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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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