1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Yes. I would write $\frac{2x+\varepsilon}2$ as $x+\frac\varepsilon2$. $\endgroup$ – MJD Jan 21 '15 at 18:52
1
$\begingroup$

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  
                                  / \  
                                 +   +
$\endgroup$
  • $\begingroup$ 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 '15 at 19:20
  • 1
    $\begingroup$ @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 '15 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.