Is a Metric space $(X,d)$ with $X=\{x\}$ an open set?

Question

I've recently started to study functional analysis using "Introduction to functional Analysis" of Edwin Kreyszig. In this book there is a theorem that states that every metric space $X$ is an open set. But what about a metric space that contains only one element? Such space couldn't be a open set because $d(x,x)=0$, and the definition of open set requires $d(x,y)=r$ where $r>0$.

The one point set ${x}$ consists of all points $y$ such that $d(x,y)<0.47$. — André Nicolas, Apr 22 '12 at 18:37
Where in the definition does it require an $y\in X$ so that $d(x,y)=r$, $r>0$? In this case any open ball centered at $x$ would equal the singleton ${x}$. — T. Eskin, Apr 22 '12 at 18:47
the book uses the same definition of open set that is in the wikipedia: http://en.wikipedia.org/wiki/Open_set#Metric_spaces so i don't understand why a set with only one element (and a metric) is an open set. — Don Ismael, Apr 22 '12 at 19:06
The definition doesn't say anything about $d(x,y) = r, r > 0$. It says: "given any point $x$ in $U$, there exists a real number $\epsilon > 0$ such that, given any point $y$ in M with $d(x, y) < \epsilon$, $y$ also belongs to $U$." So, as @AndréNicolas suggested, take $\epsilon = 0.47$. There is only one possible $y$, namely $x$, and this is in the set, so you're done. — Robert Israel, Apr 22 '12 at 19:27
As an aside, it doesn't make sense to ask "Is $(X,d)$ an open set?" There are no such things as "open sets" -- instead there are things such as "open sets of $T$" (for appropriate kinds of $T$). You really meant to ask "Is $X$ an open set of $(X,d)$?" — , Apr 23 '12 at 06:59
On a side-note, do you really want to delve into functional analysis without understanding completely what's going on in general topology? That's asking for trouble, I think. — Najib Idrissi, Apr 23 '12 at 07:05
And thus $0.47$, despite being afflicted with an incurably dull personality, had made it into the Academy of Pop Stars, where all the most popular mathematical constants had learned their trade. Much work would have to make up for its lack of a particular talent. But at the Academy, they would eventually find a fitting role for it... — Bruno Joyal, Apr 23 '12 at 07:11

score 2 · Answer 1 · answered Apr 23 '12 at 08:10

Well, a set $X$ is a metric space when there exists a map $d \colon X \times X \to [0,+\infty)$ such that

$d(x,y)=d(y,x)$
$d(x,y)=0$ implies $x=y$
$d(x,y)\leq d(x,z)+d(z,y)$

for all $x$, $y$, $z \in X$. I can't see any reason why $\operatorname{card}X>1$.

Is a Metric space $(X,d)$ with $X=\{x\}$ an open set?

1 Answers1