Cardinal of P(S) (Adam's question at Yahoo Answers)

  • Context: MHB 
  • Thread starter Thread starter Fernando Revilla
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on proving that for any positive integer n, the cardinality of the power set P(S) is equal to 2^n, where |S| = n. The proof utilizes the binomial theorem, demonstrating that the sum of the binomial coefficients from 0 to n equals 2^n. This conclusion is derived from the equation |P(S)| = ∑(k=0 to n) C(n, k), confirming that the total number of subsets of a set S with n elements is indeed 2^n.

PREREQUISITES
  • Understanding of set theory and cardinality
  • Familiarity with the binomial theorem
  • Knowledge of binomial coefficients, denoted as C(n, k)
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the binomial theorem in detail
  • Explore combinatorial proofs related to set theory
  • Learn about different types of sets and their properties
  • Investigate applications of power sets in computer science
USEFUL FOR

Mathematicians, computer scientists, students studying discrete mathematics, and anyone interested in combinatorial analysis and set theory.

Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
Here is the question:

Show for any n ϵ ℤ+ with |S| = n, then |P(S)| = 2n.?

Here is a link to the question:

Show for any n

I have posted a link there to this topic so the OP can find my response.
 
Physics news on Phys.org
Hello Adam,

The number of subsets of $S$ with $k$ elements is $\displaystyle\binom{n}{k}$. So, using the binomial theorem: $$\begin{aligned}\left|\mathcal{P}(S)\right|&= \displaystyle\binom{n}{0}+\displaystyle\binom{n}{1}+\displaystyle\binom{n}{2}+\ldots+ \displaystyle\binom{n}{n}\\&=\displaystyle\binom{n}{0}1^n\cdot 1^0+\displaystyle\binom{n}{1}1^{n-1}\cdot1+\displaystyle\binom{n}{2}1^{n-2}\cdot1^2+\ldots+\displaystyle\binom{n}{n}1^0 \cdot 1^n\\&=(1+1)^n\\&=2^n\end{aligned}$$
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K