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Exercise 83 - Chapter 3

Let \( (P, \le) \) be a preorder, and suppose that \( x, y \in P \). Show that their product and their meet are equivalent, \( (x \times y) \cong (x \wedge y) \).

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  • 1.

    By definition of product (Def. 3.81), there are morphisms\(\pi_1:x\times y\to x\), \(\pi_2:x\times y\to 2\). So, \(x\leq(x\times y)\) and \(y\leq(x\times y)\). This means that \(x\times y\) is an upper bound for \(x\) and \(y\). By the universal property of products, any \(Z\) such that there are morphisms \(f:Z\to x\) and \(g:Z\to y\) has a unique morphism into \(x\times y\). Thus, \(x\times y\) is a least upper bound for \(x\) and \(y\), i.e., it equals \(x\wedge y\).

    Comment Source:By definition of product (Def. 3.81), there are morphisms\\(\pi_1:x\times y\to x\\), \\(\pi_2:x\times y\to 2\\). So, \\(x\leq(x\times y)\\) and \\(y\leq(x\times y)\\). This means that \\(x\times y\\) is an upper bound for \\(x\\) and \\(y\\). By the universal property of products, any \\(Z\\) such that there are morphisms \\(f:Z\to x\\) and \\(g:Z\to y\\) has a unique morphism into \\(x\times y\\). Thus, \\(x\times y\\) is a least upper bound for \\(x\\) and \\(y\\), i.e., it equals \\(x\wedge y\\).
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