by Stuart Armstrong 348 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 348 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. reply
 by Stuart Armstrong 348 days ago | link Yes, you’re right. reply

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