Prove S is a linear ordering on A∪{b} Assume that (A,R) is a linearly ordered set, and that b∉A. Define S on A∪{b} by xSy iff x,y∈A and xRy, OR x∈A and y=b, OR x=y=b. Prove that S is a linear order on A∪{b}. 
If A were the set of real numbers and R were usual ordering on real numbers, what could be said of b?
Proof: Let x, y ∈ A∪{b}. Then x,y∈A OR x∈A, y=b OR x=y=b.
Case 1: If x, y ∈ A, then xRy or yRx, since A is a linearly ordered set. Thus xSy or ySx.
Case 2: If x∈A, y=b, then .......
Case 3: If x=y=b, then xSy and ySx. 
If A were the set of real numbers and R were usual ordering on real numbers, then b=∞?
Am I doing it right? How do I proceed case 2?
 A: To prove that $S$ is a linear order, we need to know three things:


*

*$S$ is reflexive

*$S$ is transitive

*For all $a,b\in A$, $aSb$ or $bSa$


We already know that $R$ satisfies these properties.
In your question, you proved most of number 3 (all you need to add is that, in case 2,  $xSy$, so clearly one of $xSy$ and $ySx$ holds).
For the first point, we know that for all $a\in A, aSa$ (because $aRa$), and we know that $bSb$, so, for all $x\in A\cup\{b\}, xSx$ (so $S$ is reflexive).
For the second point, we know that, for all $x,y,z \in A\cup \{b\}$, if $xSy$ and $ySz$, then $xSz$. If all three of $x,y$ and $z$ are in $A$, then this holds because $R$ is transitive, so we only need to prove more in the case that $x=b$, $y=b$, or $z=b$.
If $x=b$, then for $xSy$ and $ySz$ to hold, $y$ and $z$ must both also equal $b$, so, since $x=z=b$, $xSz$.
If $x\in A$, and $y=b$, then $xSy$, and for $ySz$ to hold, $z$ must equal $b$, so $xSz$.
Finally, if $x\in A$, $y\in A$, and $z=b$, then $xSz$
Therefore, $S$ is reflexive, trasitive, and total, and so is a linear order. You are correct that if $A=\Bbb{R}$, then $b$ is conceptually "$\infty$."
