Show that a topological space is connected iff every non-empty proper subset has a non-empty boundary

Show that a topological space is connected iff every non-empty proper subset has a non-empty boundary

Let A be a subset of the topological space X. Recall that the boundary of A, denoted by dA, is defined as complement of the interior of A in the closure of A. As the interior of A is always contained in the closure of A, we see that dA is empty if and only if the interior of A equals the closure of A. Therefore dA is empty if and only if A is closed and open.

We will show that a topological space is disconnected if and only if it contains a non-empty, proper subset with empty boundary. If the space is disconnected, let A be a connected component. A is open and closed and therefore has empty boundary. On the other hand, if A is a non-empty, proper subset of X with empty boundary, then it is closed and open. Therefore it is a union of connected components, and the complement of A is open. X may be written as the disjoint union of A and its complement, so X is disconnected.