Commutative Algebra and Algebraic Geometry (6): Presheaf and Sheaf I

Category of Presheaf on a Topological SpaceSheaf and Ringed SpaceExamples of Presheaves, Sheaves, and Ringed SpacesConstant Function PresheafLocally Constant Function SheafRegular Function Sheaf on Affine VarietyPresheaf of ModulesSheaf of ModulesAn interesting application of sheaf: Sequence in analysis as a Ring-Valued Sheaf

Category of Presheaf on a Topological Space

Let $(X,\tau )$$(X,\tau)$ be a topological space. We could view the lattice of open sets as a category.

i.e. $\mathrm{Ob}(\mathrm{Op}(X)):=U\in \tau$$\mathrm{Ob}(\mathrm{Op}(X)) := U \in \tau$, $\mathrm{Mor}(V,U):=V\subseteq U$$\mathrm{Mor}(V,U) := V \subseteq U$. This category is isomorphic to the category whose objects are open sets again and $\mathrm{Mor}(V,U)$$\mathrm{Mor}(V,U)$ is the inclusion map $i:V\to U$$i:V \to U$.

It is easy to see this forms a category, and $\mathrm{\varnothing }$$\emptyset$ is the initial object, $X$$X$ is the final object. For $U,V\in \tau$$U, V \in \tau$, their product is $U\cap V$$U \cap V$ and their coproduct is $U\cup V$$U \cup V$.

Definition. Let $\mathcal{C}$$\mathcal{C}$ be a category, then the category of $\mathcal{C}$$\mathcal{C}$-valued presheaves is defined by the functor category:

$$\begin{array}{}\text{(1)}& \mathrm{Fun}(\mathrm{Op}(X{)}^{\mathrm{op}},\mathcal{C})\end{array}$$

Hence a $\mathcal{C}$$\mathcal{C}$-valued presheaf on $(X,\tau )$$(X,\tau)$ is just a functor from $\mathrm{Op}(X{)}^{\mathrm{op}}$$\mathrm{Op}(X)^{\mathrm{op}}$ to $\mathcal{C}$$\mathcal{C}$. For two $\mathcal{C}$$\mathcal{C}$-valued presheaves over $X$$X$, $\mathcal{F},\mathcal{G}$$\mathcal{F}, \mathcal{G}$, a morphism between them is a natural transformation. i.e.

$$\begin{array}{}\text{(2)}& \begin{array}{c}\begin{array}{ccc}\mathcal{F}(U)& \stackrel{{\eta }_U}{\to }& \mathcal{G}(U)\\ \mathcal{F}(i)\downarrow & & \downarrow \mathcal{G}(i)& \\ \mathcal{F}(V)& \stackrel{{\eta }_V}{\to }& \mathcal{G}(V)\end{array}\end{array}\end{array}$$

For a presheaf $\mathcal{F}$$\mathcal{F}$ over $X$$X$, we usually use the notation ${\mathrm{res}}_V^U$$\mathrm{res}_{V}^U$ for the image of the morphism $\mathcal{F}(V\subseteq U)$$\mathcal{F}(V \subseteq U)$ and call ${\mathrm{res}}_V^U$$\mathrm{res}_{V}^U$ the restriction map from $U$$U$ to $V$$V$. The elements $s$$s$ in $\mathcal{F}(U)$$\mathcal{F}(U)$ are called sections, and for $V\subseteq U$$V \subseteq U$, we use the notation $s{|}_V$$s|_{V}$ to refer to ${\mathrm{res}}_V^U(s)$$\mathrm{res}_{V}^U(s)$.

Sheaf and Ringed Space

Definition. A sheaf is a presheaf that requires the following property.

