Euclidean metric in $\mathbb{R}^n$; the singleton is not open in such a metric space

Question

I am trying to prove this but just don't see it. We are talking about openness in the metric sense, yes?

So, my attempt is

Let $x \in \mathbb{R}^n$ and $d$ represent the Euclidean metric, $d(x,y)=\sqrt{\sum(x_i-y_i)^2}$. An open ball around a point $x$ is defined for some $r>0$ to be $B_r(a)=\{y \in \mathbb{R}^n:d(x,y)<r\}$. Take the singleton set $U=\{u\} \subseteq \mathbb{R}^n$. An open ball around any point on $U$ (which is essentially, just around the one and only point, $u$), is $B_r(u)=\{v \in U:d(u,v)<r\}=\{u\}=U$. So an open ball in any point of $U$ is $U$ itself; $d(u,u)=0 < r$, $\forall r>0$. So by the Euclidean metric, the open ball is essentially the sphere $S^n$ with radius $0$. But then clearly, $U \subseteq U$ meaning that $B_r(u) \subseteq U$.

then I conclude from the definition of an open set $U$ (in metric space $(X,d)$),

$U$ is open in $X$ if and only if for every $u \in U$, $\exists r>0$ such that the open ball $B_r(u) \in U$

(am I right...?)

which tells me that, from the last line I have deduced from my attempt that a singleton is actually open in $\mathbb{R}^n$ with the Euclidean metric.

I admit I am prone to misunderstandings of ideas and definitions; topology and analysis, to me, is the pure embodiment of abstractness which is essentially about poking the human brain.

If there are any mistakes (I am sure there is) please state them and correct them explicitly. Any such help would support me learn better, thank you.

Answer 1

You have just proved that {x} ,as a subset of the metric subspace {x} of the n-dimensional euclidean space, is an open set. Saying it in another way you have proved a particular case of the fact that if (X,d) is a metric space, X itself is open!

Answer 2

You first say (correctly, except that you wrote $ a $ for $ x $ at one point) that an open ball around $ x $ in $ \mathbb R ^ n $ is $$ B _ r ( x ) = \{ y \in \mathbb R ^ n : d ( x , y ) < r \} $$ for some $ r > 0 $. But then later you write that an open ball around $ u $ is $$ B _ r ( u ) = \{ v \in U : d ( u , v ) < r \} \text . $$ Besides changing some names ($ x $ to $ u $ and $ y $ to $ v $), you made a change here that matters; you changed $ \mathbb R ^ n $ to $ U $. This is your mistake.

An open ball around $ u $ in $ \mathbb R ^ n $ is $$ B _ r ( u ) = \{ v \in \mathbb R ^ n : d ( u , v ) < r \} \text , $$ and that must contain more than just $ u $ (because $ r > 0 $, as long as $ n > 0 $; for example, if $ v = u + \frac 1 2 r e _ 1 $ where $ e _ 1 $ is the first standard basis vector in $ \mathbb R ^ n $).

The set that you wrote, which is really $$ B _ r ( u ) \cap \{ u \} = \{ v \in U : d ( u , v ) < r \} \text , $$ is an open ball around $ u $ in $ \{ u \} $; that is, it tells you whether $ \{ u \} $ is open in the subspace $ \{ u \} $. (As the answer by @AleD remarked, this is a special case of the more general fact that any metric space is open in itself.) But the question is whether $ \{ u \} $ is open in $ \mathbb R ^ n $, so you need to use open balls in $ \mathbb R ^ n $.

Answer 3

The statement in your proof that $B_r(u)=\{v \in U:d(u,v)<r\}=\{u\}=U$ is wrong in $\mathbb{R}^n$. Think of the line: the ball is an open interval $(u-r,u+r)$. If $r=0$ this is empty, if $r>0$ it's uncountable and contains much more than just $u$.

All you have managed to prove is that $\{u\}\subset B_r(u)$, but equality of the two sets is false.

Answer 4

Your definition of an open ball of radius $r>0$ around a point in $U$ is incorrect. It's the set of all points in $\mathbb R^n$, not $U$, that are less than $r$ away from the point.

Using this definition, we see that he open ball or radius $r>0$ around $u \in \{u\}$ consists of infinitely many points, and hence cannot be contained in $\{u\}$.

Answer 5

You wrote:

then I conclude from the definition of an open set $U$ (in metric space $(X,d)$),

$U$ is open in $X$ if and only if for every $u \in U$, $\exists r>0$ such that the open ball $\require{enclose}\enclose{updiagonalstrike}{B_r(u) \in U}$

I crossed out, because it must have been $B_r(u)\subset U$ even if $X=U=\{u\}$ is a singleton set as you considered in your thinking. Since, $\{u\}\in\{u\}$ is a mistake in set theory.

Answer 6

Singletons in Euclidean space must be closed. Why?

We know that unions of open sets are once again open. But if singletons are open then their unions are open. Every set is just a union of its points.

But we know there are sets that exist which are not open.