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sav08:homework10 [2008/04/30 16:10]
piskac
sav08:homework10 [2008/04/30 16:56]
vkuncak
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 Let $(A,​\sqsubseteq)$ be a [[:partial order]] such that every set $S \subseteq A$ has the greatest lower bound. ​ Prove that then every set $S \subseteq A$ has the least upper bound. Let $(A,​\sqsubseteq)$ be a [[:partial order]] such that every set $S \subseteq A$ has the greatest lower bound. ​ Prove that then every set $S \subseteq A$ has the least upper bound.
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 ===== Problem 3 ===== ===== Problem 3 =====
  
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 Define $iter(x) = \sqcup \{ f^n(x) \mid n \in \{0,​1,​2,​\ldots \}\}$. ​ (This is in fact equal to $\lim_{n\to\infty} f^n(x)$.) Define $iter(x) = \sqcup \{ f^n(x) \mid n \in \{0,​1,​2,​\ldots \}\}$. ​ (This is in fact equal to $\lim_{n\to\infty} f^n(x)$.)
  
-Compute $iter(0)$ (if you can, prove that the computed value is correct by definition of $iter$). ​ Is $iter(0)$ a fixpoint of $f$?  Is $iter(iter(0))$ a fixpoint of $f$? Is $f$ an $\omega$-continuous function?+Compute $iter(0)$ (prove that the computed value is correct by definition of $iter$, that is, that the value is indeed $\sqcup$ of the set of values).  Is $iter(0)$ a fixpoint of $f$?  Is $iter(iter(0))$ a fixpoint of $f$? Is $f$ an $\omega$-continuous function?
  
 === Optional part d) === === Optional part d) ===
  
 Define a monotonic function $f : A \to A$ such that, for every natural number $k$, the value $iter^k(0)$ is not a fixpoint of $f$.  (It may be difficult to draw $f$.) Define a monotonic function $f : A \to A$ such that, for every natural number $k$, the value $iter^k(0)$ is not a fixpoint of $f$.  (It may be difficult to draw $f$.)
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