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 $Q$.

  • $\begingroup$ Thank you for making this crystal clear to me. So really, I could take any $r > 0$ and say that $B(x, r) \subset X$. $\endgroup$ Jun 6 '20 at 16:51

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