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sav08:lattices [2011/05/03 16:46] vkuncak |
sav08:lattices [2012/04/01 21:04] evka |
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| **Lemma:** In a lattice every non-empty finite set has a lub ($\sqcup$) and glb ($\sqcap$). | **Lemma:** In a lattice every non-empty finite set has a lub ($\sqcup$) and glb ($\sqcap$). | ||
| - | **Proof:** is by ++| **induction!** ++ \\ | + | **Proof:** is by induction!\\ |
| Case where the set S has three elements x,y and z:\\ | Case where the set S has three elements x,y and z:\\ | ||
| Let $a=(x \sqcup y) \sqcup z$. \\ By definition of $\sqcup$ we have $z \sqsubseteq a$ and $ x \sqcup y \sqsubseteq a $.\\ | Let $a=(x \sqcup y) \sqcup z$. \\ By definition of $\sqcup$ we have $z \sqsubseteq a$ and $ x \sqcup y \sqsubseteq a $.\\ | ||
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| Note: if you know that you have least upper bounds for all sets, it follows that you also have greatest lower bounds. | Note: if you know that you have least upper bounds for all sets, it follows that you also have greatest lower bounds. | ||
| - | **Proof:** ++|by taking the least upper bound of the lower bounds. Converse also holds, dually.++ | + | **Proof:** by taking the least upper bound of the lower bounds. Converse also holds, dually. |
| **Example:** Every subset of the set of real numbers has a lub. This is an axiom of real numbers, the way they are defined (or constructed from rationals). | **Example:** Every subset of the set of real numbers has a lub. This is an axiom of real numbers, the way they are defined (or constructed from rationals). | ||