Let $U$$U$ be an open set of $X$$X$, and for any open covering $(U_i{)}_{i\in I}$$(U_i)_{i \in I}$, and a family $s_i\in \mathcal{F}(U_i)$$s_i \in \mathcal{F}(U_i)$,

$\mathrm{\forall }i,j\in I$$\forall i,j \in I$ such that $s_i{|}_{U_i\cap U_j}=s_j{|}_{U_i\cap U_j}⟹\mathrm{\exists }!s\in \mathcal{F}(U)=s_i$$s_i|_{U_i \cap U_j} = s_j|_{U_i \cap U_j} \implies \exists! s \in \mathcal{F}(U) = s_i$ such that $s{|}_{U_i}$$s|_{U_i}$ for all $i$$i$. Here $\mathrm{\exists }!$$\exists!$ means exists uniquely.

Equivalently, we can break it into two conditions.

Let $U$$U$ be an open set in $X$$X$ and an open covering $(U_i{)}_{i\in I}$$(U_i)_{i \in I}$.

$†.$$\dagger.$ For $s,t\in \mathcal{F}(U)$$s,t \in \mathcal{F}(U)$, $s{|}_{U_i}=t{|}_{U_i}⟹s=t$$s|_{U_i} = t|_{U_i} \implies s = t$.

$†.$$\dagger.$ For a family $s_i\in \mathcal{F}(U_i)$$s_i \in \mathcal{F}(U_i)$, if $s_i{|}_{U_i\cap U_j}=s_j{|}_{U_i\cap U_j}$$s_i|_{U_i \cap U_j} = s_j|_{U_i \cap U_j}$, then there exists an $s\in \mathcal{F}(U)$$s \in \mathcal{F}(U)$ such that $s{|}_{U_i}=s_i\mathrm{\forall }i\in I$$s|_{U_i} = s_i \forall i \in I$.

The category of sheaves is the fully faithful subcategory of $\mathrm{Fun}(\mathrm{Op}(X{)}^{\mathrm{op}},\mathcal{C})$$\mathrm{Fun}(\mathrm{Op}(X)^{\mathrm{op}}, \mathcal{C})$. Usually, we use $\mathrm{Sh}(X)$$\mathrm{Sh}(X)$ to refer to the category of sheaves over $X$$X$.

If the category $\mathcal{C}$$\mathcal{C}$ admits all limits, then we can use equalizer to define a sheaf as follows:

Proposition. Let $\mathcal{F}$$\mathcal{F}$ be a $\mathrm{Set}$$\mathrm{Set}$-valued sheaf over $X$$X$, then $\mathcal{F}(\mathrm{\varnothing })=\{\ast \}$$\mathcal{F}(\emptyset) = \{*\}$.

Proof. Let $\mathrm{\varnothing }=(U_i{)}_{i\in I}$$\emptyset = (U_i)_{i \in I}$ be any open covering of $\mathrm{\varnothing }$$\emptyset$, it is easy to see $\mathrm{\forall }i\in I,U_i=\mathrm{\varnothing }$$\forall i \in I, U_i = \emptyset$. Firstly, notice that $\mathcal{F}(\mathrm{\varnothing })\ne \mathrm{\varnothing }$$\mathcal{F}(\emptyset) \ne \emptyset$; otherwise, we do not have ${\mathrm{res}}_{\mathrm{\varnothing }}^U$$\mathrm{res}_{\emptyset}^U$. Secondly, assume that $\mathcal{F}(\mathrm{\varnothing })$$\mathcal{F}(\emptyset)$ is not isomorphic to $\{\ast \}$$\{*\}$. Consider an open covering $X=(U_i{)}_{i\in I}$$X = (U_i)_{i \in I}$ such that $U_i\ne \mathrm{\varnothing }\mathrm{\forall }i\in I$$U_i \ne \emptyset \forall i \in I$, and let $s,t\in \mathcal{F}(X)$$s, t \in \mathcal{F}(X)$ such that $s{|}_{\mathrm{\varnothing }}\ne t{|}_{\mathrm{\varnothing }}$$s|_{\emptyset} \ne t|_{\emptyset}$ but $s{|}_{U_i}=t{|}_{U_i}$$s|_{U_i} = t|_{U_i}$. The sheaf axiom tells us that $s=t$$s = t$, which is a contradiction. $◻$$\square$

Corollary. For $\mathrm{Grp},\mathrm{Ring},\mathrm{Mod}(R)$$\mathrm{Grp}, \mathrm{Ring}, \mathrm{Mod}(R)$ valued sheaves over $X$$X$, $\mathcal{F}(\mathrm{\varnothing })=0$$\mathcal{F}(\emptyset) = 0$​.

