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I found a question and two answers that both are very complete but I cannot understand the meaning behind some of the symbols that are used. They are different from what I am familiar with.

The question can be found here.

Because questions must be self-contained, I will quote the section of the answer (from Brian M. Scott) that I am confused with.

Let $\langle P,\le\rangle$ be a partial order. Let $$\mathscr{L}=\big\{\langle X,\preceq_X\rangle:X\subseteq P\text{ and }\preceq_X\text{ is a linear order on }X\text{ extending }\le\big\}\;.$$

For $\langle X,\preceq_X\rangle,\langle Y,\preceq_Y\rangle\in\mathscr{L}$ define $\langle X,\preceq_X\rangle\sqsubseteq\langle Y,\preceq_Y\rangle$ iff $X\subseteq Y$ and $\preceq_Y\upharpoonright(X\times X)=\preceq_X$; clearly $\langle\mathscr{L},\sqsubseteq\rangle$ is a partial order. Let $\mathscr{C}$ be a chain in $\langle\mathscr{L},\sqsubseteq\rangle$.

What is the meaning of $\preceq_Y\upharpoonright(X\times X)=\preceq_X$? I understand that this must be stating that $\preceq_Y$ extends $\preceq_X$ by the context of the question but I still do not understand what the symbols each mean and how to interpret them.

Can someone also explain what the upper bound of the chain is? I understand that it must involve some sort of union of all the elements in the chain but it is not clear to me how to construct this union or if I can even define such a union.

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user157227
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1 Answers1

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The $\LaTeX$ code \restriction for $\restriction$ itself provides a clue to the meaning of $\preceq_Y\upharpoonright(X\times X)$ and it is $$\preceq_Y\upharpoonright(X\times X):=\preceq_Y\cap \,(X\times X).$$

Do not forget that binary relations, in particular partial orders, are just sets of ordered pairs.

The obvious choice for an upper bound of $\mathscr C$ is $\cup \,\mathscr C$. To see that $\cup \,\mathscr C$ is an upper bound of $\mathscr C$ (in $\langle\mathscr{L},\sqsubseteq\rangle$), you must check that $\cup \,\mathscr C\in \langle\mathscr{L},\sqsubseteq\rangle$ and that for all $\langle A, \preceq _{A}\rangle\in \mathscr C$ it holds that $\langle A, \preceq _{A}\rangle\sqsubseteq \cup \, \mathscr C$. Note that $\cup \,\mathscr C$ can be written as an ordered pair whose each entry is itself a union of sets.

Git Gud
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  • I think my entire issue is that I did not know that $\preceq_Y$ is a set of ordered pairs. My book has not mentioned this. I assume it can be defined as the set of all $(a,b)$ where we want to say $a \leq b$. Is this correct? This makes the union and the upper bound of the chain entirely more sensible. On another note, \upharpoonright also provides a $\upharpoonright$, if you were wondering. – user157227 Aug 25 '14 at 18:48
  • A tangential question, why use $\restriction$ when we can simply use $\cap$? – user157227 Aug 25 '14 at 18:49
  • You are correct in your guess at the definition. There's one detail missing. We say $(a,b)\in \leq$ instead of $a\leq b$ and this is so for all binary relations. I noticed that you used a different $\LaTeX$ code for the symbol and that's why I made a point of letting you know about the alternative. Why use $\restriction$? I'm just guessing here, but I'd say because it is for the meaning of the symbol itself 'restriction'. We're going from something big, to something smaller. – Git Gud Aug 25 '14 at 18:57
  • Thanks for the help, it all makes sense now. – user157227 Aug 25 '14 at 19:02
  • @user157227 You're welcome. – Git Gud Aug 25 '14 at 19:05