Why does the interval $(0,1)$ demonstrate that $\mathbb R$ is not a well ordered set?

Question

Why is real numbers not a well ordered set?

I taught since if we take in the case negative real numbers we don't know what is the least element but I saw some explanation on other same type of questions in stack exchange

One of them stated that in " (0,1) we don't have any least element" how is this ...i didn't got it

Answer 1

If you give me any number in $(0,1),$ say $0.00001,$ I can find a number that is less than it, say $0.000005.$ In fact, for this particular set, this will always work... if you give me $x,$ I give you back $x/2,$ and $x/2<x.$ Therefore $(0,1)$ has no least element.

(The precise technical definition of a least element is an element $x\in S$ such that for all $y\in S,$ $x\le y.$ The fact that for any $x\in(0,1),$ I can find a $z\in(0,1)$ such that $z<x$ is exactly the negation of the proposition that there is a least element.)

Since well ordering means any subset has a least element, and $(0,1)\subset \mathbb R,$ this means the standard ordering on the reals is not a well order.

It is true that perhaps it is easier to see that the negative reals, or even $\mathbb R$ itself has no least element, but $(0,1)$ works just fine. Note that there are also a lot of examples of subsets that do have a least element, like $[0,1]$ or $\{300\}$, but that doesn't matter since to be a well order, every subset needs to have a least element.

Answer 2

A set is referred to as well-ordered if and only if every non-empty subset has a least element. By the conventional definition of ordering, the set of the real numbers on the open interval $(0,1)$ is not well ordered for two reasons:

$(1)$ note that the interval DOES NOT include $0$

$(2)$ there are infinitely many real numbers that get closer and closer to $0$

Hence, for any real number $a \in (0,1)$ that you call the least element, I will find another one given by $\frac{a}{2}$.

HOWEVER, according to the well-ordering theorem, any set can be well-ordered depending on how you define the ordering. For example, $\mathbb{Z}$ is not well-ordered according to the conventional ordering of (<). But, 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, a given set may or may not be well-ordered depending on the context and the definitions we are working with.