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# Exercise 83 - Chapter 3

edited June 2018

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

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\$$.