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A subclass $C$ of a partially ordered class is called $\textit{convex}$ if it satisfies the following condition: If $a\in C$ and $b\in C$ and $a\leq x\leq b$ then $x\in C$. Let $A$ and $B$ be partially ordered classes, let $f:A \rightarrow B$ be an increasing function, and let $C$ be a convex subclass of B. Prove that $f^{-1}(C)$ is a convex subclass of $A$.

Let $a, b\in f^{-1}(C)$ by definition of inverse image of $C$, then:

$a \in f^{-1}(C) \implies a\in A \land f(a)\in C$

$b \in f^{-1}(C) \implies b\in A \land f(b)\in C$

Suppose that $a \leq b \implies f(a) \leq f(b)$

But the problem is, how i can get the "$x$" for obtain $a\leq x \leq b$

sango
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  • Convexity says if $a,b \in C$ and $a \leq x \leq b$, then $x \in C$. So you need to start by assuming $a, b \in f^{-1}(C)$ and $a \leq x \leq b$. Then show $x \in f^{-1}(C)$. – kccu Nov 04 '16 at 04:00
  • @kccu Thanks man, you was make me feel like an idiot. – sango Nov 04 '16 at 04:08

1 Answers1

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Let $a,b \in f^{-1}(C)$ and $a\leq x \leq b$ with $x\in A$. The definition of inverse image says:

$a \in f^{-1}(C) \implies a\in A \land f(a)\in C$ and $b \in f^{-1}(C) \implies b\in A \land f(b)\in C$

By hypothesis $a\leq x \leq b \implies a\leq x \land x\leq b$, by hypothesis $f$ is a increasing function, then:

$a\leq x \land x\leq b \implies f(a)\leq f(x) \land f(x)\leq f(b) \implies f(a) \leq f(x) \leq f(b)$

$f(a),f(b)\in C$ and $C$ is convex, then: $f(x)\in C$

As $f(x) \in C \land x\in A$, by definition of inverse image $x\in f^{-1}(C)$

Then, $f^{-1}(C)$ is a convex subset of $A$

sango
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