# 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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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.

Let A be a nonempty subset of X, and let M be its upper bound, so that x <= M, for all x in A. x cannot possibly equal M, for any x in A, since then it couldn't not satisfy the least upper bound property, since then for this very x there would be some upper bound M1 such that M1 < M and M1 < x. So, x < M, for all x in A.

Let's see what A looks like. It is an interval, either of the form <-∞, d> or <e, f>, or some union of such intervals. Does A U X\A disconnect X? Well, A is nonempty, open, X\A is nonempty too, since it contains the upper bounds for A. Can X\A be written (assume now that A = <-∞, f> for simplicity, the other cases are quite obvious too) as <f, +∞>? I claim that f is no upper bound for A. Since if it were, f1, would be too, and then A would be equal <-∞, f1>, which it isn't, specially because iii) holds. Does this work?

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Any thoughts?