The correct title must be: Proof of Theorem 90 of Appendix of Kelley's book General Topology or, Proving a theorem about $\mathcal R$-sections or Prove of this prove that proof aboutTheorem 90 of Appendix of Kelley's book General Topology .

Proposition. Let $(X,\mathcal R)$ be an ordered set (or class). If $Y\subset X$ is an $\mathcal R$-section and every element $y\in Y$ is also an $\mathcal R$-section of $X$, then $\bigcup Y$ and $\bigcap Y$ are $\mathcal R$-sections of $X$.

EDIT 1: The statement was misunderstood by me. The real porposition is:

Let $I$ non-empty (possibly $I$ is a set). If every element $y\in I$ is an $\mathcal R$-section of $X$, then $\bigcup I$ and $\bigcap I$ are also.

So the comment of @WilliamElliot is unnecesary, because in any momment we have said $I\subset X$. And probably it may be false. In my opinion, $I$ plays the role of an index set.

Addendum 1. This statement corresponds to the Appendix of J. L. Kelley General Topology, Theorem 90 (p. 264).

EDIT 2: Below is the rest of my question. Whit the new remarks it has no sense. My problem is that I can't understand why (or how) an element $y\in Y$ can be an $\mathcal R$-section of $X$ if $y$ is not a subset of $X$ (as least as far as I know). It would have sense for me if the text said $\{y\}$, which is actually a subset of $X$.

The same applies to $\bigcup Y$ and $\bigcap Y$: by DEFINITION, both are sets (classes) formed by elements of sets in $Y$:

$$ \bigcup Y= \{x:\exists y\in Y \mbox{ such that } x\in y\}\\ \bigcap Y= \{x|x\in y \;\forall y\in Y\}. $$

Again, to be an $\mathcal R$-section, we should consider $\{\bigcup Y\}$ and $\{\bigcap Y\}$.

Addendum 2. (Real questions) Once I have understood the proposition, the result seems me evident. However, I don't know how how to prove it rigorously. I know I have a collection of sets $\{y:y\in I\}$ such that

$$ \mbox{if } y_1,y_2\in Y \mbox{ with } y_1\neq y_2 \Longrightarrow y_1\subset y_2 \mbox{ or } y_2\subset y_1 , $$

So I actually have a descendent chain of sets

$$ y\supset y' \supset y'' \supset \cdots $$

1.- Proof for $\bigcup I$. With the above in mind, I think $bigcup I = y$ which is an $\mathcal R$-section by hypothesis.

2.- Proof for $\bigcap I$. I think the definition of $\mathcal R$-section avoids there is some $y\in I$ such that $y=\emptyset$. So I think all them has a common element, the least element $y_0$, and thus $\bigcap I$ is non-empty. Now, let $x\in X$ and $y\in\bigcap I$ with $x\mathcal R y$ and suppose $x\notin \bigcap I$. That implies that $x$ is less than the least element $y_0$. But then, $x$ wouldn't below either to $y\in I$, which is a contradiction, because $y$ was an $\mathcal R$-section$. So $\bigcap I$ is an $\mathcal R$-section.

Questions about this proof:

1.- Is the proof of $bigcup I$ correct?

2.- Is the proof of $\bigcap I$ correct?

3.- In 2., do you think I need to proove $y_0\in \bigcap I$? Is there a shorter proof?

4.- If all the inclusions are proper, then is it possible to show that $\bigcap I=\{y_0\}$?

Definition. Let $\mathcal R$ be a well-order in $X$. An $\mathcal R$-section is $Y\subset X$ such that if $x\in X$ and $y\in Y$ with $x\mathcal R y$, then $x\in Y$.

Informally, a set (class) $Y$ is said to be an $\mathcal R$-section if there is no element in $X\setminus Y$ that precedes the elements of $Y$.

Final Remark. I have prefered keep the old question and add the edits and corrections, because I think that is more proper. Sorry if it is a bad idea.

Thanks a lot for your patience.

  • $\begingroup$ There seems to be a lot of typos here. And what is a $\mathcal{R}$-section of an ordered set $(X,\mathcal{R})$? $\endgroup$ – amrsa Jan 29 '18 at 14:06
  • $\begingroup$ @amrsa: See the edit $\endgroup$ – Dog_69 Jan 29 '18 at 22:48
  • 1
    $\begingroup$ Take at look at the construction of the ordinals. 0 is empty, 1 = {0}, 2 = {0,1}, 3 = {0,1,2}, etc. $\endgroup$ – William Elliot Jan 30 '18 at 2:53
  • $\begingroup$ So a section is what is otherwise called an order-ideal; notice however that in Wikipedia, Ideal (order theory) requires more conditions. $\endgroup$ – amrsa Jan 30 '18 at 11:36
  • $\begingroup$ I also think you should see if William Elliot's suggestion applies (this depends on the context of in which it is being used). I don't like this kind of language abuse (although it makes perfect sense when studying ordinals), but perhaps $y$ is, in this context supposed to mean $\{x \in X : x \leq y\}$, or similar... $\endgroup$ – amrsa Jan 30 '18 at 11:40

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