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Let $\geq$ be a partial order on a set $X$. For each $x\in X$, let $F(x)\equiv \{a\in X:a\geq x\}$ and $\mathcal{F}\equiv\{F(x):x\in X\}$. I want to show that $F:X\to\mathcal{F}$ is a bijection.

First I have shown that it is an injection. Take $x,x'\in X, x\neq x'$. Suppose by contradiction $F(x)=F(x').$ Since $x\in F(x),x\in F(x')\implies x\geq x'$. Similarly, $x'\geq x$. Thus, $x=x'$, which is a contradiction. thus, $F(x)\neq F(x')$.

I'm having trouble showing that $F$ is a surjection. How do I do that?

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  • To me it looks like $\mathcal{F}$ is defined as the range of the function in question? What am I missing? What field is this from? – martinkleins Nov 07 '23 at 04:40

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I guess I know what you are concerning. It's not that obvious that for any $F\in \mathcal F$ you can always find an element in X such that f(x)=F. I think the safer way is to find the ``smallest'' element for any given F, which means that any $y\in F, y\ge x$, then $F=f(x)$. Is that right?

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