Separating Points

What It Means to Separate Points

In the Stone-Weierstrass argument, point separation is the condition that lets an algebra distinguish any two different locations in its domain. It is a local capability: for each pair, some function in the algebra must take different values at the two points. The function that works can depend on the pair. This is the starting point for the Two-Point Interpolation Theorem from Stone-Weierstrass Motivation, and it will also let us construct functions with prescribed values on finite sets.

Let \(K\) be a set and let \(\mathcal{F}\) be a family of real-valued functions on \(K\). The definition does not require \(K\) to be compact or \(\mathcal{F}\) to be an algebra. Those additional hypotheses matter in some applications, but not for the meaning of separation itself.

Definition: The family \(\mathcal{F}\) separates points of \(K\) if, for every pair of distinct points \(x,y\in K\), there is a function \(f\in\mathcal{F}\) such that \(f(x)\ne f(y)\). If no such function exists for some distinct \(x,y\), the family does not separate those points.

If \(K\) has at most one point, every family separates points of \(K\) vacuously: there are no distinct pairs to test. For larger sets, the quantifiers are important. We require a suitable function for each pair, not one function that distinguishes all pairs at once. Indeed, a single function separates every pair precisely when it is injective on \(K\); a family may separate points even when none of its individual functions is injective.

First Tests for Separation

For a subset \(K\) of the real line, the coordinate function \(e(x)=x\) immediately separates points: if \(x\ne y\), then \(e(x)\ne e(y)\). Thus any family containing this function separates points. More generally, a family may separate points through a different coordinate or observable, even when it does not contain the identity function.

A useful test arises when an algebra is generated by one function. Suppose \(g:K\to\mathbb{R}\), and let \(\mathbb{R}[g]\) denote the family of functions \(P\circ g\), where \(P\) ranges over real polynomials. This is a unital subalgebra of the real-valued functions on \(K\): sums and products correspond to sums and products of polynomials, and the constant functions are included.

Proposition: The algebra \(\mathbb{R}[g]\) separates points of \(K\) if and only if \(g\) is injective on \(K\).

Proof. If \(g\) is injective, then for distinct \(x,y\in K\) we have \(g(x)\ne g(y)\). The function \(g\) itself belongs to \(\mathbb{R}[g]\), so it separates this pair. Hence the algebra separates points.

Conversely, suppose \(g\) is not injective. Then there are distinct \(x,y\in K\) with \(g(x)=g(y)\). For every polynomial \(P\),

(P\circ g)(x)=P(g(x))=P(g(y))=(P\circ g)(y).

Thus no element of \(\mathbb{R}[g]\) distinguishes this pair, so the algebra does not separate points. This proves both directions. \(\square\)

Worked Example: A Function That Separates on One Domain but Not Another

Consider \(g(x)=x^2\). On \(K=[0,1]\), this function is injective. To check this, take \(x,y\in[0,1]\). If \(x^2=y^2\), then \((x-y)(x+y)=0\). Since \(x+y\geq0\), either \(x=y\) or \(x+y=0\); in the latter case nonnegativity forces \(x=y=0\) as well. Thus \(x=y\), and \(\mathbb{R}[g]\) separates points on \([0,1]\).

On \(K=[-1,1]\), however, \(g(-1)=1=g(1)\), although \(-1\ne1\). Every function in the algebra has the form \(P(x^2)\), so for every polynomial \(P\),

$$ P((-1)^2)=P(1)=P(1^2). $$

The algebra therefore fails to separate \(-1\) and \(1\). The same formula \(x^2\) has different separation behavior because the domain has changed.

Worked Example: Even Functions on a Symmetric Set

Let \(K=[-2,2]\), and let \(\mathcal{E}\) be the family of even functions on \(K\), meaning functions \(f\) satisfying \(f(-x)=f(x)\) for every \(x\in K\). For any \(f\in\mathcal{E}\), the two distinct points \(1\) and \(-1\) satisfy

$$ f(1)=f(-1). $$

Consequently, \(\mathcal{E}\) does not separate points of \(K\). This conclusion applies to the entire family, not just to one even function. By contrast, if we restrict the even function \(x^2\) to \([0,2]\), it is injective there, so the algebra of polynomials in \(x^2\) separates points on that restricted domain.

