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Archive for February 20th, 2009

## 1.3 Exercise 11

"Show that an element in an ordered set has at most one immediate successor and at most one immediate predecessor.  Show that a subset of an ordered set has at most one smallest element and at most one largest element."

(Taken from Topology by James R. Munkres, Second Edition, Prentice Hall, NJ, 2000. Page 29.)

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SOLUTION

The element $a \in A$ ordered can have no more than one immediate successor, for, suppose there were. This means that $(a, b)$ and $(a, b')$ are both empty. Also, $a < b$ and $a < b'$ by definition and because the set is ordered.  In particular, however, either $b < b'$ xor $b' < b$ because the set is ordered. Suppose $b < b'$.  Taking into account all the conditions, $a < b < b'$ and $(a, b')$ is nonempty, a contradiction.  Next suppose $b' < b$.  Again, the conditions suggest $a < b' < b$, and $(a, b)$ is nonempty, another contradiction.

We argue that an ordered set can have at most one immediate predecessor analogously.

To show that a subset of an ordered set has at most one smallest element, we argue again by contradiction.  Suppose otherwise, so that $a \in A_0 \subset A$, an ordered set, and also $a' \in A_0$ are smallest elements.  In particular, this means $a \leq x \verb| | \forall x \in A_0$, but also $a' \leq x \verb| | \forall x \in A_0$.  Since the set is ordered, however, either $a < a'$ xor $a' < a$.  Suppose it is the first case, but then $a'$ is not less than all elements in $A_0$, because $a$ belongs to it.  It cannot be a smallest element too.  Suppose it is the second case, but then again $a$ is not less than all elements in $A_0$, because $a'$ belongs to it.  It cannot be a smallest element too.

We argue the fact that an ordered set has at most one largest element analogously.

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