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Suppose $C(X)$ is the set of functions $f:X \rightarrow \mathbb{R}$.

A bijection $T:C(X) \rightarrow C(Y)$ is called order isomorphism if $f \leq g$ if and only if $Tf \leq Tg$.

$f \leq g$ means that $f(x) \leq g(x)$ for all $x \in X$. Similarly, $f < g $ means that $f(x) < g(x)$ for all $x \in X$.

Question: If $f < g$, then $Tf < Tg$?

The statement seems correct but I don't know how to prove it.

If the statement above is not correct, then is it possible to obtain $f \geq 0 \Rightarrow Tf \geq 0$?

Idonknow
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  • What is $A(X)$? – anomaly Jun 26 '15 at 03:52
  • What is meant here by $f \le g$ and $f < g$? Is it that $f(x)\le g(x)$ and $f(x) < g(x)$ for all $x$, resp.? Or are you imposing some other order on $A(X)$? Also why introduce $A(X)$ in the first place? It seems like you're just restating $C(X)$.. – Cameron Williams Jun 26 '15 at 04:03

2 Answers2

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First of all, one usually defines $f < g$ to mean $f \leq g$ and $f \neq g$. This is different from your definition, which is a little confusing. For the sake of clarity, I will write $f \prec g$ whenever $f(x) < g(x)$ holds for all $x \in X$. (Now we have $f \preceq g$ if and only if $f \prec g$ or $f = g$. Note that $f \preceq g$ implies $f \leq g$, but not the other way around.)

For your first question, the answer is yes. Note that $C(X)$ is a lattice and that we have $$ (f \wedge g)(x) = \min(f(x), g(x)),\qquad\text{for all $x\in X$}. $$ Furthermore, note that an order isomorphism between lattices automatically is a lattice isomorphism. In other words, we have $T(f \wedge g) = Tf \wedge Tg$ for all $f,g\in C(X)$. Now suppose that $f,g\in C(X)$ are given with $f \prec g$ but $Tf \not\prec Tg$, then there is some $y_0\in Y$ such that $Tf(y_0) = Tg(y_0)$ holds. We define $h \in C(Y)$ by $$ h(y) = \begin{cases} Tf(y) + 1,&\quad\text{if $y = y_0$}; \\ Tf(y),&\quad\text{if $y \neq y_0$}. \end{cases} $$ Now set $h' = T^{-1}(h)$, then we have $$ T(h' \wedge g) = h \wedge Tg = Tf. $$ Since $T$ is injective, it follows that $h' \wedge g = f$ holds. Since we have $f(x) = \min(h'(x), g(x)) < g(x)$ for all $x\in X$, it follows that $h' = f$ must hold. But now we have $T(h') = h \neq Tf$, which is a contradiction.

For the second question, the answer is no. Suppose that $X = Y$ holds and consider the function $T : C(X) \to C(Y)$ given by $f \mapsto f - 1$, that is: $$ (Tf)(x) = f(x) - 1. $$ Then clearly $T$ is an order isomorphism, but we have $T(0) \not\geq 0$.

Josse van Dobben de Bruyn
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You are correct:
$$ \def\imp{\quad\Longrightarrow\quad} f < g \imp f \le g \imp Tf \le Tg. $$ But, if $Tf = Tg$, then $f=g$ (since $T$ is an injection), which contradicts the fact that $f < g$.

Sammy Black
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  • We can have $f \leq g$ without $f = g$ or $f < g$ when $f$ and $g$ are functions. – Ben Grossmann Jun 18 '15 at 03:57
  • Oh, right: $f \le g$ means that $f(x) \le g(x)$ for all $x \in X$, huh? – Sammy Black Jun 18 '15 at 06:02
  • I don't think your proof is correct as what @Omnomnomnom mentioned. – Idonknow Jun 25 '15 at 13:33
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    Indeed, the proof is mistaken for the reason mentioned above: $\le$ is an order on the set of functions, but it is not the order associated to the preorder $<$. Had it been so, your reasoning would have been correct. The notation tricked you. Rather than getting downvoted, you'd better delete your answer. – Alex M. Jun 25 '15 at 14:01