Is an open interval (in one-dimensional R) a strictly convex set?

I am teaching math economics. The class and I are struggling with whether an open point set of one-dimensional R is a strictly convex set. The interval in question is X=(a,b). We are working with a definition of strict convexity that says that X is strictly convex if the convex combination of any two elements of X formed using lambda=(0,1) consists of points that are interior points of X. By this definition, and according to my understanding of the definitions of epsilon neighbourhoods, interior points, and open sets, I believe that the open interval X=(a,b) is both convex and strictly convex. Again, this is for point sets of one-dimensional R only that I am asking. Thank you.

• You are correct in your understanding. – Mike Earnest Sep 26 '17 at 20:43

If $X$ is $any$ interval of $\Bbb R$ and $u,v \in X$ with $u<v$ then $(u,v)\subset X$ (otherwise $X$ would not be an interval ). And $(u,v)$ is an open subset of $\Bbb R$ so $(u,v)\subset Int (X),$ because $Int (X)$ is the union of $all$ subsets of $X$ that are open subsets of $\Bbb R.$
For each $w\in (u,v)$ there is a (unique) $r\in (0,1)$ such that $w=(1-r)u+rv$ by elementary algebra: $$w=(1-r)u+rv\iff w-u=r(v-u)\iff r=(w-u)/(v-u)\quad \text {(as } v-u\ne 0)$$.... and $0<w-u<v-u$ so $0<(w-u)/(v-u)<1.$