Simple topology proof **Claim:** In a metric space $(X, d)$ the set $X$ is open. **Proof:** For each point $a ∈ X$, every open ball centered at $a$ is contained in $X$. Thus $X$ is open. * * * Could someone expound...--prophetes.ai

Simple topology proof Claim: In a metric space $(X, d)$ the set $X$ is open. Proof: For each point $a ∈ X$, every open ball centered at $a$ is contained in $X$. Thus $X$ is open. * * * Could someone expound on the proof? Is it supposed to be a trivial proof, where we know that every open ball centered at $a$ is contained in $X$? If so, how do we know that?

Recall that in a metric space $(X,d)$ the open ball centred at $a \in X$ of radius $r \in \mathbb{R}, r>0$ is by definition (!)

$$B_X(a,r) = \\{x \in X: d(x,a) < r \\}$$

which is a subset of $X$. It's indeed a triviality from the definition.

E.g. if $X = [0,1]\subseteq \mathbb{R}$ with $d(x,y) = |x-y|$ ,then $B_X(0,\frac{1}{4}) = [0,\frac{1}{4})$; one has to stick to your space, as it were. And so $[0,1]$ is open in $(X,d)$.