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A similar question is this one.

I proved that if $X$ is a totally ordered set, then an element of $X$ has at most one immediate successor and at most one immediate predecessor.

Initially when I read the statement to prove I thought of sets like $\{0,1\}$, where $1$ has no immediate successor. Then this set is an example of having elements with $0$ immediate successor or predecessor: all finite totally ordered sets have this characteristic.

Now, $\mathbb R$ is a totally ordered set, can I say that the elements of $\mathbb R$ have no immediate successor because of:

$$x\prec \frac{2x+\varepsilon}{2}\prec x+\varepsilon,$$ $\forall x\in\mathbb R,\varepsilon >0$? Similarly for immediate predecessors?

Is there a set (not necessarily totally ordered) that have elements with more than one immediate successor or predecessor?

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    $\begingroup$ Yes. I would write $\frac{2x+\varepsilon}2$ as $x+\frac\varepsilon2$. $\endgroup$
    – MJD
    Jan 21, 2015 at 18:52

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Yes, using the arithmetic mean is a fine way to demonstrate that no $x\in\Bbb R$ has an immediate predecessor or successor. Here’s a picture of a partially ordered set with an element $e$ that has two immediate predecessors ($+$) and two immediate successors ($*$):

                                 *   *  
                                  \ /  
                                   e  
                                  / \  
                                 +   +
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  • $\begingroup$ is there any name for such sets? or is'having more than one immediate successors or predecessors' not a very interesting feature (yet)? $\endgroup$
    – Vladimir Vargas
    Jan 21, 2015 at 19:20
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    $\begingroup$ @Vladimir: I don’t know of one; it doesn’t seem to be a very interesting feature in and of itself. $\endgroup$
    – Brian M. Scott
    Jan 21, 2015 at 19:22

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