Connectedness is product-closed

This article gives the statement, and possibly proof, of a topological space property satisfying a topological space metaproperty
View all topological space metaproperty satisfactions | View all topological space metaproperty dissatisfactions
|

Property "Page" (as page type) with input value "{{{property}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process.Property "Page" (as page type) with input value "{{{metaproperty}}}" contains invalid characters or is incomplete and therefore can cause unexpected results during a query or annotation process.

Statement

Property-theoretic statement

The property of topological spaces of being connected is a product-closed property of topological spaces.

Verbal statement

An arbitrary product of connected spaces is connected, in the product topology.

Definitions used

Connected space

Further information: connected space

Product topology

Further information: product topology

Proof

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.

Proof details

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

To prove: $X$ is a connected space

Proof: First, observe that for fixed values of other coordinates, the inclusion from $X_{i}$ to $X$ is a subspace embedding, so its image is connected. In other words, every slice is connected. Further, using slices along different $X_{i}$, we see that if two points of $X$ differ in only finitely many coordinates, they are in the same connected component.

Now, suppose $X$ is a disjoint union of two nonempty open subsets $U$ and $V$. Then, by the above, it is clear that if two points of $X$ differ in only finitely many coordinates, they must either both lie in $U$ or both lie in $V$.

Now, since $U$ is open, it contains a basis element -- say, an open subet $W$ given by $\times W_{i}$ where finitely many of the $W_{i}$ are not the whole space. Let's call these finite ones $i_{1},i_{2},\dots ,i_{n}$. For any point of $X$, we can thus find a point of $W$ that differs from it in at most these $n$ coordinates. By the above, any point of $X$ must be in $U$, contradicting the assumption that $U$ and $V$ are nonempty open subsets whose disjoint union is $X$.

References

Textbook references

• Topology (2nd edition) by James R. Munkres, ^{More info}, 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