Hacker News new | comments | ask | show | jobs | submit login

> - All functions are continuous, and all sets automatically carry the "right" topology. This means, for instance, that constructing the Real numbers in the standard ways (Cauchy sequences or Dedekind cuts) also endows them with the desired topology (the Euclidean topology in this case).

I don’t understand how it is possible to have such an axiom. Why can you not construct the function $f : x \mapsto I(x > 2)$ where $I$ is 1 for true, 0 for false. And then prove that the preimage of $(-1,0.5)$ is $(-\infty,2]$ which is not open?

Technically, f is not well defined. Or rather, indicator functions aren't well defined.

In your case it is not provable that (x > 2 or x =< 2), that would require exactly the law of the excluded middle.

This is explained in one of the later slides.

I’ll accept that indicator functions are not well defined in general however I don’t understand why f should be ill-defined. As I understand it, negation is still allowed in intuitionist mathematics, but you don’t get that (not not X -> X). You still get that (not not not X -> not X), and that (X and not X -> bot). Furthermore it is only a general law of the excluded middle that is not allowed and it is quite possible to prove that e.g. (x : R -> (x > 2 and not x < 2 and not x = 2) or (x <= 2 and not x > 2)).

Does it break down somewhere in the definition of continuity?

If `x` is a representation of a real number, how do you compute if it is above or below 2? Suppose you start computing its digits, and you get 2.0000000, you don't know if you will get a non-0 digit later if you keep computing.

Guidelines | FAQ | Support | API | Security | Lists | Bookmarklet | Legal | Apply to YC | Contact