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Math Help - 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 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)").
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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 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:


    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.

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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 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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