James Munkres Topology: Theorem 16.3

Posted 2021-01-26 peabody

Theorem 16.3 If (A) is a subspace of (X) and (B) is a subspace of (Y), then the product topology on (A imes B) is the same as the topology (A imes B) inherits as a subspace of (X imes Y).

Comment: To prove the identity of two topologies, we can either show they mutually contain each other or prove the equivalence of their bases. Because a topological basis has smaller number of elements or cardinality than the corresponding topology, proof via basis is more efficient.

Proof: Let (mathcal{C}) be the topological basis of (X) and (mathcal{D}) be the basis of (Y). Because (A subset X) and (B subset Y), the subspace topological bases of them are (mathcal{B}_A = {C cap A vert forall C in mathcal{C} }) and (mathcal{B}_B = {D cap B vert forall D in mathcal{D} }) respectively according to Lemma 16.1.

Due to Lemma 15.1, the basis of the product topology on (A imes B) is

[
mathcal{B}_{A imes B} = { (C cap A) imes (D cap B) vert forall C in mathcal{C}, forall D in mathcal{D} }.
]

Meanwhile, the basis of the product topology on (X imes Y) is

[
mathcal{B}_{X imes Y} = { C imes D vert forall C in mathcal{C}, forall D in mathcal{D} }.
]

Restricting (mathcal{B}_{X imes Y}) to the subset (A imes B), the basis of the subspace topology on (A imes B) is

[
egin{aligned}
ilde{mathcal{B}}_{A imes B} &= { (C imes D) cap (A imes B) vert forall C in mathcal{C}, forall D in mathcal{D} } \\
&= { (C cap A) imes (D cap B) vert forall C in mathcal{C}, forall D in mathcal{D} },
end{aligned}
]

which is the same as that of the product topology on (A imes B). Hence, this theorem is proved.

The above process of proof can be illustrated as below.

Remark: The above two routes for generating topology on (A imes B) must lead to the same result, otherwise, the theory itself is inappropriately proposed. A theory must be at least self-consistent before its debut in reality.