Proposition. Let $\mathcal{F}$$\mathcal{F}$ be a $\mathrm{Set}$$\mathrm{Set}$-valued sheaf on $X$$X$. If $U\cap V=\mathrm{\varnothing }$$U \cap V = \emptyset$, then $\mathcal{F}(U\cup V)=\mathcal{F}(U)\times \mathcal{F}(V)$$\mathcal{F}(U \cup V) = \mathcal{F}(U) \times \mathcal{F}(V)$. This is also true for arbitrary pairwise disjoint open sets. i.e. $\mathcal{F}(\coprod _{i\in I}{U}_i)=\prod _{i\in I}\mathcal{F}(U_i)$$\mathcal{F}(\coprod_{i \in I} U_i) = \prod_{i \in I} \mathcal{F}(U_i)$.

Proof. Define $\psi :\mathcal{F}(U\cup V)\to \mathcal{F}(U)\times \mathcal{F}(V)$$\psi: \mathcal{F}(U \cup V) \to \mathcal{F}(U) \times \mathcal{F}(V)$ by $s⟼(s{|}_U,s{|}_V)$$s \longmapsto (s|_U, s|_V)$. We need to prove that $\psi$$\psi$ is bijective.

$†.$$\dagger.$ Injective: Let $(s{|}_U,s{|}_V)=(t{|}_U,t{|}_V)$$(s|_U, s|_V) = (t|_U, t|_V)$, then $s{|}_U=t{|}_U$$s|_U = t|_U$ and $s{|}_V=t{|}_V$$s|_V = t|_V$.

$†.$$\dagger.$ Surjective: Since $U,V$$U, V$ is an open covering of $U\cup V$$U \cup V$, for every $(s_i,s_j)$$(s_i, s_j)$, $s_i{|}_{U\cap V}=s_j{|}_{U\cap V}$$s_i|_{U \cap V} = s_j|_{U \cap V}$ since $U\cap V=\mathrm{\varnothing }$$U \cap V = \emptyset$. By the sheaf condition, there exists an $s\in \mathcal{F}(U\cup V)$$s \in \mathcal{F}(U \cup V)$. $◻$$\square$

Definition. Let $X$$X$ be a topological space and ${\mathcal{O}}_X$$\mathcal{O}_X$ be a $\mathrm{Ring}$$\mathrm{Ring}$-valued sheaf over $X$$X$. Then we call ${\mathcal{O}}_X$$\mathcal{O}_X$ the structure sheaf over $X$$X$, and $(X,{\mathcal{O}}_X)$$(X, \mathcal{O}_X)$ a ringed space.

Examples of Presheaves, Sheaves, and Ringed Spaces

Fix a topological space $X$$X$ here.

Continuous Function Sheaf

Let $Y$$Y$ be another topological space and define $\mathcal{F}(U):=\{f:U\to Y\mid f\text{ is continuous}\}$$\mathcal{F}(U) := \{f: U \to Y \mid f \text{ is continuous}\}$. This is a sheaf.

It is easy to see there exists a unique function $f$$f$ glued from an open covering. To see that $f$$f$ is continuous as well, consider:

$$\begin{array}{}\text{(3)}& f^{-1}(V)=\{x\in X\mid f(x)\in V\}=\bigcup _{i\in I}\{x\in U_i\mid f(x)\in V\}=\bigcup _{i\in I}\{x\in U_i\mid f_i(x)\in V\}=\bigcup _{i\in I}{f}_i^{-1}(V)\end{array}$$

Here is a kind of example of a ringed space.

Math Essays: Topological Ring and Ringed Space (marco-yuze-zheng.blogspot.com)

This example is really important and we will use it later.

