Do elements of (strictly) totally ordered sets have unique neighbours?

Question

The real numbers form a totally ordered set, which according to Wikipedia is also strictly totally ordered, i.e.

• transitive: $a < b, b < c \Rightarrow a < c$
• semiconnex: if $a \neq b$ then $a < b$ or $b < a$
• antireflexive: $a < a$ is false
• asymmetric: if $a < b$ is true then $b < a$ is false

Following a discussion under the answer to my last question if in a strictly totally ordered set such as $\mathbb{R}$, every element $x$ has a unique neighbor: the element which is larger (smaller) than $x$, and smaller (larger) than every other element $c$ such that $c > x$ ($c < x$).

Since these relationships can be determined for all $ x ∈\mathbb{R}$, I suspect the problem lies in having to determine an infinite number of relations.

If this was true, then a smallest positive real number would exist.

Answer 1

No, in e.g. $\Bbb R$ or $\Bbb Q$ no element has "unique neighbours": for $0$ e.g. if $x>0$ there is always $x'$ so that $0 < x' < x$. The order is a so-called dense order.

Answer 2

What would you want the successor to $0$ be?

Do you want to say the successor of $0$ is $1$? If so, I can disprove that by showing you $1/2$.

Do you want to say the successor of $0$ is $.001$? If so, I can disprove that by showing you $.0005$.

Do you want to be sneaky and write down a number on a hidden piece of paper, and not tell me what it is, but instead just tell me "I've just written down a number $x > 0$ which is the successor to $0$ !!!", I can still disprove it by showing you the mathematical expression $\frac{x}{2}$. You can easily prove the following implication: $$0 < x \quad\implies\quad 0 < \frac{x}{2} < x $$ It follows that $x$ is not the successor to $0$, no matter what value you assign to it.

The conclusion is that $0$ has no successor in the real line.

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Let me add that the Wikipedia axioms of a total order are fine for a general definition of a total order, but the specific total order on the real line $\mathbb R$ satisfies some additional axioms not on that list, namely:

• Trichotomy: For every $x \in \mathbb R$, exactly one of the following holds: $x = 0$; or $x > 0$; or $-x > 0$.
• Closure of positive numbers under addition: For every $x,y \in \mathbb R$, if $x > 0$ and $y > 0$ then $x+y > 0$
• Closure of positive numbers under multiplication: For every $x,y \in \mathbb R$, if $x > 0$ and $y > 0$ then $xy > 0$.

You'll find these additional axioms here on wikipedia, for example. You can see at the same link that there is another very important inequality axiom called the completeness axiom, but the above three inequality axioms are already enough to prove the implication I wrote earlier, from which one concludes that $0$ has no successor in the real line.

If you want to work with a total order that does not satisfy these additional inequality axioms, then the order you are working with is not the order on the real numbers.