For weak partially ordered set $ R(f) = \{ \langle A,B \rangle | f(A) \subseteq f(B) \} $ , show function $ T(f) $ is one-to-one.

Question

Problem:
Let $ f \in P(\mathbb{N}) \rightarrow P(\mathbb{N}) $ be one-to-one. We define the weak partialy ordered set $ R(f) = \{ \langle A,B \rangle | f(A) \subseteq f(B) \} $.
Let $ F \subseteq P(\mathbb{N}) \rightarrow P(\mathbb{N}) $ be the set of one-to-one functions that satisfy the above.
Prove/Disprove: The function $ T(f) = R(f)$ for all $ f \in F $, is one-to-one.

Attempt: I think it is disproof, here's what I did; I tried taking $ f_1 (A) = A \cup \{ 0 \} , f_2(A) = A \cup \{ 0,1 \} $, for all $ A \in P( \mathbb{N} ) $. And tried to showing that $ R(f_1) = R(f_2) $ (but note that $ f_1 \neq f_2 $ ). However , I could only show that $ R(f_1) \subseteq R(f_2) $ but not $ R(f_2) \subseteq R(f_1) $. After that failed attempt I thought maybe I'm wrong and actually the statement above is true. But then after assuming $ R(f_1) = R(f_2) $ I had difficulty showing $ f_1 = f_2 $ , obviously I need to use the fact that $ R(f) $ is a partially ordered set, but I don't exactly see how ( I thought anti-symmetry property would be useful but was unable to use it in a way that'd help me ). Can you please help?

I meant $ f: P(\mathbb{N}) \rightarrow P(\mathbb{N}) $. What I wrote is another notation ( I think a less familiar one but I didn't know that ). — hazelnut_116, Apr 27 '21 at 12:57

score 2 · Accepted Answer · answered Apr 27 '21 at 22:24

Let $E=\{2n:n\in\Bbb N\}$, and let $\varphi:\Bbb N\to E:n\mapsto 2n$. Let

$$f:\wp(\Bbb N)\to\wp(\Bbb N):A\mapsto\varphi[A]\,,$$

and let $g$ be the identity map on $\wp(\Bbb N)$. Then for all $A,B\subseteq\Bbb N$ we have $f(A)\subseteq f(B)$ iff $A\subseteq B$ iff $g(A)\subseteq g(B)$, so $R(f)=R(g)$, but clearly $f\ne g$: e.g., $f(\Bbb N)=E\ne\Bbb N=g(\Bbb N)$.

For weak partially ordered set $ R(f) = \{ \langle A,B \rangle | f(A) \subseteq f(B) \} $ , show function $ T(f) $ is one-to-one.

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