Exploring the Properties of C and U in $\mathcal{P} (M_n)$

[roam]

Homework Statement

Let [tex]n \geq 1[/tex] be a positive integer and let [tex]M_n = \{ 1,...,n \}[/tex] be a set with n elements. Denote by [tex]\mathcal{P} (M_n)[/tex] the set of all subsets of M_n. For example [tex]\mathcal{P} (M_2) = \{ \{ \emptyset \}, \{ 1 \}, \{ 2 \}, \{ 1,2 \} \}[/tex].

Show that [tex]C=(\mathcal{P} (M_n) , \cap)[/tex] and [tex]U=(\mathcal{P} (M_n) , \cup)[/tex] each has an identity. Decide whether C and U are groups.

The Attempt at a Solution

For [tex]C=(\mathcal{P} (M_n) , \cap)[/tex], let [tex]a,b,c \in \mathcal{P} (M_n)[/tex]

• Associativity: [tex](a \cap b) \cap c = a \cup (b \cap c)[/tex] [tex]\checkmark[/tex]
  
• Identity: is the empty set => [tex]a \cap \emptyset = a[/tex] [tex]\checkmark[/tex]
  
• Inverse: I can't see what's the inverse of this group! for an element a we need an inverse b such that [tex]a \cap b = \emptyset[/tex]. I think this is only true when a & b are completely distinct but I'm not sure...

Similarly [tex]U=(\mathcal{P} (M_n) , \cup)[/tex] satisfies the associativity and I think its identity is also [tex]\emptyset[/tex]. But what is the inverse??

I need help finding the inverses, and please let me know if the rest of my working is correct.
Any help is really appreciated.

 

[Hurkyl]

If you don't understand it for all n, try special cases. Maybe continue with your n=2 example. That looks like a very small finite set -- you should be able to answer any particular question at all with a few seconds effort.

 

[Dick]

First try and think clearly. I don't think a intersect {} is a, if that's what you mean to say. But you are right that there is a problem with inverses. They aren't groups.

 

[roam]

First try and think clearly. I don't think a intersect {} is a, if that's what you mean to say. But you are right that there is a problem with inverses. They aren't groups.

Oops, I made a mistake! So, for [tex]C=(\mathcal{P} (M_n) , \cap)[/tex]

The identity for each element is itself, right? Because [tex]a \cap a = a[/tex]. Therefore the inverse for every element is also itself. So it's a group?

And for [tex]U=(\mathcal{P} (M_n) , \cup)[/tex]

[tex]A \cup \emptyset = A[/tex]. Therefore [tex]\emptyset[/tex] is the identity.

But what is the inverse? I need "b" such that [tex]a \cup b = \emptyset[/tex]. Even if there is no inverse, I guess I have to give some kind of explanation

 

[HallsofIvy]

Oops, I made a mistake! So, for [tex]C=(\mathcal{P} (M_n) , \cap)[/tex]

The identity for each element is itself, right? Because [tex]a \cap a = a[/tex]. Therefore the inverse for every element is also itself. So it's a group?

No, there is a unique identity for a group, not a different "identity" for each member. Is there a single set, x, such that [itex]a\cap x= a[/itex] for every a? (Look at [itex]M_n[/itex] itself.)

And for [tex]U=(\mathcal{P} (M_n) , \cup)[/tex]

[tex]A \cup \emptyset = A[/tex]. Therefore [tex]\emptyset[/tex] is the identity.

But what is the inverse? I need "b" such that [tex]a \cup b = \emptyset[/tex]. Even if there is no inverse, I guess I have to give some kind of explanation

If set A had and inverse, B, say, then you would have to have [itex]A\cup B= \emptyset[/itex]. How do the number of members of A and [itex]A\cup B[/itex] compare?

 

[roam]

No, there is a unique identity for a group, not a different "identity" for each member. Is there a single set, x, such that [itex]a\cap x= a[/itex] for every a? (Look at [itex]M_n[/itex] itself.)

Yes, there is a single set, whenever n=1 then M_n has only one element. So the inverse of 1 is 1, it's also the identity. But this only true when n=1! So, [tex]C=(\mathcal{P} (M_n) , \cap)[/tex] not a group?

If set A had and inverse, B, say, then you would have to have [itex]A\cup B= \emptyset[/itex]. How do the number of members of A and [itex]A\cup B[/itex] compare?

The only member of [tex]A\cup B[/tex] is the empty set whereas the number of members of A is greater than or equal to 1. But what does that prove?

 

[HallsofIvy]

No, there is a unique identity for a group, not a different "identity" for each member. Is there a single set, x, such that [itex]a\cap x= a[/itex] for every a? (Look at [itex]M_n[/itex] itself.)

Yes, there is a single set, whenever n=1 then M_n has only one element. So the inverse of 1 is 1, it's also the identity. But this only true when n=1! So, [tex]C=(\mathcal{P} (M_n) , \cap)[/tex] not a group?

? For any subset, A, of M_n, [itex]A\cap M_n= A[/itex].

If set A had an inverse, B, say, then you would have to have [itex]A\cup B= \emptyset[/itex]. How do the number of members of A and [itex]A\cup B[/itex] compare?

The only member of [tex]A\cup B[/tex] is the empty set whereas the number of members of A is greater than or equal to 1. But what does that prove?

It proves that the can't be equal! And since they must be the same set in order that B be "inverse" to A, there cannot be an inverse.