# When are the zero sets of two continuous functions in the Stone-Čech compactification included in one another?

Let $$X$$ be a topological space (feel free to add some simplifying assumptions here, like “completely regular” provided at least the case of finite-dimensional manifolds is covered). Let $$f,g \in C^*(X)$$ where $$C^*(X)$$ denotes the ring of bounded continuous real-valued functions on $$X$$. Denote $$f^\beta, g^\beta \colon \beta X\to \mathbb{R}$$ the continuous functions corresponding to $$f,g\colon X\to \mathbb{R}$$ under the canonical isomorphism $$C(\beta X) \cong C^*(X)$$ where $$\beta X$$ is the Stone-Čech compactification of $$X$$, and let $$Z(f^\beta),Z(g^\beta)$$ be their zero-sets, i.e., $$Z(f^\beta) = \{p\in\beta X : f^\beta(p)=0\}$$.

Question: how can we tell whether $$Z(f^\beta) \subseteq Z(g^\beta)$$ merely by looking at $$f,g$$ (without looking at the Stone-Čech compactification)?

To give an idea of the flavor of criteria I'm looking for, let me observe that:

Fact: $$Z(f^\beta) = \varnothing$$ iff $$f$$ is bounded away from $$0$$ on $$X$$ (i.e. $$\exists \varepsilon>0. |f|\geq\varepsilon$$). (Proof: if $$|f|\geq\varepsilon$$ then clearly $$|f^\beta|\geq \varepsilon$$ so $$Z(f^\beta)=\varnothing$$. But conversely, if $$Z(f^\beta)=\varnothing$$, since $$\beta X$$ is compact, the image of $$|f^\beta|$$ is a compact subset of $$\mathbb{R}_{>0}$$, so it is lower bounded by some $$\varepsilon>0$$.) Equivalently, this means that $$f$$ is invertible in $$C^*(X) \cong C(\beta X)$$.

Based on the above fact, I first thought that $$Z(f^\beta) \subseteq Z(g^\beta)$$ meant $$|g|\leq C|f|$$ on $$X$$ for some constant $$C$$, but this is not correct (take $$X=\mathbb{R}$$ and $$f\colon x\mapsto \min(1,x^2)$$ and $$g\colon x\mapsto \min(1,|x|)$$: then $$f^\beta$$ and $$g^\beta$$ both vanish only at $$0$$ but we don't have $$|g|\leq C|f|$$). Maybe something like “$$Z(f)\subseteq Z(g)$$ and there exists $$K\subseteq X$$ compact and $$C$$ such that $$|g|\leq C|f|$$ outside $$K$$”?

Motivation: thinking about the PS in this answer made me ask whether, for $$X$$ a manifold and $$f$$ bounded continuous on $$X$$, we can find $$g$$ bounded and smooth on $$X$$ such that $$Z(f^\beta) = Z(g^\beta)$$ (and to find a criterion for the latter, looking at $$Z(f^\beta) \subseteq Z(g^\beta)$$ first seems natural).

Lemma: Suppose that $$X$$ is a compact Hausdorff space. Let $$f,g:X\rightarrow[0,\infty)$$ be continuous functions. Then the following are equivalent:

1. $$Z(f)\subseteq Z(g)$$.

2. For all $$\epsilon>0$$, there exists a $$\delta>0$$ where $$f^{-1}[0,\delta]\subseteq g^{-1}[0,\epsilon]$$.

3. There exists a function $$u:[0,\infty)\rightarrow[0,\infty)$$ that is continuous at the point $$0$$ with $$u(0)=0$$ and where $$g\leq u\circ f$$.

4. There exists a continuous function $$v:[0,\infty)\rightarrow[0,\infty)$$ with $$v(0)=0$$ and where $$g\leq v\circ f$$.

5. There exists a continuous bijection $$w:[0,\infty)\rightarrow[0,\infty)$$ with $$w(0)=0$$ and where $$g\leq w\circ f$$ (the mapping $$w$$ is necessarily a homeomorphism and increasing).

