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by Stuart Armstrong 568 days ago | Scott Garrabrant likes this | link | parent

Can you argue that \(X\) must have a semi-metric compatible with the topology by using \(d(x,y)=sup_{z\in X}|h(x,z)-h(y,z)|\)?

I’m wondering if you can generalise this to some sort of argument that goes like this. Using X, project down via \(\pi\) from \(X^0=X\) to \(X^1=X^0/d\). Let \(\phi\) be our initial surjection; it’s now a bijection between \(X^1\) and maps from \(X^0\) to \([0,1]\).

If the projection is continuous, then every map from \(X^1\) to \([0,1]\) lifts to a map from \(X^0\) to \([0,1]\). Restricting to the subset of maps that are lifts like this, and applying \(\phi^{-1}\), gives a subset \(X^2 \subset X^1\). We now have a new equivalence relationship, maps from \(X^1\) that are equal to each other on \(X^2\). Project down from \(X^2\) by this relationship, to generate \(X^3\). Continue this transfinitely often (?) to generate a space \(X'\) where \(\phi\) is a homeomorphism, and find a contradiction?

This feels dubious, but maybe worth mentioning…



by Alex Mennen 567 days ago | Scott Garrabrant and Stuart Armstrong like this | link

I haven’t checked that argument carefully, but that sounds like it should give you \(X'\) with a continuous bijection \(\phi:X'\rightarrow[0,1]^{X'}\), which might not necessarily be a homeomorphism.

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by Stuart Armstrong 567 days ago | link

Yes, you’re right.

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