$y-x^2 \ge 0$ is a convex set in $\mathbb{R}^2$ with respect to which order?

Question

I am reading topology for the first time .I am stuck in a portion in Munkres which talks about order topology(section - $14$). Whatever I could understand from it is that when we define a convex set it is necessary to define an order.

Definition of convex set :$Y$ is called a convex set if $Y$ subset of $X$ ($X$ is an ordered set) then if $a < b$ of $Y$ holds then the entire interval $(a,b)$ of points of $X$ lies in $Y$.

Take the graph $y - x^2 \ge 0$ this is a convex set with respect to the definition that I know -Pick any two points in the set and join them using a line we see that the line is present inside the graph.

I am really confused that this happens with respect to which order ?This set is not convex with respect to dictionary order. Then what order did we use when we defined the set to be convex?

Answer 1

Usually in the context of subsets of $\Bbb R^n$ (or any real vector space), we have the following definition:

A subset $S \subset \Bbb R^n$ is said to be convex, if for all $x, y \in S$ and $t \in [0, 1]$, it is the case that $tx + (1 - t)y \in S$.

In other words, given two points in $S$, the line segment joining the two points is contained in $S$.

This is different from convexity as defined in Munkres, which is defined for totally ordered sets (or "simply ordered sets").