Basic Neighborhoods in the Pointwise Topology

Neighborhoods Record Finitely Many Values

A function can be tested at many inputs, but a basic open neighborhood in the pointwise topology only restricts its values at finitely many of them. The restrictions are imposed simultaneously: a function belongs to the neighborhood only if it satisfies every one of those finite constraints. This description gives a practical way to test whether a proposed function is in the closure of a family.

Let $X$ be a set, with no topology required on $X$, and let $(Y,\mathcal{T}_Y)$ be a topological space. Write $Y^X$ for the set of all functions $f:X\to Y$. For each $x\in X$, the coordinate projection is the map $\pi_x:Y^X\to Y$ given by $\pi_x(f)=f(x)$. The pointwise topology on $Y^X$ is the product topology of these coordinate spaces.

Definition: A basic open set in the pointwise topology has the form $$ \bigcap_{i=1}^{n}\{g\in Y^X:g(x_i)\in U_i\}, $$ where $n$ is a nonnegative integer, $x_1,\ldots,x_n$ are distinct points of $X$, and each $U_i$ is open in $Y$. When $n=0$, the intersection is $Y^X$. A basic open neighborhood of $f$ is such a set with $f(x_i)\in U_i$ for every $i$.

Thus, a neighborhood can require a function to take values in specified open sets at specified inputs, while leaving all other inputs unrestricted. It need not require exact equality at the constrained inputs: the sets $U_i$ may be any open neighborhoods of the corresponding values $f(x_i)$. Conversely, every open neighborhood of $f$ contains a basic open neighborhood of $f$. These basic sets therefore suffice for testing whether every neighborhood of $f$ meets a given family.

Theorem (Finite-constraint closure test): Let $A\subseteq Y^X$ and $f\in Y^X$. Then $f\in\overline{A}$ if and only if, for every finite list of distinct points $x_1,\ldots,x_n\in X$ and open neighborhoods $U_i$ of $f(x_i)$ in $Y$, there is $g\in A$ such that $g(x_i)\in U_i$ for every $i$.

Proof. Suppose $f\in\overline{A}$. The set $\bigcap_{i=1}^{n}\{g:g(x_i)\in U_i\}$ is a basic open neighborhood of $f$. By the neighborhood test for closure, it intersects $A$. A function $g$ in that intersection belongs to $A$ and satisfies all the stated constraints.

Conversely, suppose every such finite collection of constraints can be satisfied by some $g\in A$. Every open neighborhood $V$ of $f$ contains a basic open neighborhood of $f$, specified by some finite list and open sets $U_i$. By the assumption, that basic neighborhood contains a member of $A$. Hence $V\cap A\ne\varnothing$. Since every open neighborhood of $f$ meets $A$, the neighborhood test gives $f\in\overline{A}$. $\square$

The empty cases are included in this statement. If $A=\varnothing$, no basic neighborhood meets $A$, including the whole space, so its closure is empty. If $X=\varnothing$, there is exactly one function from $X$ to $Y$, and the only finite constraint list is the empty one. The condition then says precisely that $A$ is nonempty, as required for that unique function to belong to $\overline{A}$.

Using the Finite-Constraint Test

Worked Example: Finite-Support Functions Are Dense

Let $X$ be any set and give $\mathbb{R}$ its standard topology. Let $A\subseteq\mathbb{R}^X$ consist of the functions $g$ that are zero except at finitely many points. We show that $\overline{A}=\mathbb{R}^X$, including when $X$ is empty.

Take any $f\in\mathbb{R}^X$ and any basic neighborhood of $f$, specified by distinct inputs $x_1,\ldots,x_n$ and open sets $U_i$ with $f(x_i)\in U_i$. Define $g:X\to\mathbb{R}$ by $$ g(x)= \begin{cases} f(x),&x\in\{x_1,\ldots,x_n\},\\ 0,&x\notin\{x_1,\ldots,x_n\}. \end{cases} $$ The function $g$ is zero outside a finite set, so $g\in A$. For every $i$, $g(x_i)=f(x_i)\in U_i$. Therefore $g$ lies in the chosen neighborhood. The finite-constraint test shows $f\in\overline{A}$.

This works for every $f\in\mathbb{R}^X$, so $\overline{A}=\mathbb{R}^X$. If the constraint list is empty, the same construction gives the zero function, which belongs to $A$. If $X=\varnothing$, the unique function has empty support and is itself in $A$. These cases agree with the conclusion.

Check: At every constrained input, the constructed function equals the target exactly. Thus it satisfies the original open-set requirement, not merely an approximate version of it.

Worked Example: The Closure of the Constant Functions

Give $X=\mathbb{R}$ no role other than as a set of inputs, and give the output space $\mathbb{R}$ its standard topology. Let $A\subseteq\mathbb{R}^{\mathbb{R}}$ be the family of constant functions. We show that $\overline{A}=A$.

Every member $f\in A$ belongs to its own closure, since every open neighborhood of $f$ contains $f$, which is in $A$. Now let $f$ be nonconstant. There are inputs $x,y\in\mathbb{R}$ with $f(x)\ne f(y)$. Set $d=|f(x)-f(y)|$, so $d>0$, and let $r=d/3$. Consider the basic neighborhood of $f$ requiring $$ |g(x)-f(x)|<r \quad\hbox{and}\quad |g(y)-f(y)|<r. $$ These are open interval constraints at two distinct inputs. If a constant function $g$ belonged to this neighborhood, then $g(x)=g(y)$. But the triangle inequality would give $$ d=|f(x)-f(y)| \le |f(x)-g(x)|+|g(y)-f(y)| <2r=\frac{2d}{3}<d, $$ which is impossible. This neighborhood misses $A$, so $f\notin\overline{A}$. Hence the closure contains exactly the constant functions.

