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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?

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  • $\begingroup$ Do you mean $(\mathbb R^+)^2$? $\endgroup$ – W. Wongcharoenbhorn 2 days ago
  • $\begingroup$ Of course, open sets will not contain their boundaries. But this is not sufficient to characterize open sets. Consider $\Bbb Q$ in the standard $\Bbb R$. Easy to know that $\overline {\Bbb Q} = \Bbb R$, i.e. $\Bbb Q \cup \partial \Bbb Q = \Bbb R$, which means $\partial \Bbb Q \supseteq \Bbb R \setminus \Bbb Q$, hence $\Bbb Q \not\supseteq \partial \Bbb Q$. However $\Bbb Q$ is not open. $\endgroup$ – xbh 2 days ago
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    $\begingroup$ It doesn't make sense to say that a set is open, unless it is described as a subset of some topological space. (It doesn't necessarily have to be a metric space.) Similarly, we can't talk about the boundary without knowing the topology. Here, we aren't even told what the set of points is, let alone the topology. What is the space that the $M_\alpha$ belong to? $\endgroup$ – saulspatz 2 days ago
  • $\begingroup$ @W.Wongcharoenbhorn Yes, what I meant above is that $x$ and $y$ are strictly positive real numbers. $\endgroup$ – Xuan 2 days ago
  • $\begingroup$ @saulspatz Here, $M_a$ is a set of points in a plane, so $M_a \in \mathbb{R}^2_{++}$. But for $S$, I'm not clear what topological space it is in. $\endgroup$ – Xuan 2 days ago
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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.

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  • $\begingroup$ Thanks for your comment! When we specify that $S \subset \mathbb{R}P^1$, which is the space of lines through the point $(b,b)$ in $\mathbb{R}^2$. Is there a way to prove that this set is open? $\endgroup$ – Xuan yesterday
  • $\begingroup$ @Xuan, does that help? $\endgroup$ – memerson 12 hours ago

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