Exercise 6.1.1

Give an example of a linearly ordered set ( L , < ) and an initial segment S of L which is not of the form { x x < a } , for any a L .

Answers

We claim that L = 𝑹 and S = { x L x 0 } with the usual order meet the criteria.

Proof. First, clearly L = 𝑹 is a linearly ordered set. So consider any a S and any x < a so that we have

x < a 0 .

Hence x S also so that by definition S is an initial segment of 𝑹 . Now suppose that S does have the form

S = { x L x < a }

for some a L . Since 0 0 clearly 0 S by the original definition so that by the above 0 < a . But now consider a 2 , which is clearly in L = 𝑹 . By the above a 2 < a since a > 0 so a 2 S but we also have 0 < a 2 (hence it is not true that a 2 0 ) since 0 < a so that by the original definition a 2 S . Since we have a contradiction it must be that S cannot be expressed in such a form. □

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2024-07-15 11:42
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