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In the book of Topology by Munkres, at page 153, it is given that

If $L$ is a linear continuum in the order topology, then $L$ is connected and so are intervals and rays in $L$.

And in the proof,

We prove that if $Y$ is a convex subspace of $L$, then $Y$ is connected.

And indeed the author uses the convexity of $Y$ in the proof.However, in the theorem there is no mention of convexity, so I do not understand why does the author assume the convexity of a subspace of a linear continuum while proving the connectivity of that linear continuum ?

Our
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  • I don't have that book, but it seems to me that "every convex subspace of $L$ is connected" is just a more concise way of saying "$L$ is connected and so are intervals and rays in $L$" which is what he wants to prove, since that's what the theorem says. So I don't understand what's bothering you. What am I missing? – bof Jan 25 '18 at 10:54
  • @bof Is the fact that $L$ is linear continuum implies it is and its intervals convex ? – Our Jan 25 '18 at 10:56
  • @bof Made some research, and ashamed that I couldn't see $L$, its intervals and rays are convex :) – Our Jan 25 '18 at 11:00

2 Answers2

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You say 'in the theorem there is no mention of convexity'.

But a statement about continuum, intervalls and rays in it is more or less exactly that: a Statement about convex objects.

And he is not assuming convexity, but using their common property as tool for the proof.

It is obvious that $L$ is convex, and it's discussed here: Linear continuum is convex

Max Punck-Institut
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Intervals and rays are order-convex, clearly. The whole space is trivially so.

In a connected ordered space, the opposite is also true: all order convex subsets of $X$ are either $X$, an (open or closed) ray, or an (open, closed, half-open) interval.

Henno Brandsma
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