You can also consider these concepts in the category of smooth manifolds, $\mathrm{Man}$$\mathrm{Man}$. Consider the ring object in $\mathrm{Man}$$\mathrm{Man}$ and $C^{\mathrm{\infty }}(M)$${C}^{\infty}(M)$ for a smooth manifold. Then, the derivation is an endomorphism of the sheaf.

Constant Function Presheaf

Let $E$$E$ be a set and let $\mathcal{F}(U)$$\mathcal{F}(U)$ be the constant function over $E$$E$. This is not a sheaf in general since if $|E|\ge 2$$|E| \ge 2$, then for disjoint $U,V$$U, V$, we could have $\mathcal{F}(U)=a\ne b=\mathcal{F}(V)$$\mathcal{F}(U) = a \ne b = \mathcal{F}(V)$ but we could not find a constant function $f$$f$ on $U\cup V$$U \cup V$ such that $f{|}_U=a$$f|_U = a$ and $f{|}_V=b$$f|_V = b$. However, if $E=\{\ast \}$$E = \{*\}$ is a singleton set, then we can construct a constant presheaf by $\mathcal{F}(U)=\{\ast \}$$\mathcal{F}(U) = \{*\}$ for all open sets $U$$U$. It is easy to see that this is a sheaf as well. This is the final object in $\mathrm{Sh}(X)$$\mathrm{Sh}(X)$. If you let $\{\ast \}$$\{*\}$ be the zero ring, then $(X,\mathcal{F})$$(X, \mathcal{F})$ is a ringed space. Also, for $|E|\ge 2$$|E| \ge 2$, a constant function presheaf is a sheaf if and only if $X$$X$ is irreducible.

Locally Constant Function Sheaf

Let $Y$$Y$ be a discrete space, then the sheaf of continuous functions from $X$$X$ to $Y$$Y$ is just the sheaf of locally constant functions.

Regular Function Sheaf on Affine Variety

Let $X$$X$ be an affine variety with the Zariski topology. Then define the ring of regular functions on $U$$U$ as

$$\begin{array}{}\text{(4)}& \mathcal{O}(U):=\{\frac{f(x)}{g(x)}\mid f,g\in A(X),g\notin I(X)\}\end{array}$$

Obviously, this is a sheaf and gives us a ringed space $(X,{\mathcal{O}}_X)$$(X, \mathcal{O}_X)$.

Presheaf of Modules

For each $x\in X$$x \in X$, assign an $R$$R$-module $M_x$$M_x$. Then define ${\displaystyle \mathcal{F}(U)=\underset{x\in U}{⨁}{M}_x}$$\displaystyle \mathcal{F}(U) = \bigoplus_{x \in U} M_x$. This gives us an example of an $\mathrm{Mod}(R)$$\mathrm{Mod}(R)$-valued presheaf. This is not a sheaf in general. For example, consider $\mathbb{N}$$\mathbb{N}$ with the discrete topology, then $(\{x\}{)}_{x\in \mathbb{N}}$$(\{x\})_{x \in \mathbb{N}}$ is an open covering. Then ${\displaystyle \mathcal{F}(\mathbb{N})=\prod _{x\in \mathbb{N}}{M}_x\ne \underset{x\in \mathbb{N}}{⨁}{M}_x}$$\displaystyle \mathcal{F}(\mathbb{N}) = \prod_{x \in \mathbb{N}} M_x \ne \bigoplus_{x \in \mathbb{N}} M_x$. Since ${\displaystyle \mathbb{N}=\coprod _{x\in \mathbb{N}}\{x\}}$$\displaystyle \mathbb{N} = \coprod_{x \in \mathbb{N}} \{x\}$ and by the previous proposition.

