I'm trying to find a way to identify a unique point in the domain of a function, but I'm having trouble doing this:

My first thought was choosing 0 if the function is defined at 0, if not, the smallest positive number if it's defined at any positive numbers, and if not, the largest negative number. But I quickly realized this doesn't work for all functions, such as 1/x.

My second thought was picking some interval (x,y) in the domain of the function and choosing the midpoint (x+y)/2. But that just leads to the same problem, how do I identify the endpoints of the interval?

After some thought, I came up with the following method of choosing an interval:

  1. Start with x = the smallest integer such that the function is defined at some number in (x,x+1)
  2. Look at the intervals (x,x+1/2) and (x+1/2,x+1). Choose the first that contains some number at which the function is defined. Repeat this until the function is defined on the whole interval.
  3. Pick the midpoint of the chosen interval.

But then I realized there's a problem: step 2 will only ever end if the function is defined on some interval. It won't work for functions that are only defined for rational numbers.

That's easy, I can just pick the rational number with the largest denominator. But what if the function is only defined for some irrational numbers?

Now I'm stuck. How can I do what I want?

  • $\begingroup$ This is closely relate. $\endgroup$ – mathcounterexamples.net Dec 10 '18 at 20:18
  • $\begingroup$ The function isn't really relevant here, all that matters is its domain. So you are asking if there is a process that given an arbitrary non-empty subset $X$ of $\Bbb{R}$ chooses an $x$ such that $x \in X$. This is a special case of the axiom of choice and it is known to be independent of the other axioms of set theory (which means there is no process of the sort that you envisage to do the job). As $2^{\omega}$ and $\Bbb{R}$ have the same cardinality, your question is essentially the same as one mathcounterexamples.net cites. $\endgroup$ – Rob Arthan Dec 10 '18 at 20:28
  • $\begingroup$ What exactly is the issue? If you are given a non-empty domain why can't you just say "pick any point"? What exactly are trying to do? It does sound like you are trying to come up with a choice algorithm, which as Rob Arthan points out, is axiomatic. $\endgroup$ – fleablood Dec 10 '18 at 20:32
  • $\begingroup$ I'm trying to find a way that always picks the same point in the function, not a random point every time. I don't really understand the links you showed me. What does "independent of the other axioms of set theory" mean? Are you saying it's impossible to do what I'm looking for? $\endgroup$ – user625011 Dec 10 '18 at 20:36
  • $\begingroup$ @user625011 Yes, what you try to do is not possible without using a kind of axiom of choice. This is not trivial. You should ask yourself « what can of construction do I authorize myself » to build the element I’m looking for? $\endgroup$ – mathcounterexamples.net Dec 10 '18 at 20:49

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