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Let's start off by recalling the definition of an open set in a metric space:

A set $A$ in a metric space $(X,d)$ is open if for each point $x\in A$ there is a number $r\gt0$ such that $B_r(x)\subset A$

Where $B_r(x)$ denotes the open ball of radius $r$ at a point $x$,

$$B_r(x)=\{y\in X:d(x,y)\lt r\}$$

Supposing that our space is some form of the reals, $\mathbb R^n$, does this not mean that there are an infinite number of open balls, and hence an infinite number of points, within A?

This is my reasoning so far: take any point $x\in A$, and say that for some $r$ we have some $B_r(x)$ centered at $x$. Then, surely, there would be some point $x_1\in B_r(x)$ such that $d(x,x_1)\lt r$. But then, must not $x_1$ be in $A?$ And so would there not also exist some other open ball, $B_{r\,'}(x_1)$ with radius $r'$ centered at $x_1$, and then so on and so forth for the other points within that radius?

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    $\begingroup$ I don't know what you mean by "some form of the reals." One answer makes use of $\Bbb Z$, where singletons are open and closed. But if you only look at open subsets of $\Bbb R^n$ with the subspace topology, then the story is different. $\endgroup$
    – pjs36
    Jul 28 '15 at 15:18
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In any metric space, there are an infinite number of ways to write down balls with a given center. But some of the balls might actually be the same. For instance, in the "discrete metric" $d(x,y)=0$ if $x=y$ and $1$ otherwise, all balls $B_r(x)$ for $r \leq 1$ are the same (they are just $\{ x \}$) while all balls $B_r(x)$ for $r>1$ are also the same (they are the whole space). In particular, if we put the discrete metric on a finite set, this gives a counterexample to your claim.

In $\mathbb{R}^n$, balls of different radii are distinct. But this might not be true in a subset of $\mathbb{R}^n$. It certainly isn't true in a discrete subset of $\mathbb{R}^n$. But it also fails in other subsets of $\mathbb{R}^n$. For instance it fails for bounded sets (since all balls whose radius is larger than the diameter are the same). It would also fail for a lot of sets with a bounded connected component, such as $\{ x \in \mathbb{R}^n : \| x \| \leq 1 \text{ or } \| x \| \geq 2 \}$.

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It may be the case that these open balls are actually equal to each-other despite having different radii. For instance, if we equip $\mathbb{Z}$ with the usual metric, then $B_{1/2}(0) = B_{1/3}(0)$, for instance.

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It is not clear to me what you are asking exactly.

Given a ball $B=B(x;r)$ in $\Bbb R^n$ all the balls $B(x;\rho)$ with $\rho<r$ are inside $B$ and so in whatever open set $A$ the ball $B$ is a subset of. This is entirely trivial.

A bit less trivial question would be if you can find infinitely many disjoint balls inside an open set $A\subseteq\Bbb R^n$, but I have no clue that this is your actual question.

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For example if $X= \{0, 1, 2\}$ with $d(x,y)$ defined as $d(x,y)=|x-y|$ then there are only $2^3=8$ subsets so at most 8 "open balls".

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  • $\begingroup$ I suspect this is the sort of thing the OP was trying to rule out by saying that the space should be "some form of the reals." I have faith that they could correctly reason that spaces with finitely many points have finitely many open balls in general :) $\endgroup$
    – pjs36
    Jul 28 '15 at 15:33
  • $\begingroup$ Please start using LaTeX, see this guide: meta.math.stackexchange.com/questions/5020/… Almost all of your posts need heavy editing $\endgroup$
    – Hirshy
    Aug 5 '15 at 8:37

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