Proof: $$5\rightarrow 4,4\rightarrow 3.$$ These are trivial.

$$4\rightarrow 5.$$ Suppose that $$v:[0,\infty)\rightarrow[0,\infty)$$ is a bijection with $$v(0)=0$$ and $$g\leq v\circ f$$. Then let $$w:[0,\infty)\rightarrow[0,\infty)$$ be the function defined by letting $$w(x)=v(x)+x$$. Then $$g\leq w\circ f$$ and $$w(0)=0$$.

$$3\rightarrow 4.$$ There is some continuous $$v:[0,\infty)\rightarrow[0,\infty)$$ where $$v(0)=0$$ and where either $$\min(\max(g),u(y))\leq v(y)$$ for all $$y$$. Therefore, for all $$x$$, we have $$g(x)\leq u(f(x))$$, so $$g(x)\leq\min(\max(g),u(f(x)))\leq v(f(x)).$$

$$5\rightarrow 2$$. Suppose that $$\epsilon>0$$. Then set $$\delta=w^{-1}(\epsilon)$$. If $$x\in f^{-1}[0,\delta]$$, then $$f(x)\leq\delta$$, so $$g(x)\leq w(f(x))\leq w(\delta)=\epsilon.$$ Therefore $$x\in g^{-1}[0,\epsilon]$$.

$$2\rightarrow 1$$. If $$\epsilon>0$$, then there is a $$\delta>0$$ where $$Z(f)\subseteq f^{-1}[0,\delta]\subseteq g^{-1}[0,\epsilon]$$. Therefore, $$Z(f)\subseteq\bigcap_{n=1}^{\infty}g^{-1}[0,1/n]=Z(g)$$.

$$1\rightarrow 3.$$ Suppose that $$Z(f)\subseteq Z(g)$$. If $$Z(f)=\emptyset$$, then $$\min(f)>0$$, so just select a suitable function $$u$$ with $$u(y)\geq\max(g)$$ whenever $$y\geq\min(f)$$. Now, assume $$Z(f)\neq\emptyset$$.

Let $$u:[0,\infty)\rightarrow[0,\infty)$$ be the mapping such that $$u(y)=\max\{g(x)\mid f(x)\leq y\}=\max g[f^{-1}(-\infty,y]].$$ Then $$u(0)=\max\{g(x)\mid f(x)\leq 0\}=0$$. Furthermore, if $$x_{0}\in X$$, then $$u(f(x_{0}))=\max\{g(x)\mid f(x)\leq f(x_{0})\}$$. Therefore, $$g(x_{0})\leq u(f(x_{0}))$$. I claim that $$u$$ is upper semicontinuous (and therefore continuous at $$0$$).

Suppose that $$y\in u^{-1}(-\infty,c)$$. Then $$\max g[f^{-1}(-\infty,y]]=\max\{g(x)\mid f(x)\leq y\}=u(y), so $$g[f^{-1}(-\infty,y]]\subseteq(-\infty,c)$$. Therefore, $$f^{-1}(-\infty,y]\subseteq g^{-1}(-\infty,c)$$. By compactness, there is some $$z>y$$ where $$f^{-1}(-\infty,z]\subseteq g^{-1}(-\infty,c)$$. However, if $$s, then $$f^{-1}(-\infty,s]\subseteq g^{-1}(-\infty,c)$$, so $$g[f^{-1}(-\infty,s]]\subseteq(-\infty,c)$$. Therefore, $$u(s)=\max(g[f^{-1}(-\infty,s]]) as well. Therefore, since there is a neighborhood $$U$$ of $$y$$ where $$u(s) whenever $$s\in U$$, the function $$u$$ is upper-semicontinuous.

Q.E.D.

Theorem: Let $$X$$ be a completely regular space with compactification $$C$$. Suppose that $$f,g:X\rightarrow[0,\infty)$$ are continuous functions that extend to continuous maps $$\overline{f},\overline{g}:C\rightarrow[0,\infty)$$. Then the following are equivalent:

1. $$Z(\overline{f})\subseteq Z(\overline{g})$$.

2. For all $$\epsilon>0$$, there exists a $$\delta>0$$ where $$f^{-1}[0,\delta]\subseteq g^{-1}[0,\epsilon]$$.

3. There exists a function $$u:[0,\infty)\rightarrow[0,\infty]$$ that is continuous at the point $$0$$ with $$u(0)=0$$ and where $$g\leq u\circ f$$.

4. There exists a continuous function $$v:[0,\infty)\rightarrow[0,\infty)$$ with $$v(0)=0$$ and where $$g\leq v\circ f.$$

5. There exists a continuous bijection $$w:[0,\infty)\rightarrow[0,\infty)$$ with $$w(0)=0$$ and where $$g\leq w\circ f$$.

Now suppose that $$X$$ is a locally compact regular space with compactification $$C$$, and suppose that $$f,g:X\rightarrow[0,\infty)$$ and $$f,g$$ extend to mappings $$\overline{f},\overline{g}:C\rightarrow[0,\infty)$$. The following result characterizes when $$Z(\overline{f}|_{C\setminus X})\subseteq Z(\overline{g}|_{C\setminus X})$$, so one can use the following result to produce more characterizations of when $$Z(\overline{f})\subseteq Z(\overline{g})$$ when $$X$$ is locally compact.

