I've come across a similar idea before. For instance, one can define $\mathbb{R}^n$ as the set of all n-tuples of real numbers. From this, we could say that $\underline {v_1} \in \mathbb{R}^n$ means $\underline {v_1}$ is some n-tuple of real values.
I suppose we could take more n-tuples in this fashion, $\underline{v_2},\underline{v_3},\underline{v_4}$, and so on. From this, we could define $V=\{{\underline {v_1}, \underline{v_2},\underline{v_3},\underline{v_4} \}}$. In other words, $V \subseteq \mathbb{R}^n$.
But then I came across the definition of a convex set:
A set $X\subseteq\mathbb{R}^n$ is convex if for all $x_1,x_2\in X$ and all $p\in [0,1]$,
$px_1 + (1-p)x_2 \in X$
This definition is then followed up by a more "intuitive version": A set $X$ is convex if you can take any two points in $X$ and draw a straight line between them, and every point on the line is also in $X$. This is accompanied by images like this:
Then I try and reconcile that with my idea of what a subset of $\mathbb{R}^n$ was before I came across the definition of a convex set.
$X$ must be some set similar to $V$, consisting of some number of n-tuples of real values. If that's the case, then perhaps that means $x_1$ and $x_2$ are n-tuples.
I imagine if you want to create the image of the green set, then $X$ is a set of 1-tuples (just a set of regular numbers I suppose) -- as in a subset of $\mathbb{R}$ ($\mathbb{R}^n$ with $n=1$). In which case $x_1$ and $x_2$ are just numbers like $4$ and $11$ -- a pair of numbers that satisfy the definition of a convex set.
In that case, I expect I don't need to modify my understanding of what a subset of $\mathbb{R}^n$ is.
If my thoughts on these concepts are correct, then I suppose there isn't a problem. Nevertheless, I want to be sure, so I thought I'd ask the question: what does it mean to be a subset of $\mathbb{R}^n$?
Well, the single element $x$ and the single element $y$ and the set of all points in the "green set" (of which $x$ and $y$ are members of) are all in the same "space".
In this drawing that space is $\mathbb R^2$ and $x = (a,b)$ and $y=(c,d)$ for some values of $a,b,c,d$ and $GREEN= \{(x,y)|x\in \mathbb R; y\in \mathbb R;$ for some but not all possible values$\}$. $GREEN$ is such that as $(a,b) \in GREEN$ and $(c,d) \in GREEN$ then for every possible value $p: 0\le p \le 1$ that the point $(p*a + (1-p)*c, p*b+(1-p)*d)$ is in $GREEN$.
If the space in $\mathbb R^n$ and $n\ge 1$ then $GREEN$ is set of $n$-tuples so that of any $X= (x_1,x_2, ....,x_n)$ , $Y=(y_1,y_2, .... y_n)\in GREEN$ then for any $p: 0\le p \le 1$ then the $n-tuple$ $pX+ (1-p)Y := (px_1+ (1-p)y_1,px_2+ (1-p)y_2,........,,px_n+ (1-p)y_n)$.
So yes.... a subset is a collection of $n$-tuples and the definition of "convex" in no way implies anything else.
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Now if you want to see $x$ as a one-tuple, and $n=1$ that that would make $GREEN$ a subset of $\mathbb R^1 = \mathbb R$. And $GREEN$ has the property that if $a,b \in GREEN$ then for every $p; 0\le p \le 1$ then $pa+(1-p)b \in GREEN$.
Note that if we assume, without lack of generality, that $a < b$ then $a = pa + (1-p)a < pa + (1-p)b < pb + (1-p)b = b$. So if $GREEN \subset \mathbb R$ then $GREEN$ is convex if for every $a,b \in GREEN; a< b$ then all $[a,b]\subset GREEN$. In other words in $R^1$ convex means nothing more or less than $GREEN$ is a (possibly infinite, maybe closed/maybe open) interval.