Specific element in an infinite set

Question

I've had an idea for some proof involving infinite sets. Consider the infinite set $X$ and the element $x \in X$. According to the well-ordering theorem, any set can be well-ordered.

Would it be a valid to assign some index number $n$ to $x$ (we now write $x_n$), such that for all $x$ in $X$ we can say that $x_i < x_n$ if $i<n$ and $x_i > x_n$ if $i>n$? Both the elements and the ordering in and of itself are of arbitrary kind.

Edit: thanks for all the answers. Clarification: As part of a bigger problem I'm trying to show that there is a bijection $f:X\rightarrow X$ $\backslash \{x\}$ for the infinite set $X$ and the element $x \in X$ (which we call $x_n$ in the light of the above). It was my idea to define a function that maps some $x_i$ to itself if $i<n$ and any $x_i$ to $x_{i+1}$ if $i>n$ to circumvent $x_n$. So it is not necessary for me to find any least elements, use the standard ordering $\geq$ or use $\mathbb{N}$ as the index set.

Answer 1

I don't believe the ordering you've defined here, that is, an order of (<), is sufficient to ensure that any infinite set will suddenly have a least element. For example, take the set of the integers $\mathbb{Z}=\{...,-3,-2,-2,0,1,2,3,...\}$. Under the ordering you have described, which $i$ indexes the least element? For any index $i$ that you assign to the least element, I will find another index given by $i-1$ such that $x_{i-1} < x_i$. Even if you define a function $f:\mathbb{N} \rightarrow \mathbb{Z}$ such that $f$ places the integers in a one-to-one correspondence with the natural numbers, which is not difficult to do, the set of the integers still lacks a least element under the ordering of (<) you have defined.

From what I understand, in order to take an infinite set and make it well ordered, you have to change your definition of the ordering from the conventional (<). For example, in order to make $\mathbb{Z}$ well ordered, we define a new ordering, say ($\prec$), such that for any arbitrary $a,b \in \mathbb{Z}$, we have $a \prec b$ whenever $|a| \le |b|$. If we define the ordering in this manner then we can order the set of integers as follows: $\{0, 1 ,-1, 2,-2,3, -3,...\}$ and thus make the set well-ordered. Under this new ordering that has been defined, the least element is now $0$ because for every $a \in \mathbb{Z}$, if $a \ne 0$, then $|0| = 0 \lt |a|$. And as you can probably see, every nonempty subset will also have a least element under this new ordering.

So to answer your question, no, I don't think the ordering you're describing will be adequate prove something about all infinite sets being well ordered.

Answer 2

As Aniruddha Deshmukh pointed out, you can't index $\ X\ $ with indices limited to a countable set, such as $\ \mathbb{N}\ $, unless $\ X\ $ is itself countable. However, any well-ordered set is order equivalent to some ordinal number $\ \omega\ $, which means that $\ X\ $ can be indexed by the set of ordinal numbers $\ \nu < \omega\ $ in such a way that $\ x_{\nu_1} < x_{\nu_2}\ $ if and only if $\ \nu_1 < \nu_2\ $. Moreover, for any given $\ x\in X\ $ and given $\ \nu<\omega\ $, it's possible to choose the indexing in such a way that $\ x_\nu = x\ $. In particular, you could, if you wish, choose $\ \nu\ $ to be any of the finite ordinals $\ 0,1,2, \dots\ $. If $\ X\ $ is uncountable, however, you can't do this for all $\ x\in X\ $, because, in that case, there must always be an uncountable subset of $\ X $ whose members will have to have infinite ordinals as indices.