Theorem: Suppose that $$X$$ is a non-compact locally compact regular space with compactification $$C$$. Let $$f,g:X\rightarrow[0,\infty)$$ be bounded continuous functions, and let $$\overline{f},\overline{g}:C\rightarrow[0,\infty)$$ be the continuous extensions of $$f,g$$ to the domain $$C$$. Then the following are equivalent.

1. $$Z(\overline{f}|_{C\setminus X})\subseteq Z(\overline{g}|_{C\setminus X})$$.

2. For each $$\epsilon>0$$, there exists a $$\delta>0$$ and a compact $$K\subseteq X$$ such that if $$x\in X\setminus K$$, then $$f(x)<\delta\rightarrow g(x)<\epsilon.$$

3. There exists a continuous bijection $$u:[0,\infty)\rightarrow[0,\infty)$$ with $$u(0)=0$$ and a function $$A:X\rightarrow[0,\infty)$$ where $$A^{-1}[\epsilon,\infty)$$ is compact for each $$\epsilon>0$$ and where $$g\leq A+(u\circ f)$$.

Proof: $$2\rightarrow 1$$. Suppose that $$c_{0}\in Z(f|_{C\setminus X})$$. Therefore, let $$(x_{d})_{d\in D}$$ be a net that converges to $$c_{0}$$. Then for each $$\epsilon>0$$, there is some $$\delta>0$$ and compact set $$K\subseteq X$$ where if $$x\in X\setminus K$$ and $$f(x)\leq\delta$$, then $$g(x)\leq\epsilon$$. Then there is some $$d_{0}\in D$$ where if $$d\leq d_{0}$$, then $$x_{d}\not\in K$$ and where $$f(x_{d})\leq\delta$$. In this case, we have $$g(x_{d})\leq\epsilon$$. Therefore, we conclude that $$g(x_{d})_{d\in D}\rightarrow 0$$, so since $$g$$ is continuous, we know that $$g(c_{0})=0$$, and therefore $$c_{0}\in Z(g|_{C\setminus X})$$. Thus, $$Z(f|_{C\setminus X})\subseteq Z(g|_{C\setminus X})$$.

$$1\rightarrow 3$$. By the above results, we know that there is a continuous mapping $$u:[0,\infty)\rightarrow[0,\infty)$$ such that $$\overline{g}|_{C\setminus X}\leq u\circ\overline{f}|_{C\setminus X}$$. Therefore, let $$A:X\rightarrow[0,\infty)$$ be the mapping defined by $$A=\max(0,g-(u\circ f))$$. Then $$g=g-(u\circ f)+(u\circ f)\leq A+(u\circ f),$$ and $$\overline{A}(c)=0$$ whenever $$c\in C\setminus X$$, so $$A^{-1}[\epsilon,\infty)$$ is a compact subset of $$X$$ whenever $$\epsilon>0$$.

$$3\rightarrow 2$$. Suppose that $$g\leq A+(u\circ f)$$. Then for all $$\epsilon>0$$, there is a $$\delta>0$$ with $$u(\delta)<\epsilon$$. Therefore, suppose that $$c\in C\setminus X$$, and $$\overline{f}(c)\leq\delta$$. Then $$\overline{g}(c)\leq\overline{A}(c)+(u\circ\overline{f})(c)=u(\overline{f}(c))\leq u(\delta)<\epsilon.$$ In particular, there is no $$c\in C\setminus X$$ with $$\overline{f}(c)\leq\delta$$ and $$\overline{g}(c)\geq\epsilon$$. Therefore, if we set $$K=\overline{f}^{-1}[0,\delta]\cap\overline{g}[\epsilon,\infty)$$, then $$K$$ is a closed subset of $$C$$, so $$K$$ is compact, but $$K$$ is also a subset of $$X$$. Therefore, if $$x\in X\setminus K$$, then $$f(x)\leq\delta\rightarrow g(x)<\epsilon$$.

Q.E.D.

I'm reasonably convinced (but don't have time to check it carefully right now) that $$Z(f^\beta)\subseteq Z(g^\beta)$$ if and only if, for every $$\varepsilon>0$$, $$f$$ is bounded away from zero on $$\{x\in X:|g(x)|\geq\varepsilon\}$$. (Note that this agrees with the Fact in the question, by taking $$g$$ to be identically $$1$$.)