# Thread: Little isomorphism problem

1. ## Little isomorphism problem

I have this kind of exercise:

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

Let $S = ran(F)$. Show that F is the isomorphism between structures $$ and $$.

Okey, my problem is how to mark function between A and S. I think I can't do it like $F:A\rightarrow S$, 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 $\forall x, y \in A(x <_A y \leftrightarrow "function"(x) \subset "function(y)")$.

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

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

Let $S = ran(F)$. Show that F is the isomorphism between structures $$ and $$.

Okey, my problem is how to mark function between A and S. I think I can't do it like $F:A\rightarrow S$, 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 $\forall x, y \in A(x <_A y \leftrightarrow "function"(x) \subset "function(y)")$.
I'm assuming that $\text{ran}(F)=F(A)$.

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

Spoiler:

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

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

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

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

Let $S = ran(F)$. Show that F is the isomorphism between structures $$ and $$.

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

Can you write it all out?