Stuck at order preservation of mapping $\phi : P \rightarrow Q$

Question

I had to show that this is order preserving:

$ P=\{ \mathcal{P}(S); \subseteq \}; |S|>1; \; Q=2 $ and $\phi$ from $P \rightarrow Q$ is defined by $\\ \phi (U)= \left\{ \begin{array}{ll} 1 & U=S \\ 0 & U \neq S\\ \end{array} \right. $

My solution:

$ U,V \subseteq P $ such that $\phi (U) = \epsilon; \; \phi(V)=\delta \\$ $\\ \Leftrightarrow \forall x\in S; \epsilon = 1 \Rightarrow \delta=1 \\ \Leftrightarrow \phi(U) \leq \phi(V)$

I do not understand where to use $0$ in this? According to my solution, this is order preserving.

Answer 1

For the order to not be preserved, we need to have two sets where $U\subseteq V$ but $\phi(U)\not\le\phi(V)$.

The latter is only possible when $\phi(U)=1$ and $\phi(V)=0$. But the former means that $U=S$, which, with $U\subseteq V$ implies $V=S$ and so $\phi(V)=1$ too - a contradiction!

Answer 2

I’m afraid that I can’t make any sense of your solution. You need to show that if $U,V\subseteq S$ and $U\subseteq V$, then $\varphi(U)\le\varphi(V)$. There are basically just two cases:

• If $V=S$, then $\varphi(V)=1$; $\varphi(U)$ is either $0$ or $1$, so $\varphi(U)\le 1=\varphi(V)$ no matter what $U$ is.
• If $V\ne S$, then $\varphi(V)=0$. Since $U\subseteq V$, we know that $U\ne S$, and therefore $\varphi(U)=0$, too. Thus, once again we have $\varphi(U)\le\varphi(V)$.