# Show a metrix space is connected iff for every non-empty proper subset the boundary is non-empty

Show a metric space X is connected iff \forall non-empty proper subset A \subset X the boundary (set of points in X whose neighborhoods contain points from both A and the complement) is non-empty.

• What definition of connected are you using? – Tim Raczkowski Nov 4 '15 at 16:36
• if it cannot be written as the union of two disjoint, non-empty closed (or open) subsets – user2684794 Nov 4 '15 at 16:52

If $X$ is disconnected, so $X = A \cup B$ where $A$ and $B$ are disjoint non-empty closed (and open) subsets. What is $\operatorname{Bd}(A)$?
Suppose $A$ is non-empty and suppose $\operatorname{Bd}(A) = \emptyset$; as the boundary of $A$ is the difference between closure and interior of $A$, we have that these coincide and so $A$ is clopen. So $X = A \cup \ldots$ proves $X$ is disconnected.
• Indeed. And $X \setminus A$ is also... – Henno Brandsma Nov 4 '15 at 17:47
• $B = X \setminus A$, one could say. But closed and open is what I meant. – Henno Brandsma Nov 4 '15 at 17:49