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Thread: Little isomorphism problem

  1. #1
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    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 $\displaystyle \forall x, y \in A(x <_A y \leftrightarrow "function"(x) \subset "function(y)")$.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Ester View Post
    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.

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  3. #3
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    Quote Originally Posted by Ester View Post
    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?
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