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Disclaimer: I'm aware this is a duplicate of Prove that open half planes are open sets.

However, I didn't find an adequate answer on that page, and it is a year old. The top answer for the question by graydad is too restrictive in my opinion because it relies on the metric function being the Euclidean distance, but I'm pretty sure a more general proof should be possible without relying on the exact computation of the metric function. (Couldn't we arrive at the same conclusion that the half-plane is open even by using any arbitrary metric like the French Metro Metric, or the Taxicab norm, or any other conceivable metric on $\mathbb{R}^2$?)

The problem is to prove that the half-plane $H_a = \{(x,y) \in \mathbb{R}^2 : x>a\}$ for any $a \in \mathbb{R}$ is an open set using any arbitrary metric function on $\mathbb{R}^2.$

I am assuming definitions of open and metric function as given in chapter 2 of Rudin:

Rudin, Walter. Principles of Mathematical Analysis (International Series in Pure and Applied Mathematics). 3rd ed. McGraw-Hill, 1976. ISBN: 9780070542358.

In particular, the definition of open set is:

A set $E$ is open if every point in $E$ is an interior point.

A point $p$ is an interior point of $E$ if there is a neighborhood $N$ of $p$ such that $N \subset E$.

A neighborhood of a point $p$ is a set $N_r(p)$ consisting of all points $q$ such that $d(p,q)<r$. The number $r$ is called the radius of $N_r(p)$.

and the definition of metric is:

A function that associates with any two points $p$ and $q$ a real number $d(p,q)$ such that:

  • $d(p,q)>0$ if $p \ne q$ else $d(p,q)=0$

  • $d(p,q)=d(q,p)$

  • $d(p,q)≤d(p,r)+d(r,q)$ for any $r \in X$.

(Also any proof should use only the most basic principles... Specifically material from the first two chapters of Rudin. That's why the other answer by Aloizio Macedo in the original post was not helpful for me.)

Edit: I removed my attempted proof using relative topology because it was too long and not helpful.

In the comments Yujie Zha suggested I do a direct proof using just the definition of open set. However, I also tried this with no success. There is probably an easy way to write a direct proof for this but I don't see it...

Here's roughly my train of thought:

Let $p=(x,y)$ be a point in $H_a$. Let $N_r(p)$ be a neighborhood of $p$ with radius $r=x−a$. (Now I would like to say $N_r(p) \subset H_a$, but how? Is it by the triangle inequality somehow?) Suppose there is a point $q=(x′,y′) \in N_r(p)$ that is not in $H_a$. Then $x′<a$. But... how do I arrive at the necessary contradiction?

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    $\begingroup$ Why do you even want to go into relative topology? The proof could be done through definition of open sets directly by showing each point in the set is an interior point. $\endgroup$ – Jay Zha Mar 14 '17 at 0:53
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    $\begingroup$ If you must prove that $H_a$ is open, you must define what a basic open set looks like since you must show that for every $x\in H_a$, we have that there exists an open set $U$ such that x\in U\subseteq H_a$; it suffices to prove the statement using basic open sets, then. $\endgroup$ – Clayton Mar 14 '17 at 0:54
  • $\begingroup$ @YujieZha I'm just grasping at straws because I tried a direct proof... But without knowing the metric function I couldn't make much progress? Maybe I should show where I'm stuck on that direct proof first. $\endgroup$ – spenceryue Mar 14 '17 at 1:18
  • $\begingroup$ @Clayton Sorry, I should mention I was assuming the definition of open set given in Rudin, which is roughly: a set $E$ is open if every point in $E$ is an interior point. A point $p$ is an interior point of E if there is a neighborhood $N$ of $p$ such that $N \subset E$. A neighborhood of a point $p$ is a set $N_r(p)$ consisting of all points $q$ such that $d(p,q)<r$. The number $r$ is called the radius of $N_r(p)$. $\endgroup$ – spenceryue Mar 14 '17 at 1:27
  • $\begingroup$ @YujieZha Let $p=(x,y)$ be a point in $H_a$. Let $N_r(p)$ be a neighborhood of $p$ with radius $r=x-a$. Now I would like to say $N_r(p)$ is contained in $H_a$, but how? Is it by the triangle inequality somehow? (Suppose there is a point $q=(x',y') \in N_r(p)$ that is not in $H_a$. Then $x'<a$. But... how do I arrive at the necessary contradiction?) $\endgroup$ – spenceryue Mar 14 '17 at 1:38
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Your comments do, perhaps, clarify what you are asking somewhat. All of the metrics you mention specifically in your comment (the Euclidean metric; the French metro metric; the taxicab metric) determine the exact same collection of open subsets of $\mathbb{R}^2$ (this is a good exercise). So, open-ness of a half plane with respect to any one of those three metrics is equivalent to open-ness of that half plane with respect to any other of those three metrics. This collection of open sets is known as the "standard topology on $\mathbb{R}^n$", as you will learn if you read a book or take a course on topology. So, to prove open-ness with respect to any one of these metrics, feel free to pick one of those three metrics, and prove open-ness just for that one.

Now, you also ask "how to show that a half-plane is open with a general metric on $\mathbb{R}^2$".

Well, that's false: a half-plane is not open with an arbitrary metric on $\mathbb{R}^2$.

Here's a counterexample. Let $f : \mathbb{R}^2 \to \mathbb{R}^2$ be the bijection which equals the identity except that $f(0,1)=(0,-1)$ and $f(0,-1)=(0,1)$. Define a metric by $$d(p,q) = |f(p)-f(q)| $$ To put this another way, take your points $p,q$, find their images under $f$, and apply the Euclidean metric to their images. A subset $A \subset \mathbb{R}^2$ is open with this metric if and only if $f(A)$ is open with the Eucliean metric. Hence, the half plane $H$ defined by $y>0$ is not open, because its image under $f$ equals $(H-\{(0,1)\}) \cup \{(0,-1)\}$ which is not open in the Euclidean metric.

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  • $\begingroup$ Wow, that is an interesting metric. I was under the vague impression that the triangle inequality requirement of a metric might imply that all neighborhoods formed under some metric were necessarily convex.. (and so intuitively or geometrically it would appear the half-plane is always open.) Actually the French metro metric also contradicts this impression of convexity, so I guess I should have recognized the error earlier. I suppose I will learn more about what a topology is in a later course. Thank you very much for your thorough response. $\endgroup$ – spenceryue Mar 14 '17 at 4:48

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