Why does knowledge of all closed sets determine all continuous real-valued functions (without using sequences)?

Question

Background:

I'm reading on book on Real analysis by Carothers where the author first shows that the map $x \mapsto d(x, A)$ is continuous (in fact, 1-Lipschitz), and then argues the following:

To appreciate what this has done for us, let’s make two simple observations. First, if $f : M \to \mathbb{R}$ is a continuous function, then the set $E = \{ x \in M : f(x) = 0 \}$ is closed. Conversely, if $E$ is a closed set in $M$, then $E$ is the “zero set” of some continuous real-valued function on $M$; in particular, $E = \{ x \in M : d(x, E) = 0 \}$. Thus a set $E$ is closed if and only if $E = f^{-1}(\{0\})$ for some continuous function $f : M \to \mathbb{R}$.

Conclusion: If you know all of the closed (or open) sets in a metric space $M$, then you know all of the continuous real-valued functions on $M$ (Theorem 5.1). Conversely, if you know all of the continuous real-valued functions on $M$, then you know all of the closed (or open) sets in $M$.

Theorem 5.1 states (in particular) that a function is continuous by the $\epsilon$-$\delta$ definition if and only if the preimage of every closed set is closed.

My question:

Now in the first paragraph, nothing is unclear to me but the first statement in the second paragraph is: I fail to understand why knowing all the closed sets lets one know all the continuous real-valued functions? Or at least, how can one reason this without using sequences (see below).

What I've tried:

The second statement (of the second paragraph) is clear to me: given a set, if one knows all the ways it can be mapped continuously, then one knows if the set can be mapped continuously to $\{0\}$ or not, i.e., if the set is closed or not by the remarks of the first paragraph.

But the first statement eludes me; after some time, I came to the conclusion (or dead end) that the logic behind it is that, given a function $f: M \to \mathbb{R}$, if one knows all the closed sets of $M$, one knows if the preimage (under $f$) of some closed set of $\mathbb{R}$ is not closed. But what bothers me here is that this logic requires one to know all the closed sets of $\mathbb{R}$ as well.

Is there something here that I'm missing or is it supposed to be just "obviously given" that one knows all the closed sets of $\mathbb{R}$?

However, I found a way how to circumvent this logic by using open sets and sequences: if we know all the closed sets, we know all the open sets as well, and hence, given a sequence, we know if some point is such that all the open sets that contain the point are eventually supersets of given sequence. That is, we know all the convergent sequences and their limits. And now we know if there exists an $x$ such that there exists a sequence $x_n \to x$ such that $f(x_n) \not\to f(x)$, i.e., we know whether $f$ is continuous.

But this seems to be a bit roundabout way of doing it, and it bothers me tremendously that I can't see the direct way of doing it. And, having searched through this site, and as far as I understand the stuff in this post, one can't even use this sequence-based reasoning in general topological spaces as they are not necessarily sequential.

Answer 1

If you know all of the closed (or open) sets in a metric space $M$, then you know all of the continuous real-valued functions on $M$ (Theorem 5.1). Conversely, if you know all of the continuous real-valued functions on $M$, then you know all of the closed (or open) sets in $M$.

This is put somewhat bluntly. What it means is this:

1. The continuity of a function $f : M \to \mathbb R$ can be expressed in terms of the closed (or open) subsets of $M$. We do not need the $\epsilon$-$\delta$-characterization of the continuity. Actually $f : M \to \mathbb R$ is continuous iff all preimages of closed (or open) subsets of $\mathbb R$ are closed (or open) in $M$.

2. The sets $f^{-1}(0) \subset M$ are closed for all continuous $f : M \to \mathbb R$, and each closed $C \subset M$ is the zero-set of a continuous $f : M \to \mathbb R$.

However, you are right: 1. requires to know all closed (or open) subsets of $\mathbb R$. Indeed they are supposed to be "obviously given". It seems that you consider this to be problematic.

Certainly it is practically impossible to explicitly describe all closed (or open) subsets of $\mathbb R$. However, it suffices to consider certain "basic" closed (or open) subsets of $\mathbb R$.

Open subsets.

Each open subset of $\mathbb R$ is the union of open intervals $(a,b)$. For a function $f : M \to \mathbb R$ we therefore have

$f^{-1}(U)$ is open in $M$ for all open $U \subset \mathbb R$ iff $f^{-1}((a,b))$ is open in $M$ for all open intervals $(a,b) \subset \mathbb R$.

The simple proof is left to you. Another characterization is

$f^{-1}(U)$ is open in $M$ for all open $U \subset \mathbb R$ iff $f^{-1}((a,\infty))$ and $f^{-1}((-\infty,b))$ are open in $M$ for all $a, b \in \mathbb R$.

The proof is again left to you.

Closed subsets.

The closed subsets are the complements of the open subsets. Thus we obtain

$f^{-1}(C)$ is closed in $M$ for all closed $C \subset \mathbb R$ iff $f^{-1}([b,\infty))$ and $f^{-1}((-\infty,a])$ are closed in $M$ for all $a, b \in \mathbb R$.

Proof. $f^{-1}(C)$ is closed in $M$ for all closed $C \subset \mathbb R$ iff $f^{-1}(U)$ is open in $M$ for all open $U \subset \mathbb R$ iff $f^{-1}((a,\infty))$ and $f^{-1}((-\infty,b))$ are open in $\mathbb R$ for all $a, b \in \mathbb R$. But $f^{-1}((a,\infty))$ is open in $\mathbb R$ iff $f^{-1}((-\infty,a])$ is closed in $M$ and $f^{-1}((-\infty,b))$ is open in $\mathbb R$ iff $f^{-1}([b,\infty))$ is closed in $M$.