Sheaf of Modules

For each $x\in X$$x \in X$, assign an $R$$R$-module $M_x$$M_x$. Then define ${\displaystyle \mathcal{F}(U)=\prod _{x\in U}{M}_x}$$\displaystyle \mathcal{F}(U) = \prod_{x \in U} M_x$. This gives us an example of an $\mathrm{Mod}(R)$$\mathrm{Mod}(R)$-valued sheaf. For any open set $U\subseteq X$$U \subseteq X$ and any open covering $(U_i{)}_{i\in I}$$(U_i)_{i \in I}$, if $s_i{|}_{U_i\cap U_j}=s_j{|}_{U_i\cap U_j}$$s_i|_{U_i \cap U_j} = s_j|_{U_i \cap U_j}$, then there exists a unique $s$$s$ such that $s{|}_{U_i}=s_i$$s|_{U_i} = s_i$.

An interesting application of sheaf: Sequence in analysis as a Ring-Valued Sheaf

This also works if we replace the $R$$R$-module with any set $M_x$$M_x$. For example, let $X=\mathbb{N}$$X = \mathbb{N}$ with the discrete topology again, and let $M_x=\{f(x)\}$$M_x = \{f(x)\}$ for a function from $\mathbb{N}$$\mathbb{N}$ to $\mathbb{R}$$\mathbb{R}$. But this is just a particular case of a continuous function sheaf. Notice that the functions $f:\mathbb{N}\to \mathbb{R}$$f: \mathbb{N} \to \mathbb{R}$ are just $(a_n{)}_{n=0}^{\mathrm{\infty }}$$(a_n)_{n=0}^{\infty}$. That is another example of a ring-valued sheaf in calculus. The subsequence is just a restriction of a section.

Here $\mathcal{F}(U)$$\mathcal{F}(U)$ is not a Noetherian ring in general. If $U$$U$ is infinite, then denote $I_n:=\{f\in \mathcal{F}(U)\mid f(x)=0\mathrm{\forall }x\ge n\}$$I_n := \{f \in \mathcal{F}(U) \mid f(x) = 0 \forall x \ge n\}$.

Then we have the ascending chain

$$\begin{array}{}\text{(5)}& I_n\subseteq I_{n+1}\subseteq I_{n+2}\subseteq \dots \end{array}$$

and it will not stop since $U$$U$ is infinite.

But if $U$$U$ is finite, then the set of ideals of $\mathcal{F}(U)$$\mathcal{F}(U)$ is finite, hence $\mathcal{F}(U)$$\mathcal{F}(U)$ is a Noetherian ring. Thus, we conclude that $\mathcal{F}(U)$$\mathcal{F}(U)$ is a Noetherian ring if and only if $U$$U$ is finite.

But only considering $\mathbb{N}$$\mathbb{N}$ is not exciting enough. In mathematical analysis, we always care about whether $(a_n{)}_{n=0}^{\mathrm{\infty }}$$(a_n)_{n=0}^{\infty}$ is convergent or not. Now let us find a good language to rewrite the convergence property.

Let ${\mathbb{N}}^{\ast }:=\mathbb{N}\cup \{\mathrm{\infty }\}$$\mathbb{N}^* := \mathbb{N} \cup \{\infty\}$ and define the topology on ${\mathbb{N}}^{\ast }$$\mathbb{N}^*$ as follows.

Let $U$$U$ be a subset $U\subseteq {\mathbb{N}}^{\ast }$$U \subseteq \mathbb{N}^*$, $U$$U$ is open if $U\subseteq \mathbb{N}$$U \subseteq \mathbb{N}$ or $\mathrm{\infty }\in U$$\infty \in U$ and $U={\mathbb{N}}^{\ast }-F$$U = \mathbb{N}^* - F$, where $F$$F$ is any finite subset of $\mathbb{N}$$\mathbb{N}$.

Let us check if this defines a topology on ${\mathbb{N}}^{\ast }$$\mathbb{N}^*$.

Firstly, $\mathrm{\varnothing }$$\emptyset$ is open since it is a subset of $\mathbb{N}$$\mathbb{N}$ and ${\mathbb{N}}^{\ast }$$\mathbb{N}^*$ is open as well since it is ${\mathbb{N}}^{\ast }-\mathrm{\varnothing }$$\mathbb{N}^* - \emptyset$ and $\mathrm{\varnothing }$$\emptyset$ is a finite subset of $\mathbb{N}$$\mathbb{N}$.

