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Definition.

A simply ordered set $L$ having more than one element is called a linear continuum if the following hold:

(1) $L$ has the least upper bound property

(2) If $x < y$, there exists $z$ such that $x < z < y$.

A subspace $Y$ of $L$ is said to be convex if for every pair of points $a, b$ of $Y$ with $a < b$, the entire interval $[a,b]$ of points of $L$ lies in $Y$.

I'm trying to prove the obvious fact that a linear continuum in the order topology is a convex space. I've been trying to prove this by way of contradiction using the above two properties, but have been unsuccessful so far. Can anyone help me out?

nomadicmathematician
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    I'm confused; convexity is a property of subspaces, but you're asking about the entire space itself. If $x<y$ with $x,y\in L$, then clearly $[x,y]\in L$ since by definition $[x,y]={ z\in L \mid x\leq z\leq y}$. Are you asking about a subspace of a linearly ordered set where the subspace is a linear continuum? – Hayden Jun 19 '14 at 12:10
  • Maybe you mean "linear sub-continuum"? – Peter Franek Jun 19 '14 at 12:17
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    I came to this question because in Topology by Munkres, there's a theorem 24.1, which states "If L is a linear continuum in the order topology, then L is connected, and so are intervals and rays in L." But the proof proceeds as "We prove that if Y is a convex subspace of L, then Y is connected." And he doesn't specifically show that L is connected so I guessed that this is because L can be considered a convex subspace of itself. – nomadicmathematician Jun 19 '14 at 12:19

1 Answers1

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  1. What you are trying to prove is tautologically true: $L$ is a convex subset of $L$ (for arbitrary ordered set $L$).

  2. Munkres shows that every convex subset of $L$ is connected; since $L$ is a convex subset of itself, it shows that $L$ is connected as well.

Moishe Kohan
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  • One might of another way to interpret OP's question: that a linear subcontinuum is always convex in any larger linear order. But that is not true, as we can always add some elements in between. – tomasz Jun 19 '14 at 20:30
  • @tomasz: You are right, but, given what OP is reading, I am betting on my interpretation of the question. – Moishe Kohan Jun 19 '14 at 20:32