A Metric Space is an open subset of itself Proof

Say we have a metric space M=(X,d), then we know that M is an open subset of itself,right? If I'm not mistaken we need this in order to show that M is also a topological space, so how do we prove it? Which property of the metric d does this derive from? All the books I've seen so far(Kreyszig's and W. Rudin's being two of them) say it's obvious/trivial etc, but I do like to be cautious and rigorous, plus I have to admit that it's not that obvious to me. Thanks in advance :)

• For a metric space, the whole space is also closed, and hence clopen (awesome word). What's the definition of an open set? If I pick a point, I can find a ball of positive radius that is still in the set. Well, if the set you're looking at is X, there's nowhere for the ball to not fit, so it's automatically open.
– Matt
Mar 10 '17 at 14:54
• I think this word "awesome" does not mean what you think it means. Mar 10 '17 at 15:22

You have a metric space $(X,d)$ and you want to see it as a topological space, i.e. you want a couple $(X,C)$ where $X$ is the same set as above and $C$ is a collection of subsets of $X$ which satisfies the three axioms that define a topology. The elements of $C$ are those that we call open sets. To turn your metric space into a topological space you declare that $U\in C$ if and only if $\forall x \in U$ $\exists \epsilon>0$ s.t. $B_{\epsilon}(x)\subset U$, where $B_{\epsilon}(x)=\{y\in X$ s.t. $d(y,x)<\epsilon\}$ is the ball centered in $x$ with radius $\epsilon$. To be rigorous you have to show that the family $C$ satisfies the axioms defining a topology, in particular $X\in C$. To show this fix $x\in X$ and $\epsilon>0$ then $B_\epsilon(x)$ is the $\boldsymbol{subset}$ of $X$ which point satifies $d(x,y)<\epsilon$, thus $B_\epsilon(x)\subset X$.