# Showing that $\varnothing$ and $X$ are open sets in a metric space $(X, \rho)$.

Let $$(X, \rho)$$ be a metric space. Define the open ball with center $$x_0 \in X$$ and radius $$r > 0$$ by $$B(x_0, r) = \{x \in X: \rho(x, x_0) < r\}$$ We say a subset $$E$$ of $$X$$ is an open set if for each $$x \in E$$, there is an $$r > 0$$ with $$B(x, r) \subset E$$ .

Let $$\mathcal{C}$$ be the collection of all open sets in a metric space $$(X, \rho)$$. I wish to show that $$\varnothing, X \in \mathcal{C}$$.

Now, I have done tons of Google searching and searching throughout MSE regarding this problem, but I haven't been able to get a complete explanation of how the proof works.

This is my understanding of the proof:

We have $$\varnothing \in \mathcal{C}$$ because since there are no elements in $$\varnothing$$, the claim that $$\varnothing$$ is an open set is vacuously true.

We have $$X \in \mathcal{C}$$ because...

and I'm a bit lost there. Every example I've seen that attempts to prove $$X \in \mathcal{C}$$ states that either this claim is obvious, or says that $$B(x, 1) \subset X$$. However, I'm failing to see why either of these two things would be true.

... because $$\{t\in X\,:\, Q(t)\}\subseteq X$$ for any predicate $$Q$$.
• Thank you for making this crystal clear to me. So really, I could take any $r > 0$ and say that $B(x, r) \subset X$. Jun 6 '20 at 16:51