Prove that a linear continuum L is a convex subset of itself.
Everywhere it is stated that the result is trivially true. I would like an explicit proof.
My trial:
Attempt 1. Does the fact that a linear continuum is densely ordered imply convex? That for every $a,b$ such that $a < b$, $[a,b]$ belongs to the linear continuum.
Attempt 2. Or are all sets convex sets of itself?
Definitions.
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
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 $Y$.
By [a,b], we mean all elements z in L such that a≤z≤b. So all sets are convex subsets of itself (this is trival in the sense that there is nothing to prove).
[Posting the answer from comments by Fermé somme]