Prove the set is open using the definition I am trying to prove that the following set is open $$S= \{M_{a}|a \in (0,1)\},$$ where
$$M_{a}=\{(x,y) \in \mathbb{R}^2_{++}: ax+(1-a)y=b, \text{ and } b \text{ is a fixed positive real number}\}.$$
Intuitively, $S$ is a set of lines on the $xy$-plane passing through the same point $(b,b)$. Without specifying the metric of this set, is there a way to prove that $S$ is an open set based on the idea that

an open set is a set that does not contain its boundary

My attempt is as follows:
The boundaries of $S$ are:

*

*when $a=0$, $M_a=\{(x,y) \in \mathbb{R}^2_{++}: y=b, \text{ where } b \text{ is a fixed positive real number}\}$;


*when $a=1$, $M_a=\{(x,y) \in \mathbb{R}^2_{++}: x=b, \text{ where } b \text{ is a fixed positive real number}\}$,
which are not in $S$. Therefore, we conclude that the set $S$ is open.
If the above sketch is correct (which I am not sure of), do we need to prove that $M_{a=0}$ and $M_{a=1}$ are the boundaries of $S$? And how we can prove this without specifying the metric space of this set?
 A: There are two problems with your attempt. First, while it is true that if a set is open then it doesn't contain its boundary, the converse doesn't hold. If we want to test whether or not a set is open in terms of its boundary, it is true that a set $U$ is open iff $\partial U \cap U = \emptyset$. This is because $\partial U$ is defined as $\bar U \setminus U^\circ$ (here $U^\circ$ is the interior of $U$). This works because if $U$ is open $U = U^\circ$ and so $\partial U = \bar U \setminus U$ and so $\partial U \cap U = \emptyset$. On the other hand, suppose that $U$ is not open. Then there must be some $u\in U$ which is not an element of $U^\circ$. But $u$ must be in $\bar U$ as $U\subset \bar U$, and so $\bar U\setminus U^\circ$ contains $u$.
Suppose we ignored that or replaced your criterion with this one. There is still a more fundamental problem, (as pointed out in the comments), which is that it doesn't make sense to talk about a set being open or closed unless you specify a topological space it is a part of. Even with our alternate approach, a random set doesn't have a boundary. We can only identify the boundary of a set in the context of a topological space. It is possible to specify the structure of a topological space without metric structure, but (unless you want that this set is trivially open), you still need to find $S$ as a subset of a larger topological space.
Edit: You suggest that we consider $S$ a subset of $\mathbb{R}P^1$. $\mathbb{R}P^1$ can be defined as $\\{(a,b) \in \mathbb{R}^2: (a,b)\neq (0,0)\\}/\sim$ where $(a,b) \sim (c,d)$ whenever $(a,b)$ is a scalar multiple of $(c,d)$. The question becomes, how do we identify $S$ with a subspace of $\mathbb{R}P^1$ given this definition. Well, each pair $[(a,b)]$ corresponds to the equation $ax + by = 0$ (we are just picking lines through the origin for simplicity although it doesn't affect the answer). This means that your set $S$ is $\\{[(a,1-a)]: 0 < a < 1\\}$. If we take look at the preimage of $S$ in $\mathbb{R}^2\setminus \\{(0,0)\\}$ (ie: before indentificaiton). These are the set of points $(x,y)$ where we can rescale them to be of the form $(a,1-a)$ with $a\in (0 ,1)$. A little algebra gives us that we can put any $(x,y)$ in this form by taking $\pm \frac{1}{x+y}(x,y)$ as long as neither $x$ nor $y$ is zero. As such, the preimage of $S$ under the quotient map is $\mathbb{R}^2$ without the $x$ and $y$ axes, and so it has open preimage. But by definition of the quotient topology this makes it open.
