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In How to Prove It: A Structured Approach, 2nd Edition, page 192, the author introduces the following definitions of smallest and minimal elements of partial orders:

Definition 4.4.4. Suppose R is a partial order on a set A, B $\subseteq$ A, and b $\in$ B. Then b is called an R-smallest element of B (or just a smallest element if R is clear from the context) if $\forall$x $\in$ B(bRx). It is called an R-minimal element (or just a minimal element) if $\lnot$$\exists$x $\in$ B(xRb $\land$ x $\neq$ b).

Then the author introduces the following example:

Let L = { (x, y) $\in$ ℝ x ℝ | x ≤ y }, as before. Let B = { x $\in$ ℝ | x ≥ 7}. Does B have any L-smallest or L-minimal elements? What about the set C = { x ∈ ℝ | x > 7 }?

Solution:

Clearly 7 ≤ x for every x $\in$ B, so ∀x $\in$ B(7Lx) and therefore 7 is a smallest element of B. It is also a minimal element, since nothing in B is smaller than 7, so ¬∃x ∈ B(x ≤ 7 ∧ x $\neq$ 7). There are no other smallest or minimal elements. Note that 7 is not a smallest or minimal element of C, since 7 $\not\in$ C . According to Definition 4.4.4, a smallest or minimal element of a set must actually be an element of the set. In fact, C has no smallest or minimal elements.

The part about B makes perfect sense to me, but I'm confused about:

In fact, C has no smallest or minimal elements

As far as my understanding goes, C does have an L-smallest element, which happens to be 8 (the example says that 7 is not a smallest/minimal element of C, which is obvious since 7 is not a member of C):

$$\forall x \in C(8Lx)$$

Which is obviously true since 8 ≤ 8, 8 ≤ 9, 8 ≤ 10, and so on.

Also, 8 looks like an L-minimal element of C as well, since:

$$\lnot \exists x \in C(xL8 \land x \neq 8)$$

The only element of C which is smaller or equal to 8 is 8, but 8 = 8.

Why does the author says that C has no smallest or minimal elements?

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  • $\begingroup$ Could it be the case that the author is choosing $x \in \mathscr{R}$ and not integers? $\endgroup$ – OnceUponACrinoid Jun 7 '15 at 3:07
  • $\begingroup$ B is not part of L sir $\endgroup$ – Hassan Jun 7 '15 at 3:07
  • $\begingroup$ @OnceUponACrinoid Not sure if it's what you're asking about, but R in the definition stands for an arbitrary partial order, while R in the other sections of the post stands for the set of real numbers. $\endgroup$ – jviotti Jun 7 '15 at 3:17
  • $\begingroup$ @Hassan L is a relation on ℝ, while B $\subseteq$ ℝ. $\endgroup$ – jviotti Jun 7 '15 at 3:18
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    $\begingroup$ @jviotti My point being there is no real number which can serve as a minimal element. 8 is the smallest natural number. $\endgroup$ – OnceUponACrinoid Jun 7 '15 at 3:18
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There is no real number which can serve as a minimal element. 8 is the smallest natural number.

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  • $\begingroup$ It doesn't seem that the question is restricted to natural number. $\endgroup$ – user99914 Jun 7 '15 at 3:47
  • $\begingroup$ Which was exactly my point if you read the comments above. $\endgroup$ – OnceUponACrinoid Jun 7 '15 at 3:48
  • $\begingroup$ @OnceUponACrinoid : It's unfortunate that I marked down your answer, as when doing review the comment column is not that clearly shown (please accept my apology). However , I suggest that you state clearly the misunderstanding of the OP in the answer, and say why this answer actually answers the question, which makes your answer more complete. $\endgroup$ – user99914 Jun 7 '15 at 3:56

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