Are there any sets A for which [tex](\mathcal{P}(A), \subseteq)[/tex] is totally ordered? Prove your answer.(adsbygoogle = window.adsbygoogle || []).push({});

To be courteous, I will include the definitions for partial ordering and total ordering.

A relation is a partial order if the relation is reflexive, antisymmetric, and transitive. (in this case, the notation [tex](\mathcal{P}(A), \subseteq)[/tex] just denotes that [tex] \mathcal{P}(A) [/tex] under the relation [tex] \subseteq[/tex] is a partially ordered set.

A partially ordered set A with partial order [tex] \leq [/tex] is said to be totally ordered if given any two elements a and b in A, either [tex] a \leq b [/tex] or [tex] b \leq a [/tex].

So, to attempt this problem. I tried making up examples first. The only set A that I could come up with for which [tex](\mathcal{P}(A),\subseteq)[/tex] is totally ordered is a set with one element. Just to see this, let A = {1}. In this case, [tex] \mathcal{P}(A) = [/tex]{[tex]\emptyset[/tex] , {1}}. So, if I pick any two elements, a and b. [tex] a \leq b [/tex] or [tex] b \leq a [/tex]. For example, if I'd pick [tex] \emptyset [/tex] and {1}. Then, [tex] \emptyset \subseteq [/tex]{1}. So I think it works. I don't know if there are any other sets, A, where this works or if I'm even thinking about this correctly. (once I figure out all the sets, I'll attempt to put a proof up). Thanks in advance.

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# Totally Ordered Relations

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