Separation Gives Finite Interpolation

For a unital subalgebra, separation provides more than the ability to distinguish one pair at a time. By combining pairwise separators using algebra operations, we can prescribe values at any finite list of distinct points. This is a finite interpolation result; it does not claim that every continuous function on \(K\) belongs to the algebra.

Theorem (Finite Interpolation from Point Separation): Let \(K\) be a set and let \(A\) be a unital subalgebra of the real-valued functions on \(K\) that separates points. For distinct points \(x_1,\ldots,x_m\in K\) and arbitrary real numbers \(c_1,\ldots,c_m\), there is a function \(p\in A\) such that \(p(x_i)=c_i\) for every \(i=1,\ldots,m\).

Proof. If \(m=1\), the constant function \(p(x)=c_1\) belongs to \(A\) because \(A\) is unital and is closed under scalar multiplication. It has the required value.

Now suppose \(m\geq2\). Fix distinct indices \(i\) and \(j\). Since \(A\) separates points, there is an \(a_{ij}\in A\) such that \(a_{ij}(x_i)\ne a_{ij}(x_j)\). The denominator in the following expression is therefore nonzero:

q_{ij} = \frac{a_{ij}-a_{ij}(x_j)\mathbf{1}} {a_{ij}(x_i)-a_{ij}(x_j)},

where \(\mathbf{1}\) is the constant function with value \(1\). Because \(A\) contains constants and is closed under addition and scalar multiplication, \(q_{ij}\in A\). Substitution gives

q_{ij}(x_i) = \frac{a_{ij}(x_i)-a_{ij}(x_j)} {a_{ij}(x_i)-a_{ij}(x_j)} =1, \qquad q_{ij}(x_j) = \frac{a_{ij}(x_j)-a_{ij}(x_j)} {a_{ij}(x_i)-a_{ij}(x_j)} =0.

For each \(i\), define \(r_i=\prod_{j\ne i}q_{ij}\), with the product taken over the finitely many indices \(j\in\{1,\ldots,m\}\) other than \(i\). Since \(A\) is closed under finite products, \(r_i\in A\). At \(x_i\), every factor is \(1\), so \(r_i(x_i)=1\). At \(x_k\) for \(k\ne i\), the factor \(q_{ik}\) occurs in the product and has value \(0\). Therefore \(r_i(x_k)=0\) whenever \(k\ne i\).

Set \(p=\sum_{i=1}^m c_i r_i\), which belongs to \(A\) because \(A\) is closed under finite sums and scalar multiplication. At a prescribed point \(x_k\), all terms in the sum vanish except the term with \(i=k\). Hence

p(x_k)=\sum_{i=1}^m c_i r_i(x_k)=c_k r_k(x_k)=c_k.

This holds for each \(k\), proving the theorem. \(\square\)

The construction explains why the algebra hypothesis matters. Separation supplies an individual function for each pair; subtraction and rescaling turn it into a function with values \(1\) and \(0\) at that pair; multiplication makes the function vanish at all the other prescribed points; and addition combines the resulting pieces. Theorem (Two-Point Interpolation) from Stone-Weierstrass Motivation gives the two-point conclusion directly. The construction here shows how the algebra operations extend that capability to an arbitrary finite set.

Worked Example: Prescribing Values at Three Points

Take \(K=\mathbb{R}\) and \(A=\mathbb{R}[x]\), the algebra of real polynomials. The identity function \(x\mapsto x\) separates distinct real numbers. At the distinct points \(-1,0,2\), define

L_{-1}(x)=\frac{x(x-2)}{3},\qquad L_0(x)=-\frac{(x+1)(x-2)}{2},\qquad L_2(x)=\frac{x(x+1)}{6}.

