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Let $\mathfrak{A}$ be a poset. For $a, b \in \mathfrak{A}$ we will denote $a \not\asymp b$ if only if there are a non-least element $c$ such that $c \leqslant a \wedge c \leqslant b$.

Let $\mathfrak{A}$, $\mathfrak{B}$ are posets. I call a pointfree funcoid a pair $\left( \alpha ; \beta \right)$ of functions $\alpha : \mathfrak{A} \rightarrow \mathfrak{B}$, $\beta : \mathfrak{B} \rightarrow \mathfrak{A}$ such that $$ \forall x \in \mathfrak{A}, y \in \mathfrak{B}: \left( y \not\asymp^{\mathfrak{B}} \alpha \left( x \right) \Leftrightarrow x \not\asymp^{\mathfrak{A}} \beta \left( y \right) \right) . $$

Conjecture If $\left( \alpha ; \beta \right)$ is a pointfree funcoid and $\alpha$ is a bijection $\mathfrak{A} \rightarrow \mathfrak{B}$, then $\alpha$ is an order isomorphism $\mathfrak{A} \rightarrow \mathfrak{B}$.

A weaker conjecture:

Conjecture If $\left( \alpha ; \beta \right)$ is a pointfree funcoid and $\alpha$ is a bijection $\mathfrak{A} \rightarrow \mathfrak{B}$ and $\beta$ is a bijection $\mathfrak{B} \rightarrow \mathfrak{A}$, then $\alpha$ is an order isomorphism $\mathfrak{A} \rightarrow \mathfrak{B}$.

If these conjectures are false, what additional conditions we may add to make them true? (Maybe, these are true for lattices? distributive lattices?)

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2 Answers 2

up vote 2 down vote accepted

Both conjectures are false for infinite posets. Take for instance any bijections of sets $\alpha,\beta$ ($\beta=\alpha^{-1}$ works fine) between $\mathbb{Z}$ and $\mathbb{Q}$. Because for all $m,n\in\mathbb{Z}$ and for all $r,s\in\mathbb{Q}$ we always have $m\asymp n$ and $r\asymp s$, there is really no condition imposed on $\alpha$ and $\beta$, but neither can be an order isomorphism.

Maybe the answer is affirmative if you look at finite posets.

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Let $\mathfrak{A}$ is a poset, let $\star a=\{ x\in\mathfrak{A} | x\not\asymp a \}$.

I will call a poset $\mathfrak{A}$ separable when $\star a=\star b \Leftrightarrow a=b$ for every $a,b\in\mathfrak{A}$.

Theorem Let $\left( \alpha ; \beta \right)$ is a pointfree funcoid from a separable poset $\mathfrak{A}$ to a separable poset $\mathfrak{B}$. If $\alpha$ is an injection, then $\alpha$ is an order embedding $\mathfrak{A} \rightarrow \mathfrak{B}$.

Proof Suppose $x \geqslant y$ but $\alpha (x) \ngeqslant \alpha (y)$.

Then by separability of $\mathfrak{B}$ there exist $z \not\asymp \alpha (y)$ such that $z \asymp \alpha (x)$.

Thus $\beta (z) \asymp x$ and $\beta (z) \not\asymp y$ what is impossible for $x \geqslant y$. $\Box$

Corollary Let $\left( \alpha ; \beta \right)$ is a pointfree funcoid from a separable poset $\mathfrak{A}$ to a separable poset $\mathfrak{B}$. If $\alpha$ is a bijection $\mathfrak{A} \rightarrow \mathfrak{B}$, then $\alpha$ is an order isomorphism $\mathfrak{A} \rightarrow \mathfrak{B}$.

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Err, both posets $\mathfrak{A}$ and $\mathfrak{B}$ need to be separable. Corrected. –  porton Jun 20 '12 at 17:40

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