Nonempty finite set is not open but is bounded - metric spaces I'm trying to show that if we have a non-empty finite subset of $\mathbb{R}^{2}$ it is not an open set but it is bounded.
I have managed to show it is closed. Also, I'm aware that singleton sets are closed.
Open:
Let $ A = \{x_{1}, x_{2}, \dots, x_{k}\}$ be a non-empty finite subset of $\mathbb{R}^{2}.$ To show it is not an open set, I am trying to show that if we draw a Ball around a point in $A$ it won't be an interior point.
Let $ r > 0$ and consider $B_{r}(x_{1}) = \{y \in A : d(y, x_{1}) < r\}.$ But I'm not too sure how to proceed. I've read that this shouldn't be able to contain any open balls, but I can't see why?
Bounded: Since this is a finite set, it will have a supremum and an infimum. Would the boundedness be as simple as the fact that every point in the finite set will be between these?
Thanks.
 A: Since $A$ is finite, the number
$$S = \min\{d(x_i,x_j) \mid i \ne j, x_i, x_j \in A\}
$$
is positive. For any $0 < s \le S$, letting $B(x_i,s)$ be the open ball around $x_i$ of radius $s$, it follows that $A \cap B(x_i,s) = \{x_i\}$. But surely $B(x_i,s)$ contains a point other than $x_i$ (can you explicitly describe such a point?), and therefore $B(x_i,s) \not\subset A$.
Here's the contradiction: If there exists $r > 0$ such that $B(x_i,r) \subset A$ then, letting $s = \min\{S,r\}$, it would follow that $0 < s \le S$ and that $B(x_i,s) \subset A$.
A: An open ball in $\mathbb{R}^2$ has uncountably many points, so any non-empty finite (or even countable) subset cannot contain an open ball around any of its points, and thus cannot be open.
In $\mathbb{R}^2$, there is no good notion of 'supremum' and 'infimum' in the same way there is in $\mathbb{R}^1$. So this argument doesn't work.
However, if you take the set of distances $\{d(x_1, 0), d(x_2, 0), \dots , d(x_k, 0) \}$, this does have a maximum, call it $r$.
Then $A \subset B_{r+1}(0)$, and so $A$ is bounded.
A: To prove a point, $w$, is not interior of a set $A$ the strategy is usually always be the same.
You prove that for any possible $r>0$, no matter how small, there will always be a point in $\mathbb R^2$ not in the set $A$ that is closer to $w$ than $r$.
Two things come to mind:  For any $r>0$ then $B_r(w) = \{y\in \mathbb R^2| d(w,y) < r\}$ will have an infinite number of points; all closer to $w$ than $r$.  ($\mathbb R^2$ is the entire space and for any point distance you will always be able to find points within that distance.  So there are infinitely many points in $B_r(w).)
But $A$ has only finitely many points.  So if $B_r(w)$, do matter how small, has infinite points it must have points that are not in $A$ and these points are closer to $w$ then $r$. And that will always be true no matter how small $r$ is.
So $w$ can not be an interior point and $A$ is not open.
......
The only trouble with this proof, it the picky teacher will ask you:  How do you know that for every $r$ no matter how small that $B_r(w) =\{$ all the point in $\mathbb R^2$ within a distance of $r$ of $w\}$ is infinite.   And how do you know that if it is infinite it must have points not contained in a finite set.
.......
A more practical strategy.  Is that if you that an $r>0$ and take $B_r(w)=\{$all to points closer to $w$ than the distance $r\}$ then if take an even smaller $s: r>s>0$ then all tha points of $B_s(w) =\{$ all the points within $s$ of $w\}$ will be withing $r$ of $w$ (because $s$ is an even smaller distance).
So to say a point $w$ is not an interior point of $A$ is to say in effect:  We can always find a point not in $A$ that is close to $w$ no matter how close we look.
Now $A$ is finite:  So pick a point $w$.  Consider all the points $x_i\in A$ where $x_i \ne w$ and consider the distance $d_i :=d(w,x_i)$.  There are finite number of these distances, they are all positive, and as it is a finite set there must be a minimum distance.  So take the the smallest such distance and call it $s$.
All the points in $\mathbb R^2$ that are closer to $w$ than $s$ and are not equal to $w$ are not in $A$ because the are closer to $w$ than $w$ is to any other element in $A$.  So they are not in $A$.
So THAT means $w$ can't be an interior point because you can always find a point close to $w$ that isn't in $A$.
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Bounded: Since this is a finite set, it will have a supremum and an infimum. Would the boundedness be as simple as the fact that every point in the finite set will be between these?

ALmost but no.
$A \subset \mathbb R^2$ and $\mathbb R^2 \ne \mathbb R$.  There is no order on $\mathbb R^2$ so $A$ dos not have infimum or supremum.
However if you take the set $D = \{d(x,y)| x,y \in A\}$ then that set is $D\subset \mathbb R$.  and it is finite so it has a supremum. (It actually has a maximum).  So let $d = \sup D$ and then for any two points $w,v\in A$ then $d(w,y)\in D$ so $d(w,y) < d$.
And therefore $A$ is bounded.
