Open sets and metric spaces

Question

First off I am new to working with balls in metric spaces so my might seem dumb to those who know better than I I don’t want to down voted cause of it

Ref: https://math.stackexchange.com/a/2124668/585321 In this proof r=0.5d(x,y). My question can the constant be smaller than it and prove the theorem

And in general is it possible?

Answer 1

The object is to prove in a metric space that points are closed, by showing that the complement of a point is always open.

So if $(M, d)$ is a metric space and $x \in M$, we want to show that $M \setminus \{x\}$ is open. Suppose $y \in M, y \ne x$. Then $d(x, y) > 0$.

Let $r$ be any number satisfying $0 < r \le d(x,y)$. Then $x \notin B_r(y)$, the open ball of radius $r$ about $y$. This is because $B_r(y) = \{z \in M\mid d(y,z) < r\}$, but $d(x,y) \ge r$. Hence $B_r(y) \subseteq M - \{x\}$.

Since $M - \{x\}$ contains a ball around each of its elements, it is open, and its complement $\{x\}$ is closed.

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The value of $k$ used in $r = kd(x,y)$ can therefore be any number in $(0,1]$. szw1710 for some reason decided to prove that $y$ had a ball that was bounded away from $x$, which isn't actually necessary for this problem (but is true for any $k < 1$). They just used $r = \frac 12 d(x,y)$ as a convenient choice. Afterall, they just needed one value that worked, not every possible value.