Transfinite ordinals as indices of set

Question

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 $Y_i = 2$?

Answer 1

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.

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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.

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