Set Thery hard question: Let C={1,2,3,5,8,9,10}, calculate |P(C)|

[HugeLag]

Let C={1,2,3,5,8,9,10}, calculate |P(C)|

To work this out am I meant to multiply P by the numbers in C? So would it look like the following.

|P(1,2,3,5,8,9,10)| = P1,P2,P3,P5,P8,P9,P10

My other theory of working this out is BODMAS. So brackets first. |P(1,2,3,5,8,9,10)|
However there is no intersection between the P and ( so I am kind of confused.

 

[pka]

Let C={1,2,3,5,8,9,10}, calculate |P(C)|
To work this out am I meant to multiply P by the numbers in C? So would it look like the following.
|P(1,2,3,5,8,9,10)| = P1,P2,P3,P5,P8,P9,P10
My other theory of working this out is BODMAS. So brackets first. |P(1,2,3,5,8,9,10)|
However there is no intersection between the P and ( so I am kind of confused.

Unless you will tell us what this notation means, there is no way to help you.
What is \(\displaystyle P(C)~?\) Is that the power set of the set \(\displaystyle C~? \)
If it is a power set then what the heck are \(\displaystyle P1,P2,P3,P5,P8,P9,P10~? \)
What is BODMAS and what does it have to do with set theory?

 

[Deleted member 4993]

Unless you will tell us what this notation means, there is no way to help you.
What is \(\displaystyle P(C)~?\) Is that the power set of the set \(\displaystyle C~? \)
If it is a power set then what the heck are \(\displaystyle P1,P2,P3,P5,P8,P9,P10~? \)
What is BODMAS and what does it have to do with set theory?

As far as I know, BODMAS is a synonym of PEMDAS (P is replaced by Bracket(B), O stands for Of, etc.).

 

[Steven G]

Let C={1,2,3,5,8,9,10}, calculate |P(C)|

To work this out am I meant to multiply P by the numbers in C? So would it look like the following.

|P(1,2,3,5,8,9,10)| = P1,P2,P3,P5,P8,P9,P10

My other theory of working this out is BODMAS. So brackets first. |P(1,2,3,5,8,9,10)|
However there is no intersection between the P and ( so I am kind of confused.

I would have thought that P(C) was the power set of C, but then you put those bars around P(C). As PKA said, we can't help you until you give us the definition of |P(C)|.

I am certain that what you did was not correct

 

[pka]

As far as I know, BODMAS is a synonym of PEMDAS (P is replaced by Bracket(B), O stands for Of, etc.).

Thank you for the reminder. That's arithmetic. Never seen it used in any set theory course.

I would have thought that P(C) was the power set of C, but then you put those bars around P(C). As PKA said, we can't help you until you give us the definition of |P(C)|.
I am certain that what you did was not correct

I also fairly sure that \(\displaystyle P(C)\) is for power set. Thus I think that the question is about the number of elements in the power set.
As for the notation, I have seen \(\displaystyle \#(C),~n(C),~|C|,~\|C\| \) all used to denote the number of elements in a set \(\displaystyle C \) . But what was posted does not come close, he seems no to know about the empty set.

Here is the theorem: \(\displaystyle \|\mathcal{P}(C)\|=2^{\|C\|} \).

 

[HugeLag]

Thank you for the reminder. That's arithmetic. Never seen it used in any set theory course.

I also fairly sure that \(\displaystyle P(C)\) is for power set. Thus I think that the question is about the number of elements in the power set.
As for the notation, I have seen \(\displaystyle \#(C),~n(C),~|C|,~\|C\| \) all used to denote the number of elements in a set \(\displaystyle C \) . But what was posted does not come close, he seems no to know about the empty set.

Here is the theorem: \(\displaystyle \|\mathcal{P}(C)\|=2^{\|C\|} \).

My professor confirmed with me that I need to find the number of elements in the power set! Just to let the people know who weren't sure.

By the way! Just for taking notes and understanding, the name of that theorem refers to Cantor's Theorem right?

2⁷ = 128? Just making sure this would be the correct answer?

Working:

Let C={1,2,3,5,8,9,10}, calculate |P(C)|

C = 7 elements.

I managed to figure it out. I asked my professor! Thanks a lot for providing me with that theorem! Really helped a lot in understanding and saving me time in the exam working questions like this out.