I will agree that Iwis connected and locally connected. I’m not sure if its second countable. It is not compact.
Just to be clear Iw={(x1,x2,⋯)∀i:xi∈[0,1]]} Now let Hi={x∈Iw|xi<0.6} . Clearly each Hi is open. Let G={x∈Iw|∀i:xi>0.4} And F={G,H1,H2,⋯}. Now clearly this family covers all of Iw. However, remove any Hi from F and x=(i0.9,0.9,⋯0.9,0.1,0.9,0.9,⋯) is no longer covered. So F is a family of open sets, which cover Iw and don’t have any finite subcover.
For your proof, I think that G is not open in the product topology. The product topology is the coarsest topology where all the projection maps are continuous.
To make all the projection maps continuous we need all sets in S to be open, where we define σ∈S iff there exists an i, such that pi(σ) is open in [0,1] and σ={x=(x1,x2,…)|xi∈pi(σ),0≤xj≤1 for i≠j}.
Let S′ be the set of finite intersection of these sets. For any σ′∈S′, there exists a finite set Nσ′⊂N such that if x∈σ′ and yi=xi for i∈Nσ′, then y∈σ′ as well.
If we take S′′ to be the arbitrary union of S′, this condition will be preserved. Thus G is not contained in the arbitrary unions and finite intersections of S, so it seems it is not an open sent.
I think you made a mistake here
The Hahn–Mazurkiewicz theorem states that
I will agree that Iwis connected and locally connected. I’m not sure if its second countable. It is not compact.
Just to be clear Iw={(x1,x2,⋯)∀i:xi∈[0,1]]} Now let Hi={x∈Iw | xi<0.6} . Clearly each Hi is open. Let G={x∈Iw | ∀i:xi>0.4} And F={G,H1,H2,⋯}. Now clearly this family covers all of Iw. However, remove any Hi from F and x=(i0.9,0.9,⋯0.9,0.1,0.9,0.9,⋯) is no longer covered. So F is a family of open sets, which cover Iw and don’t have any finite subcover.
Hum, Iω should be compact by Tychonoff’s theorem (see also the Hilbert Cube, which is homeomorphic to Iω).
For your proof, I think that G is not open in the product topology. The product topology is the coarsest topology where all the projection maps are continuous.
To make all the projection maps continuous we need all sets in S to be open, where we define σ∈S iff there exists an i, such that pi(σ) is open in [0,1] and σ={x=(x1,x2,…)|xi∈pi(σ),0≤xj≤1 for i≠j}.
Let S′ be the set of finite intersection of these sets. For any σ′∈S′, there exists a finite set Nσ′⊂N such that if x∈σ′ and yi=xi for i∈Nσ′, then y∈σ′ as well.
If we take S′′ to be the arbitrary union of S′, this condition will be preserved. Thus G is not contained in the arbitrary unions and finite intersections of S, so it seems it is not an open sent.
Also, Iω is second-countable. From the wikipedia article on second-countable:
I’ve figured out the difference, I was using the box topology https://en.wikipedia.org/wiki/Box_topology , while you were using the https://en.wikipedia.org/wiki/Product_topology.
You are correct. I knew about finite topological products and made a natural generalization, but it turns out not to be the standard meaning of Iw.
Thanks for introducing me to the box topology—seeing it defined so explicitly, and seeing what properties it fails, cleared up a few of my intuitions.