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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 $r>0$.

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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: 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 $(X,d)$?" – Hurkyl Apr 23 '12 at 6:59

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

  1. $d(x,y)=d(y,x)$
  2. $d(x,y)=0$ implies $x=y$
  3. $d(x,y)\leq d(x,z)+d(z,y)$

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

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