Let $f:[0,1] \to \mathbb R$ be a uniformly continuous function such that each value of $f(x)$ is greater than zero. Is its infimum greater than zero in BISH?

I believe that it is indeed the case if one assumes the Fan Theorem. But independent of it, I'm not sure.


Note: It's possible to get around this problem in practice by interpreting $f > 0$ to mean that there exists a constant $c$ such that $f(x) > c > 0$ for all $x$. This is an example of a pseudo-order. The fact that the infimum is greater than zero is then a tautology.

  • $\begingroup$ In response to your edit: yes, the infimum of $f$ now is greater than $0$, but often the new problem becomes to show that $\inf(f)> c$ ... and then we're back at square one. $\endgroup$ – Franka Waaldijk Nov 13 '19 at 13:20
  • $\begingroup$ What my paper shows is that there is no simple way to avoid the Fan Theorem if one wishes continuous functions to have certain nice properties. And I am not convinced that the non-simple ways that have been proposed so far to remediate this are really workable in all areas of mathematics. $\endgroup$ – Franka Waaldijk Nov 13 '19 at 13:23
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    $\begingroup$ To the reader wondering what "BISH" could mean: it seems that it means "the Bishop school of constructive mathematics". $\endgroup$ – Alex M. Nov 14 '19 at 9:36

In BISH the follwoing two statements are equivalent:

(i) If $f:[0,1] \to \{y\in\mathbb R\, | \,y>0\}$ is uniformly continuous, then there is $n\in\mathbb N$ such that $\forall x \in [0,1]\ [f(x)>\frac{1}{n}]$

(ii) The Fan Theorem FT

This was already proved in Julian, W.H., and Richman, F., 1984, “A uniformly continuous function on [0, 1] that is everywhere different from its infimum”, Pacific Journal of Mathematics,111: 333–340.

A simpler proof, and a lot more consequences, are given in my paper On the foundations of constructive mathematics --- especially in relation to the theory of continuous functions

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