Consider the poset \((A; \preceq)\) and \( S \subseteq A\).
\(a \in A\) is a lower (upper) bound of \(S\) if \(b \preceq a\) (\(b \succeq a) \) for all \(b \in S\)
Commit: f46c51c4 - Update deck.json
Author: Jonas B <65017752+Scr1pting@users.noreply.github.com>
Date: 2025-12-31T14:37:29+01:00
Changes: 4 note(s) changed (0 added, 4 modified, 0 deleted)
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| Text | Consider the poset \((A; \preceq)\) and \( S \subseteq A\).<br><br><div>\(a \in A\) is a {{c1::<b>lower (upper) bound</b> of \(S\)}} if {{c2::\( |
Consider the poset \((A; \preceq)\) and \( S \subseteq A\).<br><br><div>\(a \in A\) is a {{c1::<b>lower (upper) bound</b> of \(S\)}} if {{c2::\(a \preceq b\) (\(a \succeq b) \) for all \(b \in S\)}}</div> |
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| Extra | Examples: \(\mathbb{Z}, \mathbb{R}\)<div>Counterexample: \(\mathbb{Z}_m, m\) not prime</div> | <div><i>A domain of elements behaving like integers.</i></div><br>Examples: \(\mathbb{Z}, \mathbb{R}\)<div>Counterexample: \(\mathbb{Z}_m, m\) not prime</div> |
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| Text | Russell's Paradox proposes the (problematic) set \(R=\) {{c1::\(\{ A \mid A \notin A\}\)}}. | Russell's Paradox proposes the (problematic) set \(R=\) {{c1::\(R = \{ A \mid A \notin A \}\)}}. |
| Extra | <div><ol><li> <div><b>Assume R contains itself</b><b></b></div> <div>Then R should <i>not</i> contain itself (because R only contains sets that do not contain themselves).</div> <div>➜ Contradiction.</div> </li><li> <div><b>Assume R does not contain itself</b><b></b></div> <div>Then it <i>does</i> meet the rule for membership in R, so it should contain itself.</div> <div>➜ Contradiction.</div></li></ol></div><i>A barber that shaves all and only those men who do not shave themselves. Does he shave himself?</i><br> |
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| Text | The <i>Hasse diagram</i> of a poset \((A; \preceq)\) is {{c1::the directed graph whose vertices are the elements of \(A\) and where there is an edge from \(a\) to \(b\) if and only if \(b\) covers \(a\). |
The <i>Hasse diagram</i> of a poset \((A; \preceq)\) is defined as {{c1::the directed graph whose vertices are the elements of \(A\) and where there is an edge from \(a\) to \(b\) if and only if \(b\) covers \(a\).}} |