Proving a one-to-one function with a power set involved

Question

Let A = P({1,2,3,4}).

f : A → A defined by f (X ) = X ∩ {1, 4}.

Can I prove that the function is one-to-one as there is one intersection? Or since there are many 1's and 4's within the power list that repeat, does it mean that it is not a one to one function.

I know that the {1,4} will have one intersection with the power set

( {},{1},{2},{3},{4},{1,2},{1,3},

{1,4}, (intersection)

{2,3},{2,4},{3,4},{1,2,3},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4})

I am used to proving that a function is one by one by proving that if f(u)=f(v) then you must have u=v, this is confusing for me. I believe that it is a one-to-one function but I am not sure how to prove it in this context.

Answer 1

Hint:

What is $f(\{1\})$? What is $f(\{1,2\})$

Edit referring to comment:

• $f(\{1\}) = \{1\} \cap \{1,4\} = \{1\}$
• $f(\{1,2\}) = \{1,2\} \cap \{1,4\} = \{1\}$

So, you have found two elements $A,B \in P(\{1,2,3,4\})$ such that $A \neq B$ and $f(A) = f(B)$.