Prove of number of subsets of a set

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Homework Help Overview

The discussion revolves around proving the number of subsets of a set, specifically using the principle of functions from one set to another. The original poster references a formula involving combinations and seeks to understand the connection between subsets and functions mapping to binary values.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between functions and subsets, questioning how to apply the principle of functions to the context of subsets. There is an attempt to restate the principle using different variables to clarify understanding.

Discussion Status

The discussion is ongoing, with participants expressing confusion about linking the concept of subsets to the principle of functions. Some guidance has been offered regarding the need for clearer connections, but no consensus has been reached.

Contextual Notes

Participants note that the proof provided in the original post is perceived as too brief to grasp fully, indicating a potential gap in understanding the underlying concepts.

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Homework Statement


A= {a1,...am}, B= {b1,...an}. If f: A→B is a function, then f(a1) can take anyone of the n values b1,...bn. Similarly f(a2). Then there are nm such function. I understand this part.

So in my book, using this principle,
nC0 + nC1 + ... + nCn = 2n is proved.

It has taken a set A= {a1,...an}. For each subset B of A, there is a function fB : A→ {0,1} given by fB(a1) = 1 if a1 ∈ B and fB(a1) = 0 otherwise. Conversely, for every f: A→ {0,1}, B = {a1 : fB(a1) =1 } ⊂ A anf f = fB. Thus there is a one-to-one correspondense between the subsets of A and the functions f: A→ {0,1}. From the above formula, hence we have
nC0 + nC1 + ... + nCn = 2n

Homework Equations

The Attempt at a Solution

[/B]The prove given is so short for me to understand anything.
 
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rajeshmarndi said:

Homework Statement


A= {a1,...am}, B= {b1,...bn}. If f: A→B is a function, then f(a1) can take anyone of the n values b1,...bn. Similarly f(a2). Then there are nm such function. I understand this part.

So in my book, using this principle,
nC0 + nC1 + ... + nCn = 2n is proved.

It has taken a set A= {a1,...an}. For each subset B of A, there is a function fB : A→ {0,1} given by fB(a1) = 1 if a1 ∈ B and fB(a1) = 0 otherwise. Conversely, for every f: A→ {0,1}, B = {a1 : fB(a1) =1 } ⊂ A and f = fB. Thus there is a one-to-one correspondense between the subsets of A and the functions f: A→ {0,1}. From the above formula, hence we have
nC0 + nC1 + ... + nCn = 2n

Homework Equations

The Attempt at a Solution

[/B]The prove given is so short for me to understand anything.
It may help to take the principle you are applying and restate it using other variables.

Let D= {d1, ... , dm}, R= {r1, ... , rn}. If f: D→R is a function, then f(d1) can take anyone of the n values r1, ... , rn. Similarly for f(d2). Then there are nm such functions.​

Now apply this to the subset situation.
 
SammyS said:
Now apply this to the subset situation.
Sorry, that is the problem, I am unable to link the subset to the above mentioned situation.
 
rajeshmarndi said:
Sorry, that is the problem, I am unable to link the subset to the above mentioned situation.
What's stopping you? What part don't you understand? Simply saying "I don't get it" isn't very helpful.
 

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