Declaring a metric space as a universe.

Question

For instance, if I wanted to prove that every metric has a particular property, such as every metric space is open in itself. I would first consider and arbitrary metric space (X,d). Therefore declaring that (X,d) is the universe. Now I would like to show that X is open relative to itself and therefore open (because it is the universe), so I assume x$\in$X. Since X is by definition the universe and there are no intrinsic or extrincic notions in an arbitrary metric space (due to declaring the arbitrary space as the universe), an open ball is by definition $\textbf{always}$ a subset of the arbitrary metric space, therefore the set X is open in itself. Is my interpretation correct? What can I do to improve my intuition?

Answer 1

To show, for the metric space (S,d), that S is open, show that
S = $\cup${ B(x,1) : x in S }
is a union of open balls, hence open, where
B(x,1) = { y in S : d(x,y) < 1 }.

Notice by definition that the balls are all subsets of S.
So there is really no need for that universe thing, simply
use the set S of the metric space (S,d).