# [SOLVED] nonempty set, injective function

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• Feb 24th 2009, 05:48 PM
deathbyproofs
[SOLVED] nonempty set, injective function
Prove or disprove: For every nonempty set A, there exists an injective function f: A -> P(A).
So this is saying that for this function f, every element of A has an image in the power set of A, right? But not every element of the power set has to be defined by f(A), right?
I think this could be proven but honestly i cannot think of a way to go about doing it. Please help and if you have a strategy please explain where you got it/how you came up with it so I can better understand the problem. Thanks so much!
• Feb 24th 2009, 10:39 PM
Grandad
Injective function and power set
Hello deathbyproofs
Quote:

Originally Posted by deathbyproofs
Prove or disprove: For every nonempty set A, there exists an injective function f: A -> P(A).
So this is saying that for this function f, every element of A has an image in the power set of A, right? But not every element of the power set has to be defined by f(A), right?

Yes, you are quite right. So if we're going to prove the statement, all we need to do is to find a function $\displaystyle f$ that does this.

Well, the power set $\displaystyle P(A)$ is the set of all the subsets of $\displaystyle A$. So, for any $\displaystyle x \in A, \{x\} \in P(A)$. Therefore if we define $\displaystyle f$ as $\displaystyle f(x) = \{x\}, \forall x \in A$, that would do it.

Grandad