Topology MOC

Product topology

The product topology is the canonical way of defining a topology on the Cartesian product of spaces. Let {(𝑋𝛼,T𝛼)}𝛼∈𝐴 be an arbitrary collection of topological spaces with cartesian product

𝑋=∏𝛼∈𝐴𝑋𝛼

and 𝜋𝛼 :𝑋 ↠𝑋𝛼 as projections. The product topology on 𝑋 is the coarsest topology on 𝑋 for which all projections 𝜋𝛼 are continuous.¹ topology Thus it has Topological subbasis

A𝑋={𝜋−1𝛼𝑈:𝑈∈T𝛼:𝛼∈𝐴}

Further characterisations

Explicit

The explicit characterisation is a little clunky due to the quirks of uncountable cartesian products. The product topology may be defined with the following topological basis topology

B𝑋={∏𝛼∈𝐴𝑈𝛼:𝑈𝛼∈T𝛼 where 𝑈𝛼≠𝑋𝛼 for finitely many 𝛼}

Proof of basis

It follows from the first characterisation that the following forms a Topological subbasis

A𝑋={𝜋−1𝛼𝑈:𝑈∈T𝛼:𝛼∈𝐴}={∏𝛽∈𝐴{𝑈𝛽=𝛼𝑋𝛼𝛽≠𝛼:𝑈∈T𝛼:𝛼∈𝐴}

When this is completed to a Topological basis via finite intersections, one obtains the explicit characterisation above.

Universal property for the product topoloogy

For every topological space (𝑍,T𝑍) and function 𝑓 :𝑍 →𝑋, then 𝑓 is continuous iff 𝜋𝛼𝑓 :𝑍 →𝑋𝛼 is continuous for all 𝛼 ∈𝐴. topology

Proof

We will first prove that the product topology satisfies the universal property. Let {𝑋𝛼,T𝛼}𝛼∈𝐴 be topological spaces and let 𝑋 =∏𝛼∈𝐴𝑋𝛼 be the cartesian product endowed with the product topology T𝑋. Let (𝑍,T𝑍) be a topological space, and 𝑓 :𝑍 →𝑋 be a function. If 𝑓 is continuous, then so are the compositions 𝜋𝛼𝑓 of continuous functions for all 𝛼 ∈𝐴. Now suppose 𝜋𝛼𝑓 :𝑍 →𝑋𝛼 is continuous for all 𝛼 ∈𝐴. We use the method of Proving continuity with a subbasis. Let 𝑈 ∈A𝑋. Then 𝑈 =𝜋−1𝛼𝑉 for some 𝛼 ∈𝐴 and 𝑉 ∈𝑋𝛼. Since 𝜋𝛼𝑓 is continuous, 𝑓−1𝑈 =(𝜋𝛼𝑓)−1𝑉 ∈T𝑍. Thus the preïmage 𝑓−1𝑈 of every subbasic open set 𝑈 ∈A𝑋 is open, whence 𝑓 is continuous. Therefore 𝑓 is continuous iff 𝜋𝛼𝑓 is continuous for all 𝛼 ∈𝐴.

Now let T′ be a topology on 𝑋 satisfying the same universal property. In particular, let (𝑍,T𝑍) =(𝑋,T𝑋) and 𝑓 =id𝑋 :(𝑋,T𝑋) →(𝑋,T′). Then since 𝜋𝛼id𝑋 =𝜋𝛼 :(𝑋,T𝑋) →𝑋𝛼 is continuous for all 𝛼 ∈𝐴, so is id𝑋 :(𝑋,T𝑋) →(𝑋,T′), wherefore T′ is coarser than T𝑋. Now let (𝑍,T𝑍) =(𝑋,T′) and 𝑓 =id𝑋 :(𝑋,T′) →(𝑋,T′). Since id𝑋 is continuous, so too is 𝜋𝛼𝑓 for all 𝛼 ∈𝐴. But T𝑋 is the coarsest topology on 𝑋 such that 𝜋𝛼𝑓 is continuous for all 𝛼 ∈𝐴, so T𝑋 =T′.

Spaces constructed as products

• Real coördinate space as products of ℝ with the standard topology, e.g. ℝ2 =ℝ ×ℝ.
• Torus topology 𝕋1 =𝕊1 ×𝕊1

Properties

• Continuous maps from the product topology are continuous in each argument
• Canonical projections are open

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Footnotes

1. 2020, Topology: A categorical approach, pp. 30–31 ↩