For an arbitrary union of open sets, notice that it is a union of two parts: a family of $U\subseteq \mathbb{N}$$U \subseteq \mathbb{N}$ and a family of ${\mathbb{N}}^{\ast }-F$$\mathbb{N}^* - F$.

i.e., we can have the following decomposition for an arbitrary union:

$$\begin{array}{}\text{(6)}& \bigcup _{i\in I}{\mathcal{O}}_i=\bigcup _{j\in J}{U}_j\cup \bigcup _{s\in S}({\mathbb{N}}^{\ast }-F_s)\end{array}$$

The ${\displaystyle \bigcup _{j\in J}{U}_j=U}$$\displaystyle \bigcup_{j \in J} U_j = U$ is still a subset of $\mathbb{N}$$\mathbb{N}$ and ${\displaystyle \bigcup _{s\in S}({\mathbb{N}}^{\ast }-F_s)=\bigcup _{s\in S}{F}_s^c={\left(\bigcap _{s\in S}{F}_s\right)}^c={\mathbb{N}}^{\ast }-\bigcap _{s\in S}{F}_s}$$\displaystyle \bigcup_{s \in S} (\mathbb{N}^* - F_s) = \bigcup_{s \in S} F_s^c = \left( \bigcap_{s \in S} F_s \right)^c = \mathbb{N}^* - \bigcap_{s \in S} F_s$. Notice that the arbitrary intersection of finite sets is still a finite set. Hence ${\displaystyle \bigcup _{s\in S}({\mathbb{N}}^{\ast }-F_s)={\mathbb{N}}^{\ast }-F}$$\displaystyle \bigcup_{s \in S} (\mathbb{N}^* - F_s) = \mathbb{N}^* - F$ for ${\displaystyle F=\bigcap _{s\in S}{F}_s}$$\displaystyle F = \bigcap_{s \in S} F_s$.

It is easy to see that $U\cup {\mathbb{N}}^{\ast }-F={\mathbb{N}}^{\ast }-F^{\prime }$$U \cup \mathbb{N}^* - F = \mathbb{N}^* - F'$ for a finite subset $F^{\prime }$$F'$ of $\mathbb{N}$$\mathbb{N}$. Hence it is closed for arbitrary unions.

For the finite intersection, consider

$$\begin{array}{}\text{(7)}& \bigcap _{i\in [n]}{\mathcal{O}}_i=\bigcap _{j\in [m]}{U}_j\cap \bigcap _{s\in [n-m]}({\mathbb{N}}^{\ast }-F_s)\end{array}$$

Here $[k]:=\{1,2,3,\dots ,k\}$$[k] := \{1,2,3, \dots, k\}$.

If $m\ne 0$$m \ne 0$, the result is a subset of $\mathbb{N}$$\mathbb{N}$, hence open. If $m=0$$m = 0$, then ${\displaystyle \bigcap _{i\in [n]}{\mathcal{O}}_i=\bigcap _{s\in [n]}({\mathbb{N}}^{\ast }-F_s)={\left(\bigcup _{s\in S}{F}_s\right)}^c}$$\displaystyle \bigcap_{i \in [n]} \mathcal{O}_i = \bigcap_{s \in [n]} (\mathbb{N}^* - F_s) = \left( \bigcup_{s \in S} F_s \right)^c$. Hence it is open as well since the finite union of finite sets is still finite. Hence it does form a topology on ${\mathbb{N}}^{\ast }$$\mathbb{N}^*$, denoted as $\tau$$\tau$.

Proposition. $({\mathbb{N}}^{\ast },\tau )$$(\mathbb{N}^*, \tau)$ is compact.

