MHB Proving Existence of Surjective Function F from P(N)\N to P(N)

AI Thread Summary
The discussion centers on proving the existence of a surjective function F from P(N)∖N to P(N), where P(N) is the power set of natural numbers. The initial definition of F is incomplete, as it does not specify the mapping rule for sets in A and B. The condition regarding proper subsets suggests a potential restriction on F, but its role remains unclear. Ultimately, it is noted that a bijection exists between P(N)∖N and P(N) due to both sets having the same cardinality, which implies that a surjective function can be constructed. Therefore, the existence of such a surjective function F is affirmed.
Ella1
Messages
2
Reaction score
0
I'd really like some help in answering the next question...anything that might help will save my life:

F is defined this way: F:A→B where A,B⊂P(N) and P(N) is the power set of the naturals.
Let S,R∈A such that S is a proper subset of R if and only if F(S) is a proper subset of F(R)

My question is to prove whether or not there is an F from P(N)∖N to P(N) which is also a surjective function?
 
Physics news on Phys.org
Ella said:
F is defined this way: F:A→B where A,B⊂P(N) and P(N) is the power set of the naturals.
This is not a complete definition of $F$. This just fixes (to some unknown $A$ and $B$) the domain and codomain of $F$ and not the rule that establishes which sets are mapped to which sets.

Ella said:
Let S,R∈A such that S is a proper subset of R if and only if F(S) is a proper subset of F(R)
This does not define $S$ and $R$ uniquely, so I am not sure what the role of this phrase is. Perhaps it is supposed to be a restriction on $F$ and this property is supposed to hold for all $S$ and $R$.

Ella said:
My question is to prove whether or not there is an F from P(N)∖N to P(N) which is also a surjective function?
Since the question is about the existence of $F$, the previous definition of $F$ is apparently irrelevant. Without any restrictions, yes, there exists a bijection between $P(\mathbb{N})\setminus\mathbb{N}$ and $P(\mathbb{N})$ because both sets has cardinality continuum.
 
Back
Top