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Both sides previous revision Previous revision Next revision | Previous revision Next revision Both sides next revision | ||
sav08:homework10 [2008/04/30 15:54] vkuncak |
sav08:homework10 [2008/04/30 16:56] vkuncak |
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r = \{ (F_1,F_2) \mid \models F_1 \rightarrow F_2 \} | r = \{ (F_1,F_2) \mid \models F_1 \rightarrow F_2 \} | ||
\] | \] | ||
- | Check whether $r$ is reflexive, symmetric, and transitive relation. | + | Check whether $r$ is reflexive, antisymmetric, and transitive relation. |
===== Problem 2 ===== | ===== Problem 2 ===== | ||
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. | ||
- | |||
===== Problem 3 ===== | ===== Problem 3 ===== | ||
- | Let $A = [0,1] = \{ x \in \mathbb{R} \mid 0 \le x \le 1 \}$ be the interval of real numbers. Recall that, by definition of real numbers and complete lattice, $(A,\le)$ is a complete lattice with least lattice element $0$ and greatest lattice element $1$. | + | Let $A = [0,1] = \{ x \in \mathbb{R} \mid 0 \le x \le 1 \}$ be the interval of real numbers. Recall that, by definition of real numbers and complete lattice, $(A,\le)$ is a complete lattice with least lattice element $0$ and greatest lattice element $1$. Here $\sqcup$ is the least upper bound operator on sets of real numbers, also called //supremum// and denoted //sup// in real analysis. |
Let function $f : A \to A$ be given by | Let function $f : A \to A$ be given by | ||
\[ | \[ | ||
f(x) = \left\{\begin{array}{l} | f(x) = \left\{\begin{array}{l} | ||
- | \frac{1}{2} + \frac{1}{4}x, \mbox{ if } x \in [0,\frac{2}{3}) | + | \frac{1}{2} + \frac{1}{4}x, \mbox{ if } x \in [0,\frac{2}{3}) \\ |
+ | \ \\ | ||
\frac{3}{5} + \frac{1}{5}x, \mbox{ if } x \in [\frac{2}{3},1] | \frac{3}{5} + \frac{1}{5}x, \mbox{ if } x \in [\frac{2}{3},1] | ||
\end{array}\right. | \end{array}\right. | ||
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=== Part a) === | === Part a) === | ||
- | Prove that $f$ is monotonic. | + | Prove that $f$ is monotonic and injective (so it is strictly monotonic). |
=== Part b) === | === Part b) === | ||
Line 36: | Line 36: | ||
=== Part c) === | === Part c) === | ||
- | Define $iter(x) = \sqcup \{ f^n(x) \mid n \in \{0,1,2,\ldots \}$ where $\sqcup$ is the least uppser bound operator on sets of real numbers (also called 'supremum' and denoted 'sup', and in this case 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 $iter(0)$. Is $iter(0)$ a fixpoint? Compute $iter(iter(0))$. 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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