, \{a,b\}\}$

Is this correct? Or do I miss something?

4. There isn't such a set, since at least the emptyset is contained in the powerset. Is this correct?

All correct.
Did you miss something? Only the closing brace. (Tauri)
 
Klaas van Aarsen said:
All correct.
Did you miss something? Only the closing brace. (Tauri)

Great! At 2 how can we disprove that formally? (Wondering)
 
A question of my own.

Say we have [math]A = \{1, 2, 4 \}[/math] and [math]B = \{ 1, 2, 3 \}[/math]

Why wouldn't [math]2^{A \times B}[/math] have sets with the form [math]\{ a_i, b_j \}[/math] rather than just [math]\{ a_i, b_i \}[/math]?

And, since we are talking about a Cartesian product wouldn't [math]2^{ \{ 1, 2 \} }[/math] be different from [math]2^{ \{ 2, 1 \} }[/math]?

Thanks!

-Dan
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...
Back
Top