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Is $B(a;0) = \{x : d(a, x) < 0\} = \varnothing$? And if so, is it always the case?

The reason I ask is because I want to know if the open interval $(a,a) = \varnothing$ when $a \in \mathbb{R}$.

Thank you.

Kind regards, Marius

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5  
Not every question involving sets has to do with set theory. –  Asaf Karagila May 9 at 13:57

3 Answers 3

up vote 6 down vote accepted

From the definition of a metric, $d(x,y)\geq0$ for all $x,y$ and $d(x,y)=0\iff x=y$, hence the set of $x$ such that $d(a,x)<0$ is empty, as no such $x$ exists.

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The fact about $d(x,y) = 0$ iff, etc is unnecessary. We just need $d(x,y) \ge 0$ for this to hold. This is more general, as pseudometrics need not satisfy the second property. –  Henno Brandsma May 9 at 13:56
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Sure, I just wrote it because they are normally bundled together into the single axiom positive definitness. –  Daniel Rust May 9 at 13:57
    
I am used to them being separate. So we can consider pseudometrics, quasimetrics (no symmetry) etc. These are often useful! –  Henno Brandsma May 9 at 13:59

The fact that an open interval $(a,a)$ is empty has nothing to do with metrics: If there were an $x \in (a,a)$, by definition, $a < x $ and $x < a$, and then it follows that $a < a$ by transitivity of orders. And by the axioms for (strict) orders, this is false for all $a$.

That $B(a,r) = \emptyset$ for $r \le 0$ follows from $x \in B(a,r) \Rightarrow 0 \le d(a,x) < r \le 0$, which would imply $0 < 0$, likewise impossible.

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That is the usual definition, yes. (Although it is not common to talk about an open ball of radius $0$.)

You might think to define $B(a,0)={a}$ but in general this is not an open set - we want open balls to be open!

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