# A connected ordered set in the order topology

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## Homework Statement

This seems very simple, that's why I want to check it.

Let X be an ordered set in the order topology. If X is connected, then X is a linear continuum.

## The Attempt at a Solution

An ordered set is a set with an order relation "<" which is antireflexive, transitive, and satisfies, for any x, y, either x < y or x > y.

First of all, X cannot be finite. Since if it were, we could find two nonempty, disjoint open intervals whose union would equal X. So, X is infinite.

A linear continuum is a set with more than one element, with the least upper bound property, and the property that for any x, y in it, there is an element z such that x < z < y.

Clearly, an infinite set has more than one element.

Let x, y be two elements of X. If there were no element between x and y, X would be disconnected. So, there must be some element z such that x < z < y.

Let A be a non empty subset of X which is bounded above, i.e. for all x in A, x < M. If there were no elements of X between x and M, X would be disconnected. So, there exists some element x1 such that x < x1 < M. The same holds for x1 and M. Hence, for any neighbourhood of M (i.e. an open interval containing M), there exists some xn which is contained in the interval, which shows that M is the least upper bound for A.

Hence, X is a linear continuum.

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Office_Shredder
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Let A be a non empty subset of X which is bounded above, i.e. for all x in A, x < M. If there were no elements of X between x and M, X would be disconnected. So, there exists some element x1 such that x < x1 < M. The same holds for x1 and M. Hence, for any neighbourhood of M (i.e. an open interval containing M), there exists some xn which is contained in the interval, which shows that M is the least upper bound for A.
M is just an arbitrary upper bound. The probability that it's the least upper bound is approximately zero. In particular the part where you started talking about neighborhoods of M is where you went off track.

This is probably best approached by contrapositive: If X is not a linear continuum, show X is disconnected. A general rule of thumb is that the property of being connected is hard to work with (because it's a statement about non-existence), whereas "not connected" is a lot easier

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Homework Helper
OK, here's another atempt.

So, the contrapositive statement is: "If X is not a linear continuum, then X is disconnected."

X is not a linear continuum if it does not satisfy at least one of these:

i) X has more than one element
ii) X has the least upper bound property
iii) for any x, y in X there exists some z between x and y

i) trivial, since a set with only one element can't possibly connected
iii) assume this doesn't hold. then there exist x, y such that there is no z between them. then <-∞, x> U <y, +∞> disconnects X.