<p>Given a complete partial order $(D,\, {\leq})$ and a Scott-continuous function $F\colon D \rightarrow D$ and some fixed point $X = F(X)$. </p>

<p>Are there any criteria that ensure that $X = \mathsf{lfp} F$ where $\mathsf{lfp} F$ is the least fixed point of $F$?</p>

<p>Furthermore, are there any criteria that ensure that $Y \leq \mathsf{lfp} F$ for an arbitrary point $Y$?</p>

CC BY-SA 3.0
2015-11-12T17:01:32.597Z
You mean without computing the LFP? To check if y is less than LFP you can check if for some n $y < F^n(\bot)$ which seems only semidecodable in general.
Kaveh
CC BY-SA 3.0
2015-11-12T18:00:24.147Z
Yes, I mean without computing or knowing the LFP. I'm looking for something more elegant than finding such an $n$. The method need not be complete, just sound.
zander
<domain-theory>
Criteria for being below the least fixed point
zander
# Criteria for being below the least fixed point

(Asked by: username_0 on 2015-11-12)

Given a complete partial order $(D,\, {\leq})$ and a Scott-continuous function $F\colon D \rightarrow D$ and some fixed point $X = F(X)$.

Are there any criteria that ensure that $X = \mathsf{lfp} F$ where $\mathsf{lfp} F$ is the least fixed point of $F$?

Furthermore, are there any criteria that ensure that $Y \leq \mathsf{lfp} F$ for an arbitrary point $Y$?

**Tags:** domain-theory

> **Comments:**
> **username_1**: You mean without computing the LFP? To check if y is less than LFP you can check if for some n $y < F^n(\bot)$ which seems only semidecodable in general.  
> **username_0**: Yes, I mean without computing or knowing the LFP. I'm looking for something more elegant than finding such an $n$. The method need not be complete, just sound.  

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## 0 answers

