# Metric spaces in which the only open sets are empty set and the full set

What are the metric spaces in which the only open sets are the empty set$$(\emptyset)$$ and the full set$$(X)$$?

My attempt: If $$X=\emptyset$$ then the above statement is trivially true. Suppose $$(X\ne\emptyset,d)$$ is a metric space in which the only open sets are $$\emptyset$$ and $$X$$, then $$x\in B(x,r)=\{y\in X|d(x,y) for any $$r>0$$. $$B(x,r)\ne \emptyset$$, which implies $$B(x,r) = X$$. If $$X$$ contains only one element then our claim is true. But if it has more than two elements, say $$x,y\in X$$, we have two distinct points, we should be able to find an open ball around one of them that does not contain the other(Hausdorff property of metric spaces). Hence we will have an open set which is neither $$X$$ nor $$\emptyset$$. A contradiction to our statement. Hence (X,d) is a metric space which contains at most 1 element

Is the above proof correct?

• Yes, it‘s good. Jan 19, 2020 at 6:55
• "we should be able to.." yes: make that explicit for a complete proof. Jan 19, 2020 at 7:07

Yes, you're correct: if there are two points or more, there is an open ball that is not empty nor $$X$$: if $$x \neq y$$ are two points of $$X$$, set $$r=d(x,y)>0$$ and $$x \in B(x,r), y \notin B(x,r)$$. So $$\emptyset \neq B(x,r) \neq X$$.
• @Sabhrant The most boring example is $X=\{0,1\}$ with the indiscrete topology $\{\emptyset, X\}$, of course. Slightly more interesting: $X=\Bbb N$ with $\mathcal{T}=\{\emptyset\} \cup \{\Bbb N \setminus F: F \subseteq \Bbb N \text{ finite }\}$, the cofinite topology. Jan 19, 2020 at 7:05