Zero sets and cozero sets of continuous real-valued functions

Let $(X, \mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $\mathbb{R}$ carry its usual topology, the metric topology of $d_{\mathbb{R}}(s,t) = |s-t|$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). For a continuous $f : X \to \mathbb{R}$ (Continuity of a map of topological spaces at a point and globally) put

$Z(f) \;:=\; f^{-1}[\{0\}] \;=\; \{\, x \in X : f(x) = 0 \,\}, \qquad \operatorname{coz}(f) \;:=\; X \setminus Z(f) \;=\; \{\, x \in X : f(x) \ne 0 \,\} .$

$Z(f)$ is the zero set of $f$ and $\operatorname{coz}(f)$ its cozero set. A subset of $X$ is a zero set of $X$ when it is $Z(f)$ for some continuous $f : X \to \mathbb{R}$, and a cozero set of $X$ when it is the complement of one. Where the target is written $[0,1]$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) with its subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace), a continuous map $X \to [0,1]$ is the same thing as a continuous map $X \to \mathbb{R}$ with all values in $[0,1]$, by the characteristic property of a map into a subspace recorded in Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace; so nothing below depends on which of the two targets is written.

Every zero set is closed and every cozero set is open. $\{0\}$ is closed in $\mathbb{R}$: its complement $\mathbb{R} \setminus \{0\}$ is open, since a point $t \ne 0$ has the bounded open interval $(t - |t|,\ t + |t|)$ around it inside $\mathbb{R} \setminus \{0\}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$, and it is unbounded, claim 3). The preimage of a closed set under a continuous map is closed (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\overline{A}) \subseteq \overline{f(A)}$, clause (c)).

Every zero set is a $G_\delta$ and every cozero set an $F_\sigma$ ($G_\delta$ and $F_\sigma$ subsets of a topological space, agreeing with the real-line notion). Writing $\iota$ for the canonical natural of $\mathbb{R}$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field), so that $1/(n+1)$ abbreviates the inverse of $\iota(n+1)$, put

$V_n \;:=\; f^{-1}\big[\,(-1/(n+1),\ 1/(n+1))\,\big] \qquad (n \in \mathbb{N}).$

Each $V_n$ is open, being the preimage of an open interval (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\overline{A}) \subseteq \overline{f(A)}$, clause (b)). Clearly $Z(f) \subseteq \bigcap_n V_n$. Conversely, if $f(x) \ne 0$ then $\varepsilon := |f(x)| > 0$, and For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ gives a natural $k \ge 1$ with $1/k < \varepsilon$; since $k \ne 0$ it is a successor, $k = n+1$ with $n \in \mathbb{N}$ (Every nonzero natural number is a successor), so $|f(x)| > 1/(n+1)$ and $x \notin V_n$. Hence $Z(f) = \bigcap_{n} V_n$ is a $G_\delta$, and $\operatorname{coz}(f)$ is an $F_\sigma$ by complementation.

Both extremes occur. The constant maps are continuous, since the preimage of any set under a constant map is $\varnothing$ or $X$ (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f(\overline{A}) \subseteq \overline{f(A)}$, clause (b)); so $X = Z(0)$ and $\varnothing = Z(1)$ are zero sets of every space, where $0$ and $1$ denote the corresponding constant maps.

• A closed set need not be a zero set, and no witness for that is exhibited here. The zero sets of $X$ are exactly the closed sets that a continuous real-valued function can see, and a space may have very few continuous real-valued functions: in the indiscrete topology on a set with at least two points, every continuous map to $\mathbb{R}$ is constant, because a nonconstant one would pull back two disjoint intervals to two disjoint nonempty open sets. So the only zero sets there are $\varnothing$ and $X$ — which in that space is also all of the closed sets and all of the $G_\delta$ sets, the only open sets being $\varnothing$ and $X$. That space therefore illustrates the scarcity of continuous functions without separating the two classes; a space with a closed set that is not a zero set is not constructed on this page.

• Where zero sets are used on this page. They are the vocabulary of complete regularity: the defining function separating a point from a closed set $C$ places $C$ inside a zero set and the point in the corresponding cozero set. They also give the sharp form of the metric case, where every closed set is a zero set.

• The name. $\operatorname{coz}$ is the standard notation in the theory of rings of continuous functions, where the zero sets of $X$ are the closed sets the ring can detect; nothing of that theory is used here.