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I tried asking ChatGPT about a math question I thought of that I don't know the answer to: Is there a function f: R -> R such that for all real x1, x2 and y wh

by rogual 4y ago

I tried asking ChatGPT about a math question I thought of that I don't know the answer to: Is there a function f: R -> R such that for all real x1, x2 and y where x1 < x2, there exists an x where x1 < x < x2 and f(x) = y ? I was hoping it might at least point me in the right direction but, although it always attempts a proof, most of its answers contain something trivially incorrect like "A set cannot be infinite, therefore..." That said, it did give me a reasonable proof that such a function can't exist if it has to be continuous, because of a thing called the intermediate value theorem, which I hadn't heard of before. But when I asked about noncontinuous functions, it went back to bullshitting. (If anyone here does know the answer, I'd love to hear it!)

solveit 4y ago
You're looking for the Conway base 13 function (https://en.wikipedia.org/wiki/Conway_base_13_function https://en.wikipedia.org/wiki/Conway_base_13_function).
deleted 4y ago
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quasisphere 4y ago
Here's sketch how to construct such a function: Let's start by defining f on Q (the rational numbers) by first splitting Q into countable number of disjoint dense subsets A_n of R (e.g. look at reduced fractions whose denominators are of the form p^k for some prime p, for fixed p every such set is dense and for different p's they are disjoint). As rationals themselves are countable, we may then set f(x) = q_n for all x in A_n, where q_n is an enumeration fo the rationals. This construction already gives us a function such that every interval (x_1, x_2) contains a point x such that f(x) = y for every rational number y. Now, in a similar way we may consider a set of the form s + Q where s is an irrational number. Setting f(x) = s + q_n for all x in s + A_n, we get a function which also attains all numbers of the form y = s + q for some rational q on every interval. Finally, let's say that two real numbers s and t are equivalent if they differ by a rational number. By the axiom of choice we can choose a representative from every equivalence class, so that for every two representatives s and t the sets s + Q and t + Q are disjoint. Using the above construction for every representative lets you define a function with the property you wanted.

by rogual 4y ago

solveit 4y ago

You're looking for the Conway base 13 function (https://en.wikipedia.org/wiki/Conway_base_13_function https://en.wikipedia.org/wiki/Conway_base_13_function).

deleted 4y ago

[deleted]

quasisphere 4y ago

Here's sketch how to construct such a function: Let's start by defining f on Q (the rational numbers) by first splitting Q into countable number of disjoint dense subsets A_n of R (e.g. look at reduced fractions whose denominators are of the form p^k for some prime p, for fixed p every such set is dense and for different p's they are disjoint). As rationals themselves are countable, we may then set f(x) = q_n for all x in A_n, where q_n is an enumeration fo the rationals. This construction already gives us a function such that every interval (x_1, x_2) contains a point x such that f(x) = y for every rational number y. Now, in a similar way we may consider a set of the form s + Q where s is an irrational number. Setting f(x) = s + q_n for all x in s + A_n, we get a function which also attains all numbers of the form y = s + q for some rational q on every interval. Finally, let's say that two real numbers s and t are equivalent if they differ by a rational number. By the axiom of choice we can choose a representative from every equivalence class, so that for every two representatives s and t the sets s + Q and t + Q are disjoint. Using the above construction for every representative lets you define a function with the property you wanted.