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

[mathmari]

Hey!

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

All correct.
Did you miss something? Only the closing brace. (Tauri)

Great! At 2 how can we disprove that formally? (Wondering)

 

[topsquark]

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