Little isomorphism problem

**Ester** · April 27th 2010, 07:51 AM

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 $<A, < >$ and $<S, \subset >$ .

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 $\forall x, y \in A(x <_A y \leftrightarrow "function"(x) \subset "function(y)")$ .

**Drexel28** · April 27th 2010, 08:13 AM

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 $<A, < >$ and $<S, \subset >$ .

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:

**Plato** · April 27th 2010, 08:18 AM

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 $<A, < >$ and $<S, \subset >$ .

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?

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