Connectedness is product-closed

Statement
Suppose $$X_i, i \in I$$ is a family of nonempty topological spaces, all of which are connected spaces. Then, the product space $$\prod_{i \in I} X_i$$, endowed with the product topology, is also a connected space.

Related facts

 * Compactness is product-closed
 * Hausdorffness is product-closed

Proof outline
The key fact that we use in the proof is that for fixed values of all the other coordinates, the inclusion of any one factor in the product is a continuous map. Hence, every slice is a connected subset.

Now any partition of the whole space into disjoint open subsets must partition each slice into disjoint open subsets; but since each slice is connected, each slice must lie in one of the parts. This allows us to show that if two points differ in only finitely many coordinates, then they must lie in the same open subset of the partition.

Finally, we use the fact that any open set must contain a basis open set; the basis open set allows us to alter the remaining cofinitely many coordinates. Note that it is this part that crucially uses the definition of product topology, and it is the analogous step to this that would fail for the box topology.

Proof details
Given: An indexing set $$I$$, a family $$(X_i)_{i \in I}$$ of nonempty topological spaces. $$X$$ is the product of these, with the product topology

To prove: $$X$$ is a connected space. In other words, $$X$$ cannot be expressed as a disjoint union of two nonempty open subsets $$U$$ and $$V$$

Proof: We use a proof by contradiction (the contradiction starts building from Step (4)). Explicitly westart with the assumption:

ASSUMPTION: Suppose $$X$$ is a disjoint union of two nonempty open subsets $$U$$ and $$V$$.

We then try to derive a contradiction. Note that Steps (1)-(3) do not make use of the assumption, and are independently true. Also, note that the definition of product topology really gets used in Step (5). Everything until that point is also true for the box topology.

Textbook references

 * , Page 150, Theorem 23.6 (does the case of finite products), adn Page 152, Exercise 10 (outlines a proof for infinite products), Chapter 3, Section 23