Can you argue that X must have a semi-metric compatible with the topology by using d(x,y)=supz∈X|h(x,z)−h(y,z)|?
I’m wondering if you can generalise this to some sort of argument that goes like this. Using X, project down via π from X0=X to X1=X0/d. Let ϕ be our initial surjection; it’s now a bijection between X1 and maps from X0 to [0,1].
If the projection is continuous, then every map from X1 to [0,1] lifts to a map from X0 to [0,1]. Restricting to the subset of maps that are lifts like this, and applying ϕ−1, gives a subset X2⊂X1. We now have a new equivalence relationship, maps from X1 that are equal to each other on X2. Project down from X2 by this relationship, to generate X3. Continue this transfinitely often (?) to generate a space X′ where ϕ is a homeomorphism, and find a contradiction?
I haven’t checked that argument carefully, but that sounds like it should give you X′ with a continuous bijection ϕ:X′→[0,1]X′, which might not necessarily be a homeomorphism.
Can you argue that X must have a semi-metric compatible with the topology by using d(x,y)=supz∈X|h(x,z)−h(y,z)|?
I’m wondering if you can generalise this to some sort of argument that goes like this. Using X, project down via π from X0=X to X1=X0/d. Let ϕ be our initial surjection; it’s now a bijection between X1 and maps from X0 to [0,1].
If the projection is continuous, then every map from X1 to [0,1] lifts to a map from X0 to [0,1]. Restricting to the subset of maps that are lifts like this, and applying ϕ−1, gives a subset X2⊂X1. We now have a new equivalence relationship, maps from X1 that are equal to each other on X2. Project down from X2 by this relationship, to generate X3. Continue this transfinitely often (?) to generate a space X′ where ϕ is a homeomorphism, and find a contradiction?
This feels dubious, but maybe worth mentioning...
I haven’t checked that argument carefully, but that sounds like it should give you X′ with a continuous bijection ϕ:X′→[0,1]X′, which might not necessarily be a homeomorphism.
Yes, you’re right.