For a numerical instance, take $f(t)=t$, with constraints at $0$ and $1$, each of radius $1/3$. A constant $g$ in the proposed neighborhood would have $|g(0)|<1/3$ and $|g(1)-1|<1/3$. Since $g(0)=g(1)$, the triangle inequality would imply $1<2/3$, a contradiction. This verifies the separation directly in this case.

A Finite Witness Can Force a Whole Function

Sometimes a small number of evaluations determines enough about a function to exclude every member of a family from a neighborhood. The next example shows that the finite-constraint test can prove a family is closed, even though its members are functions on an infinite set.

Worked Example: The Affine Functions Are Closed

Give $\mathbb{R}$ its standard topology, and consider the family $A\subseteq\mathbb{R}^{\mathbb{R}}$ of affine functions $g(t)=a+bt$, where $a,b\in\mathbb{R}$. We prove $\overline{A}=A$.

Every $f\in A$ belongs to $\overline{A}$, because $f\in A$. For the reverse inclusion, take $f\notin A$. The values $f(0)$ and $f(1)$ determine a candidate affine function $$ h(t)=f(0)+(f(1)-f(0))t. $$ Because $f$ is not affine, there is some $t\in\mathbb{R}$ for which $f(t)\ne h(t)$. This $t$ is different from $0$ and $1$, since $h(0)=f(0)$ and $h(1)=f(1)$. Write $\delta=|f(t)-h(t)|>0$, and set $C=|1-t|+|t|$. In particular $C\ge 1$, since $|1-t|+|t|\ge |(1-t)+t|=1$. Choose $r=\delta/(2(C+1))>0$.

Take the basic neighborhood of $f$ requiring $|g(s)-f(s)|<r$ at each of the three inputs $s=0,1,t$. Any affine function satisfies $g(t)=(1-t)g(0)+tg(1)$. Thus, if $g$ belonged to this neighborhood, then $$ \begin{aligned} |g(t)-h(t)| &=\left|(1-t)(g(0)-f(0))+t(g(1)-f(1))\right|\\ &<\bigl(|1-t|+|t|\bigr)r=Cr. \end{aligned} $$ Also $|f(t)-g(t)|<r$. Consequently, $$ \delta=|f(t)-h(t)| \le |f(t)-g(t)|+|g(t)-h(t)| <(C+1)r=\frac{\delta}{2}, $$ which is impossible because $\delta>0$. The neighborhood therefore misses $A$, and $f\notin\overline{A}$. We have proved $\overline{A}=A$.

For a concrete calculation, let $f(t)=t^2$ and constrain values at $0,1,2$ to within $1/10$ of $0,1,4$, respectively. If an affine $g$ met the first two constraints, then $g(2)=2g(1)-g(0)$, so $$ |g(2)-2| =|2(g(1)-1)-g(0)| \le 2|g(1)-1|+|g(0)| <\frac{3}{10}. $$ But the third constraint gives $|g(2)-4|<1/10$, and hence $$ |g(2)-2|\ge 2-|g(2)-4|>\frac{19}{10}. $$ The bounds $ |g(2)-2|<3/10$ and $ |g(2)-2|>19/10$ cannot both hold. Thus this finite-coordinate neighborhood of $t^2$ contains no affine function.

A Common Pitfall: Checking Inputs Separately

To test whether $f\in\overline{A}$, it is not enough to show that for each individual input $x$, some member of $A$ takes a value near $f(x)$. The same member must meet all the constraints in a chosen finite list.

For example, let $A$ again be all constant real-valued functions on $\mathbb{R}$, and let $f(t)=t$. At the input $0$, the constant function $g_0(t)=0$ agrees exactly with $f$; at the input $1$, the constant function $g_1(t)=1$ agrees exactly with $f$. The incorrect conclusion would be that these separate matches show $f\in\overline{A}$. They do not: $g_0$ and $g_1$ are different functions.

The correction is to test a single neighborhood with both constraints. Require $|g(0)-0|<1/3$ and $|g(1)-1|<1/3$. A constant $g$ satisfying both would imply $$ 1=|0-1| \le |g(0)-0|+|g(1)-1| <\frac{2}{3}, $$ which is impossible. Thus this one basic neighborhood misses all of $A$, and $f\notin\overline{A}$.

Takeaway: A pointwise-topology neighborhood specifies open conditions at finitely many inputs at once. To prove closure membership, meet every such finite collection of conditions with a member of the family; to disprove it, find one finite collection that no member can meet.

Check Your Understanding

For each question, distinguish a constraint at one input from a simultaneous finite list of constraints.

Write a basic neighborhood of $f\in Y^X$ that constrains $f$'s values at three distinct inputs.
Let $A$ be the functions on $\mathbb{R}$ that vanish outside a finite set. Which construction shows that every $f\in\mathbb{R}^{\mathbb{R}}$ belongs to $\overline{A}$?
For the constant functions on $\mathbb{R}$, explain why a nonconstant $f$ can be excluded from their closure using two inputs.
Show that the function $t\mapsto t^2$ is not in the closure of the affine functions by choosing finitely many inputs and open interval constraints.
If $X=\varnothing$, how does the finite-constraint closure test decide whether the unique function $X\to Y$ belongs to $\overline{A}$?

Basic Neighborhoods in the Pointwise Topology

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