Proof. Let ${\displaystyle {\mathbb{N}}^{\ast }=\bigcup _{i\in I}{U}_i}$$\displaystyle \mathbb{N}^* = \bigcup_{i \in I} U_i$ be an open covering, then there exists a $U_i=F^c$$U_i = F^c$ for some finite subset of $\mathbb{N}$$\mathbb{N}$. Hence we only need to pick some $U_i$$U_i$ that could cover $F$$F$. Hence we get a finite subcovering of ${\displaystyle {\mathbb{N}}^{\ast }=\bigcup _{i\in I}{U}_i}$$\displaystyle \mathbb{N}^* = \bigcup_{i \in I} U_i$. $◻$$\square$

Now let us consider the continuous functions from $X=({\mathbb{N}}^{\ast },\tau )$$X = (\mathbb{N}^*, \tau)$ to $\mathbb{R}$$\mathbb{R}$.

Proposition. $f$$f$ is continuous if and only if ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}f(n)=f(\mathrm{\infty })}$$\displaystyle \lim_{n \to \infty} f(n) = f(\infty)$.

Proof.

Only if: Suppose $f$$f$ is continuous, then for any $ϵ>0$$\epsilon > 0$, there exists an open neighborhood $U$$U$ of $\mathrm{\infty }$$\infty$ such that for all $x\in U,|f(\mathrm{\infty })-f(x)|\le ϵ$$x \in U, |f(\infty) - f(x)| \le \epsilon$. But the open neighborhood of $\mathrm{\infty }$$\infty$ must have the form ${\mathbb{N}}^{\ast }-F$$\mathbb{N}^* - F$. Let $N=inf({\mathbb{N}}^{\ast }-F)$$N = \inf(\mathbb{N}^* - F)$, then $\mathrm{\forall }n\ge N$$\forall n \ge N$, $|f(\mathrm{\infty })-f(x)|\le ϵ$$|f(\infty) - f(x)| \le \epsilon$. Hence it converges to $f(\mathrm{\infty })$$f(\infty)$.

If: We only need to prove $f$$f$ is convergent at $\mathrm{\infty }$$\infty$, since at other points it is continuous automatically (discrete topology).

Assume $f(n)$$f(n)$ converges to $f(\mathrm{\infty })$$f(\infty)$. Let $(f(\mathrm{\infty })-ϵ,f(\mathrm{\infty })+ϵ)$$(f(\infty) - \epsilon, f(\infty) + \epsilon)$ be an open neighborhood of $\mathrm{\infty }$$\infty$. Then there exists an $N$$N$ such that $\mathrm{\forall }n\ge N,f(n)\in (f(\mathrm{\infty })-ϵ,f(\mathrm{\infty })+ϵ)$$\forall n \ge N, f(n) \in (f(\infty) - \epsilon, f(\infty) + \epsilon)$. Hence $f^{-1}(f(\mathrm{\infty })-ϵ,f(\mathrm{\infty })+ϵ)={\mathbb{N}}^{\ast }-\{n\in \mathbb{N}\mid n<N\}\cup S$$f^{-1}(f(\infty) - \epsilon, f(\infty) + \epsilon) = \mathbb{N}^* - \{n \in \mathbb{N} \mid n < N\} \cup S$, where $S$$S$ is the subset of $\mathbb{N}$$\mathbb{N}$ such that $f(n)\in (f(\mathrm{\infty })-ϵ,f(\mathrm{\infty })+ϵ)$$f(n) \in (f(\infty) - \epsilon, f(\infty) + \epsilon)$ but $n<N$$n < N$. Hence $f(n)$$f(n)$ converges to $f(\mathrm{\infty })$$f(\infty)$ implies $f$$f$ is continuous at $\mathrm{\infty }$$\infty$. Hence $f$$f$ is continuous. $◻$$\square$

Therefore, the ring of continuous functions from $({\mathbb{N}}^{\ast },\tau )$$(\mathbb{N}^*, \tau)$ to $\mathbb{R}$$\mathbb{R}$ is just the ring of convergent functions.

This ringed space $(X,{\mathcal{O}}_X)$$(X, \mathcal{O}_X)$ is important in the study of analysis. We will return to this example after we discuss more about sheaves, such as stalks and locally ringed spaces...