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Let $X$ be a linearly ordered set in the order topology which is connected. Show that $X$ is a linear continuum.

  • A linear continuum is a linearly ordered set $X$ with more than one element that is densely ordered (i.e., between any two members there is another), and has the least upper bound property (i.e., every nonempty subset with an upper bound has a least upper bound)

I know that $X$ has the intermediate value property because it is connected, but I don't know why it satisfies the least upper bound property.

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If $A$ is non-empty and has an upper bound $u$ but no least upper bound, consider the sets

$$A_1 = \{ x \in X \mid \exists y(x) \in A: x \le y(x) \}$$ and $$A_2 = X \setminus A_1 = \{ x \in X \mid \forall y \in A: x > y \}$$

As $A \subset A_1$, it is non-empty. As $u$ cannot be in $A$ (as an upper bound for $A$ that is in $A$ is a maximum which is also a least upper bound for $A$), $u > y$ for all $u \in A$, so $u \in A_2$, so $A_2$ is non-empty as well.

Both sets are disjoint, and cover $X$ by definition.

$A_1$ is open, as for all $x \in A_1$, $x \in (\leftarrow, y(x)) \subset A_1$, if indeed $x < y(x)$. If $x = y(x) \in A$, we know that some $x' \in A$ must exist with $x' > x$, as otherwise $x = \max(A) = \sup(A)$ which cannot be. Then $x \in (\leftarrow, x') \subset A_1$ as well.

$A_2$ is open, because if $x \in A_2$, $x$ is an upper bound for $A$, but it cannot be the smallest such (as $A$ has no smallest upper bound by assumption!), so there is a smaller upper bound $z$ for $A$ and clearly $x \in (z, \rightarrow) \subset A_2$.

This contradicts connectedness of $X$.

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  • $\begingroup$ Sorry for stupid question, but what does $(\leftarrow, y(x))$ stand for? Does it stand for $(b, y(x))$, here $b$ is the smallest element in $X$? If so, why $x \in (\leftarrow, y(x))$, because $x$ can be equal to $y(x)$ $\endgroup$ – le duc quang Oct 24 '15 at 5:06
  • $\begingroup$ $(\leftarrow,y) = \{x \in X: x < y \}$ $\endgroup$ – Henno Brandsma Oct 24 '15 at 6:30
  • $\begingroup$ Yeah, in that case, I don't understand why $x \in (\leftarrow, y(x))$, because x can be equal to $y(x)$. I think $A_1$ should be $\{x \in X | \exists y(x) \in A: x \lt y(x)\}$ $\endgroup$ – le duc quang Oct 24 '15 at 6:47
  • $\begingroup$ Edited the answer to make it more complete. It needed one extra argument. $\endgroup$ – Henno Brandsma Oct 24 '15 at 10:28
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Suppose that it doesn't satisfy the least upper bound property for a proper subset that has an upper bound. Take such a subset set that doesn't satisfy the least upper bound property, and the complement of the subset, and show that they are both open.

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    $\begingroup$ This almost works, but you have to "convexify" the set. See my answer. $\endgroup$ – Henno Brandsma Apr 14 '14 at 19:37

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