1. ## Little isomorphism problem

I have this kind of exercise:

Let < be the partial order in the set A and define function $\displaystyle F:A \rightarrow P(A)$ by determining $\displaystyle F(a) = \{x \in A \mid x \leq a\}$, when $\displaystyle a \in A$.

Let $\displaystyle S = ran(F)$. Show that F is the isomorphism between structures $\displaystyle <A, < >$ and $\displaystyle <S, \subset >$.

Okey, my problem is how to mark function between A and S. I think I can't do it like $\displaystyle F:A\rightarrow S$, because I have already defined F before.

Rest of the exercise I can handle, I just have to show that the function between A and S is bijection and that $\displaystyle \forall x, y \in A(x <_A y \leftrightarrow "function"(x) \subset "function(y)")$.

2. Originally Posted by Ester
I have this kind of exercise:

Let < be the partial order in the set A and define function $\displaystyle F:A \rightarrow P(A)$ by determining $\displaystyle F(a) = \{x \in A \mid x \leq a\}$, when $\displaystyle a \in A$.

Let $\displaystyle S = ran(F)$. Show that F is the isomorphism between structures $\displaystyle <A, < >$ and $\displaystyle <S, \subset >$.

Okey, my problem is how to mark function between A and S. I think I can't do it like $\displaystyle F:A\rightarrow S$, because I have already defined F before.

Rest of the exercise I can handle, I just have to show that the function between A and S is bijection and that $\displaystyle \forall x, y \in A(x <_A y \leftrightarrow "function"(x) \subset "function(y)")$.
I'm assuming that $\displaystyle \text{ran}(F)=F(A)$.

What are you having trouble with? Isn't it this simple:

Spoiler:

So, we must merely show that $\displaystyle F:A\to\mathcal{P}(A)$ is an order embedding.

To see that it's injective suppose that $\displaystyle F(a)=F(a')$ then $\displaystyle \left\{x\in A:x\leqslant a\right\}=\left\{x\in A:x\leqslant a'\right\}$. But, $\displaystyle a\leqslant a\implies a\in F(a)\implies a\in F(a')\implies a\leqslant a'$ similar logic implies that $\displaystyle a'\leqslant a$ from where injectivity follows.

Now, to see that $\displaystyle a\leqslant a'\implies F(a)\subseteq F(a')$ we must merely note that if $\displaystyle x\in F(a)$ then $\displaystyle x\leqslant a\leqslant a'$ and so $\displaystyle x\in F(a')$. The conclusion follows.

3. Originally Posted by Ester
Let < be the partial order in the set A and define function $\displaystyle F:A \rightarrow P(A)$ by determining $\displaystyle F(a) = \{x \in A \mid x \leq a\}$, when $\displaystyle a \in A$.

Let $\displaystyle S = ran(F)$. Show that F is the isomorphism between structures $\displaystyle <A, < >$ and $\displaystyle <S, \subset >$.

Okey, my problem is how to mark function between A and S. I think I can't do it like $\displaystyle F:A\rightarrow S$, because I have already defined F before.
I am confused by you question. You are told to show that $\displaystyle F:A \leftrightarrow S$.
Because $\displaystyle F$ defines $\displaystyle S$ it must be a surjection.
Suppose that $\displaystyle F(a)=F(b)$ implies that $\displaystyle \left\{ {x:x \leqslant a} \right\} = \left\{ {y:y \leqslant b} \right\}$.
Then note that $\displaystyle a\in F(a)~~b\in F(b)$, use anti-symmetry.
Moreover, if $\displaystyle a\le b$ then $\displaystyle a\in F(b)$.

Can you write it all out?