Let $g$ be the square function:

$$g : \mathbb{R} \rightarrow \mathbb{R}$$ $$g(x) = x^2$$

Let $f$ be a function which takes a real-valued function of one real variable (such as $g$ as input), and returns a set of that functions fixed points. For example,

$$f(g) = \{0,1\}.$$

My questions are:

  1. Is this a well-defined function?
  2. Is this a set-valued function?
  3. How would we declare its domain and co-domain?

If I were to write this in Haskell, whose notation often mimics that used in mathematics, I would write something like

f :: Float a => (a -> a) -> [a]

That is, it's a function that takes as input a function (a -> a), which itself takes a float and returns a float, and returns a list of floats. However, I don't recall seeing a similar notation in mathematics.

  • $\begingroup$ Mathematically, the set of all functions from $\mathbb R$ to $\mathbb R$ is denoted with: $\mathbb R ^ {\mathbb R}$. Thus : $f : \mathbb R ^ {\mathbb R} \to \{ 0,1 \}$. $\endgroup$ – Mauro ALLEGRANZA Sep 5 '17 at 14:26
  1. Is this a well-defined function?

Yes, for example by writing $$ f: \mathbb R^{\mathbb R} \to \mathcal P(\mathbb R)\\ f(g) = \{x \in \mathbb R \mid g(x) = x\} $$ (This also answers 3.) The set $\mathbb R^{\mathbb R}$ is the set of functions $\mathbb R \to \mathbb R$. The set $\mathcal P(\mathbb R)$ is the power set of $\mathbb R$, i.e. the set of all subsets of $\mathbb R$.

  1. Is this a set-valued function?

Yes, since it returns sets of real numbers.

Your Haskell notation is an appropriate analogue, but note that [a] is not exactly the same as the power set (since you can have repeated elements, and order matters), and note that it's not possible to compute the fixed points of a general function.

  • $\begingroup$ I know I'm not supposed to write "great, thanks"-type comments, but this was such a beautifully clear answer. Thanks a lot. $\endgroup$ – user2987808 Sep 5 '17 at 14:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.