Connectedness is product-closed
This article gives the statement, and possibly proof, of a topological space property (i.e., connected space) satisfying a topological space metaproperty (i.e., product-closed property of topological spaces)
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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.
Given: An indexing set , a family of nonempty topological spaces. is the product of these, with the product topology
To prove: is a connected space. In other words, cannot be expressed as a disjoint union of two nonempty open subsets and
Proof: We use a proof by contradiction (the contradiction starts building from Step (4)). Explicitly westart with the assumption:
ASSUMPTION: Suppose is a disjoint union of two nonempty open subsets and .
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.
|Step no.||Assertion/construction||Facts used||Given data used||Previous steps used||Explanation|
|1||Choose and fix the values of all coordinates except the coordinate. Consider the inclusion from to that sends a point of to the tuple where all other coordinates are the pre-determined ones. This is a subspace embedding and in particular, the image of is a connected subset in .|
|2||If two elements of differ in only one coordinate, they are in the same connected component.||Step (1)||Step-direct -- the image of the corresponding is a connected subset containing both.|
|3||If two elements of differ in only finitely many coordinates, they are in the same connected component.||Step (2)||If two elements differ in finitely many coordinates, we can construct a sequence of elements going from one to the other where adjacent members of the sequence differ in only one coordinate. Any two adjacent members of the sequence are in the same connected component, and thus, all members of the sequence are in the same connected component.|
|4||If two points of differ in only finitely many coordinates, they must either both lie in or both lie in .||Assumption||Step (3)||By Step (3), any two points differing in only finitely many coordinates lie in the same connected component, say . However, if one lies in and the other lies in , then itself is a union of non-empty disjoint open subsets of itself and , making it disconnected. Thus, they must both lie in the same piece.|
|5||Since is open and non-empty, it contains a basis element -- say, an open subset given by where all the are non-empty and only finitely many of the are not the whole space. Let's call these finite ones .||definition of product topology|
|6||For any point of , we can thus find a point of (and hence of ) that differs from it in at most the coordinates found in Step (5).||Step (5)||Simply replace the coordinates by coordinates of elements of and leave the remaining coordinates as they are.|
|7||Any point of is in , completing the contradiction to the assumption that is nonempty.||Steps (4), (6)||By Step (6), we can, for any point in , find an element of that differs from it in finitely many coordinates. Thus, by Step (4), the point in that we start with must also be in .|
- 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