Are open sets and closed sets fundamentally different, considering only theorems that apply to every metric space?

Question

I'm going through a real analysis course, and i noticed that every theorem we've seen about open sets can be transformed into a theorem about closed sets and vice versa. Here's an example:

Let $(X, d)$ be a metric space and consider $A \subset X$, $B \subset X$. Then:

$$ Int(A) \cap Int(B) = Int(A \cap B) $$ $$ Int(A) \cup Int(B) \subset Int(A \cup B) $$

Now, if you consider the complement of both relations:

$$ Int(A) \cap Int(B) = Int(A \cap B) $$ $$ \Rightarrow (Int(A) \cap Int(B))^c = (Int(A \cap B))^c $$ $$ \Rightarrow Cl(A^c) \cup Cl(B^c) = Cl(A^c \cup B^c) $$ $$ Int(A) \cup Int(B) \subset Int(A \cup B) $$ $$ \Rightarrow (Int(A) \cup Int(B))^c \supset (Int(A \cup B))^c $$ $$ \Rightarrow Cl(A^c) \cap Cl(B^c) \supset Cl(A^c \cap B^c) $$

You end up with a very similar theorem but with closure of sets:

$$ Cl(A) \cup Cl(B) = Cl(A \cup B) $$ $$ Cl(A \cap B) \subset Cl(A) \cap Cl(B) $$

Another example is that $Int(A)$ is the maximum open set contained on $A$, while $Cl(A)$ is the minimum closed set that contains $A$.

One last example is continuity of metric spaces. A function from a metric space $(X,d)$ to another metric space $(Y,\lambda)$ is continuous if and only if the preimage of every open set on Y is open on X. The same function is also continuous if and only if the preimage of every closed set on Y is closed on X.

This duality, at least with the theorems i've seen, seems to tell that open sets and closed sets really behave the same way.

I know that, if you consider continuous functions on real numbers, each function will have a maximum and a minimum on a closed interval, but this doesn't always apply on open intervals.

However, the real numbers along with absolute value is only one metric space, in which axioms of real numbers apply. This isn't true for every metric space.

So, i want to consider only definitions and theorems that apply to every metric space. This is (informally speaking because i don't know about mathematical logic, if i'm wrong correct me or ignore this), i want to consider only the axiomatic system of metric spaces and it's theorems, not theorems that only some models verify along with other axioms.

Since you can't use real number axioms on every metric space, the extreme value theorem isn't a counter-example to my question, it's not intrinsic to metric spaces in general.

Finally, the question is: Considering only statements that are true for every metric space, do open and closed sets actually behave the same way?

Answer 1

Suppose we limit our language drastically by only allowing ourselves to refer to sets and subsethood-comparability; so we can't say "$X$ has a single element" or "$X\subseteq Y$," but we can say "Either $X\subseteq Y$ or $Y\subseteq X$" (denote this latter relationship by $\gtreqless$ for simplicity). Then for any topological space $\mathscr{A}=(A;\tau)$, the resulting "set comparability structures" $$Open_\mathscr{A}=(\tau; \gtreqless)\quad\mbox{and}\quad Closed_\mathscr{A}=(\check{\tau}; \gtreqless)$$ (where $\check{\tau}=\{U\subseteq A: A\setminus U\in \tau\}$ is the set of $\tau$-closed sets) are isomorphic via the complement map $U\mapsto A\setminus U$. So any way of distinguishing them must involve either referring to points (= elements of $A$) directly or making a distinction between $\subseteq$ and $\supseteq$. This seems to match the notion of duality you have in mind, and so we get a positive answer to at least one version of your question.

Once we can count points or refer to $\subseteq$ as such, however, open and closed are quite easily distinguishable. For example, the statement "If every singleton is closed, then every set is closed" is not true in all metric spaces, but the statement "If every singleton is open, then every set is open" is true in all metric spaces. Relatedly, consider "There exist minimal nonempty closed sets" (true in all metric, or even $T_1$, spaces) versus "There exist minimal nonempty open sets" (false even in the usual topology on the real line).

Narratively, I'd say that the real question is whether we think in terms of sets or in terms of ways of cutting the space. The former lets us distinguish closed-ness and open-ness in most settings, but the latter is much coarser; in particular, an open set $U$ corresponds to literally the same "cut" - namely, $A=U\sqcup (A\setminus U)$ - as does its closed complement $A\setminus U$. (I've talked about a similar focus-shift at the end of this old answer, in the context of symmetry in logic with respect to negation; look for the phrase "assertion logic," and don't be too put off by the technical character of the rest of the answer.)

Answer 2

The examples in your question are covered by the general fact that the lattice of open subsets of $X$ with respect to inclusion is isomorphic to the lattice of closed sets with respect to reverse inclusion. (The isomorphism just sends each set to its complement.) Because of the reversal, $\cup$ in the first lattice corresponds to $\cap$ in the second, and vice versa.