Let C(I) be the space of continuous functions I→I. Then any element of C(I) defines a unique function Q∩I→I (the converse is not true—most functions Q∩I→I do not correspond to continuous functions I→I). Pulling C(I) back to I via Iω we define the set Y⊂I.
Thus Y maps surjectively onto C(I). However, though C(I) maps into C(Y) by restriction (any function from I is a function from Y), this map is not onto (for example, there are more continuous functions from I−{1/2} than there are from I, because of the potential discontinuity at 1/2).
Now, there are elements of I−Y that map to functions in C(Y) but not in C(I). So there’s a hope that there may exist an X with Y⊂X⊂I, C(I)⊂C(X)⊂C(Y), and X mapping onto C(X). Basically, as X `gets bigger’, its image in C(Y) grows, while C(X) itself shrinks, and hopefully they’ll meet.
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.
A small note: it’s not hard to construct spaces that are a bit too big, or a bit too small (raising the possibility that a true X lies between them).
For instance, if I is the unit interval, then we can map I onto the countable hypercube Iω ( https://en.wikipedia.org/wiki/Space-filling_curve#The_Hahn.E2.80.93Mazurkiewicz_theorem ). Then if we pick an ordering of the dimensions of the hypercube and an ordering of Q∩I, we can see any element of Iω - hence any element of I - as a function from Q∩I to I.
Let C(I) be the space of continuous functions I→I. Then any element of C(I) defines a unique function Q∩I→I (the converse is not true—most functions Q∩I→I do not correspond to continuous functions I→I). Pulling C(I) back to I via Iω we define the set Y⊂I.
Thus Y maps surjectively onto C(I). However, though C(I) maps into C(Y) by restriction (any function from I is a function from Y), this map is not onto (for example, there are more continuous functions from I−{1/2} than there are from I, because of the potential discontinuity at 1/2).
Now, there are elements of I−Y that map to functions in C(Y) but not in C(I). So there’s a hope that there may exist an X with Y⊂X⊂I, C(I)⊂C(X)⊂C(Y), and X mapping onto C(X). Basically, as X `gets bigger’, its image in C(Y) grows, while C(X) itself shrinks, and hopefully they’ll meet.
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.