I'm trying to understand how the transfinite ordinals work, so I tried to come up with an example of a set of numbers with an element at a transfinite index:

Let the totally ordered set $X = \{\frac{1}{2},\frac{3}{4},\frac{7}{8},\dots\} \cup\{1\}$ be ordered according to the standard ordering of the rational numbers. Would I be correct in assuming that $X_\omega=1$ since it is the first number in the set with an infinite index? If not, what does it mean to find $X_\omega$ and which at index would $1$ be?

Likewise, if I ordered the set $Y=\{\frac{1}{2},\frac{3}{4},\frac{7}{8},\dots\}\cup\{1+\frac{1}{2},1+\frac{3}{4},1+\frac{7}{8},\dots\}\cup\{2\}$ in the same way, would $Y_{2\omega} = 2$? If not, what would be the index $i$ where $Y_i = 2$?

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    $\begingroup$ Yes, $X_\omega=1$ and $Y_{\omega+\omega}=2.$ However, according to the convention for ordinal multiplication that I learned, $$\omega+\omega=\omega2\ne2\omega=\omega.$$ Are they teaching a different notation nowadays? What book are you using? $\endgroup$
    – bof
    Jan 14 '17 at 4:06
  • $\begingroup$ @bof I'm just using Wikipedia and SE. If $\omega+\omega \neq 2\omega$, what would $Y_{2\omega}$ mean? $\endgroup$ Jan 14 '17 at 4:12
  • $\begingroup$ Like I said, $2\omega=\omega,$ so $Y_{2\omega}=Y_\omega=\frac32.$ In the traditional notation, $\omega2=\omega+\omega,$ and $2\omega=2+2+2+\cdots=\omega.$ They may seem backwards but it makes the laws of exponentiation come out right: $\alpha^{\beta+\gamma}=\alpha^\beta\alpha^\gamma$ and $(\alpha^\beta)^\gamma=\alpha^{\beta\gamma}.$ $\endgroup$
    – bof
    Jan 14 '17 at 4:33
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    $\begingroup$ @ostrichofevil $1/4 < 1/2$, $1/8 < 1/4$, and so on, so your set has no least element and is not well ordered. However, if you reverse the ordering of just that rational numbers of $X$ so that $1/4 > 1/2$ etc but $\forall x \in X: x = 1 \lor x < 1$, then it is a well-order and order-isomorphic to $\omega + 1$. $\endgroup$
    – Dan Simon
    Jan 14 '17 at 17:27
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    $\begingroup$ A linear order is a well-order iff every non-empty subset S has a least member within S. For a set to be well-ordered by any binary relation R, the relation R must be satisfy the conditions in the def'n of a linear order, and also be a well-order. $\endgroup$ Jan 14 '17 at 23:09

Edit. I was $\TeX$ blind and I took the order in which the sets were written above, not the one inherited from the real line. I'm modifying my answer according to the intent in the OP's comment above.

Some pieces of useful information:

  • Ordinals can be regarded as types of isomorphism of well-orders. Hence two well-orders of the same “shape” have the same ordinal.
  • The ordinal sum $\alpha+\beta$ is “$\alpha$, then $\beta$”: Take any well-order of type $\alpha$, stick a well-order of type $\beta$ to the right and that new well-order has type $\alpha+\beta$.
  • The product $\alpha\cdot\beta$ is “$\alpha$, $\beta$ times”: Place copies1 of $\alpha$ in a row so that the set of such copies are in order type $\beta$. Alternatively, take a copy of $\beta$ and replace each of its elements for a copy of $\alpha$.
  • The “transfinite index” of any element $x$ in a well-order (actually, this is called its rank) is the order type of the set of elements below $x$.

So, using your examples, the well-order $$ \textstyle X = \bigl\{\overbrace{\frac{1}{2},\frac{3}{4},\frac{7}{8},\dots}^{\text{type }\omega}\bigr\} \cup \overbrace{\{1\}}^{\text{type }1} $$ is of type $\omega+1$ and the rank of $1$ is $\omega$. In $Y$, $$ \textstyle Y=\bigl\{\overbrace{\frac{1}{2},\frac{3}{4},\frac{7}{8},\dots}^{\text{type }\omega}\bigr\}\cup\bigl\{\overbrace{1+\frac{1}{2},1+\frac{3}{4}}^{\text{type }2},1+\frac{7}{8},\dots\bigr\}\cup\{2\}, $$ the rank of $1+\frac{7}{8}$ is $\omega+2$, and $Y\setminus\{2\}$ is the result of replacing each element in a copy of the ordinal $2$ (say, $\{0,1\}$) by a copy of $\omega$ (namely, $\bigl\{\frac{1}{2},\frac{3}{4},\frac{7}{8},\dots\bigr\}$ and $\bigl\{1+\frac{1}{2},1+\frac{3}{4},1+\frac{7}{8},\dots\bigr\}$). Hence its order type is $\omega\cdot2$ (which equals $\omega+\omega$) and therefore the rank of $2$ is also $\omega\cdot2$.

Finally, for your question on a set being well-ordered by $\in$, you should learn about the von Neumann ordinals.

1 A copy of $\alpha$ is (for the purposes of this answer) a well-order of type $\alpha$.

  • $\begingroup$ If you were to keep doing this, in order to get the set $ A = \{\frac{1}{2},\frac{1}{4},\frac{1}{8},\dots\}\bigcup\,\{1+\frac{1}{2},1+\frac{1}{4},1+\frac{1}{8},\dots\}\bigcup\,\{2+\frac{1}{2},2+\frac{1}{4},2+\frac{1}{8},\dots\}\bigcup\dots\bigcup\,\{\omega\}$, would the rank of $\omega$ be $ \omega^2$? $\endgroup$ Jan 14 '17 at 22:18
  • $\begingroup$ @ostrichofevil Yes, if both of us pay attention to what @DanSimon said above! I'm going to fix my answer accordingly; I suggest you do the same. Btw, use \cup with no additional spacing for binary unions. $\endgroup$ Jan 15 '17 at 0:29
  • $\begingroup$ I'll do so. Thanks for the $\LaTeX$ tip! $\endgroup$ Jan 15 '17 at 1:26

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