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 $d_{\mathbb{R}}(s,t)=\mathrm{\mid }s-t\mathrm{\mid }$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ 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

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

$Z(f)$ is the zero set of $f$ and $\mathrm{coz}(f)$ 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 $t\mathrm{\ne }0$ has the bounded open interval $(t-\mathrm{\mid }t\mathrm{\mid },t+\mathrm{\mid }t\mathrm{\mid })$ 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: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ 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 $f(\bar{A})\subseteq \overline{f(A)}$, 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 $1\mathrm{/}(n+1)$ abbreviates the inverse of $\iota (n+1)$, put

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

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 $f(\bar{A})\subseteq \overline{f(A)}$, clause (b)). Clearly $Z(f)\subseteq {\bigcap }_n{V}_n$. Conversely, if $f(x)\mathrm{\ne }0$ then $\epsilon :=\mathrm{\mid }f(x)\mathrm{\mid }>0$, and For every $\epsilon >0$ in a complete ordered field there is a natural $n\ge 1$ with $1\mathrm{/}n<\epsilon$ gives a natural $k\ge 1$ with $1\mathrm{/}k<\epsilon$; since $k\mathrm{\ne }0$ it is a successor, $k=n+1$ with $n\in \mathbb{N}$ (Every nonzero natural number is a successor), so $\mathrm{\mid }f(x)\mathrm{\mid }>1\mathrm{/}(n+1)$ and $x\mathrm{\notin }{V}_n$. Hence $Z(f)={\bigcap }_n{V}_n$ is a $G_{\delta }$, and $\mathrm{coz}(f)$ 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 $\mathrm{\varnothing }$ 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 $f(\bar{A})\subseteq \overline{f(A)}$, clause (b)); so $X=Z(0)$ and $\mathrm{\varnothing }=Z(1)$ 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 $\mathrm{\varnothing }$ 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 $\mathrm{\varnothing }$ 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. $\mathrm{coz}$ 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.