connectedness in terms of open partitions of spaces A space is usually defined to be connected if it can't be nontrivially partitioned into opens.
Can we relax the word 'opens' to 'spaces' and equality to homeomorphism?
In other words, is the following true:
A space is disconnected iff it is isomorphic to the topological disjoint union of two nonempty spaces.
How to see this?
 A: Yes, this is the same, because in a disjoint sum of two spaces, both these (copies of) spaces are open in the whole sum.
So, formally, if $X$ is homeomorphic (not isomorphic!) to $A \oplus B$, where $A$ and $B$ are non-empty spaces, there exists a homeomorphism $h: X \rightarrow S = (A \times \{0\}) \cup (B \times \{1\})$, where the latter set has the finest topology that makes both $i_A: A \rightarrow S$ and $i_B: B \rightarrow S$ continuous, with $i_A(a) = (a,0), i_B(b) = (b,1)$. This means that $O \subseteq S$ is open iff $(i_A)^{-1}[O]$ is open in $A$ and $(i_B)^{-1}[O]$ is open in $B$.
Then $A \times \{0\}$ is open in $S$ because $(i_A)^{-1}[A \times \{0\}] = A$ and $(i_B)^{-1}[A \times \{0\}] = \emptyset$. Similarly $B \times \{1\}$ is open in $S$ as well. It follows that, because $h$ is continuous and surjective that $h^{-1}[A \times \{0\}]$ and $h^{-1}[B \times \{1\}]$ form a disjoint non-trivial partition of $X$.
On the other hand, if $X = U \cup V$ where $U,V$ are non-empty open, and disjoint, a subset $O$ in $X$ is open iff $O \cap U$ and $O \cap V$ are both open in their respective subspace topologies. This is easy to check. But then $X$ is homeomorphic to $U \oplus V$ (in their subspace topology).  
