How to prove that this function is primitive recursive?

Define a function f:X->Y from a metric space X to another metric space Y. If any subset A of X satisfies f(¯A)⊂¯¯¯¯¯¯¯¯¯¯¯f(A)f(A¯)⊂f(A)¯f(\bar A)\subset \overline {f(A)}, then for any subset B of Y, f−1(intB)⊂intf−1(B)fâ&

• Here, ¯AA¯\bar A refers to the closure of A and intBintBint B the interior of B. I do not want to use the continuity of fff to prove this conclusion. Thank you!

• Answer:

  The first condition you gave, f(¯¯¯¯A)⊆¯¯¯¯¯¯¯¯¯¯¯f(A)f(A¯)⊆f(A)¯f(\overline{A}) \subseteq \overline{f(A)}, is an alternate definition of continuity. The proof of this fact is not so interesting, but can be found at https://proofwiki.org/wiki/Continuity_Defined_by_Closure. Hence, we need to show that if f is continuous, then f−1(intB)⊆intf−1(B)f−1(intB)⊆intf−1(B)f^{-1}(int B) \subseteq int f^{-1}(B). But, recall that if f is continuous, then the preimage of an open set is open. In other words, f−1(intB)f−1(intB)f^{-1}(int B) is open. This makes the inclusion easy to see: f−1(intB)f−1(intB)f^{-1}(int B) is contained in f−1(B)f−1(B)f^{-1}(B), and since the former is open, it must be contained in the interior of the latter.