Direct substitution verifies their values. At \(x=-1\), the three values are \(1,0,0\); at \(x=0\), they are \(0,1,0\); and at \(x=2\), they are \(0,0,1\). For example, \(L_{-1}(-1)=(-1)(-3)/3=1\), while \(L_0(-1)=0\) and \(L_2(-1)=0\). At \(0\), \(L_0(0)=-(1)(-2)/2=1\), and the other two expressions contain a factor of \(x\). At \(2\), \(L_2(2)=2\cdot3/6=1\), while \(L_{-1}(2)=0\) and \(L_0(2)=0\).

Thus the polynomial

p(x)=4L_{-1}(x)-L_0(x)+2L_2(x)

satisfies \(p(-1)=4\), \(p(0)=-1\), and \(p(2)=2\). Each \(L\) is a polynomial, so \(p\in A\). This is a concrete instance of finite interpolation; the values at the three nodes follow from the verified values of the three component polynomials.

Uniform Approximation Preserves Separation

Point separation is also stable under taking a uniform closure. The reason is quantitative: once a function gives two points different values, a sufficiently small uniform perturbation cannot erase the difference. This is useful when working with \(\overline{A}\), since the Stone-Weierstrass proof uses functions in that closure as well as functions in the original algebra.

Theorem (Separation Is Preserved by Uniform Closure): Let \(K\) be a nonempty compact set and let \(A\subseteq C(K)\) separate points. Then the uniform closure \(\overline{A}\) also separates points of \(K\).

Proof. Fix distinct \(x,y\in K\). Since \(A\) separates points, choose \(h\in A\) with \(h(x)\ne h(y)\), and set \(d=|h(x)-h(y)|\), so \(d>0\). Because \(h\in A\subseteq\overline{A}\), it is already an element of the closure and separates this pair. More generally, the stability estimate can be seen by choosing any \(g\in C(K)\) with \(\|g-h\|_\infty<d/3\), which is possible for members of \(\overline{A}\) approximating \(h\) (and in particular can be taken from \(A\) when needed). Then

\begin{aligned} |g(x)-g(y)| &\geq |h(x)-h(y)|-|g(x)-h(x)|-|g(y)-h(y)|\\ &\geq d-2\|g-h\|_\infty\\ &>d-\frac{2d}{3} =\frac{d}{3}>0. \end{aligned}

Thus every function within \(d/3\) in the supremum norm still separates \(x\) and \(y\). In particular, for each pair the original separator \(h\) belongs to \(\overline{A}\), so \(\overline{A}\) separates that pair. Since the pair was arbitrary, \(\overline{A}\) separates points of \(K\). \(\square\)

The robust estimate is often the more useful part of the proof: separation of a fixed pair does not disappear under a perturbation smaller than half the gap between the two values. The theorem’s closure conclusion itself also follows directly from \(A\subseteq\overline{A}\); the estimate records why separation is stable under uniform approximation and can be used when transferring a particular separator to a nearby function.

What Separation Does—and Does Not—Say

Separation is a structural hypothesis, not an approximation conclusion by itself. It says that no two points are permanently indistinguishable to every function in the family. For a unital subalgebra of \(C(K)\) on a nonempty compact set, the Real Stone-Weierstrass Theorem combines this property with the algebra operations and compactness to obtain density in \(C(K)\). The separation condition is one ingredient in that theorem, not a claim that the algebra already contains every continuous function.

A common error is to find one function that separates a chosen pair and then assume it separates all pairs. The definition only guarantees a possibly different function for each pair. A second error is to confuse a family with the algebra it generates: even if a chosen generator fails to be injective, adding other functions could restore separation. The proposition about \(\mathbb{R}[g]\) applies specifically to the algebra generated by that single function.

Question	What to check	What does not suffice
Does a family separate points?	For every distinct pair, find at least one function with unequal values.	Finding a separator for only one pair.
Does \(\mathbb{R}[g]\) separate points?	Check whether \(g\) is injective on the specified domain.	Checking injectivity on a larger or smaller domain instead.
Can values be prescribed on finitely many points?	Use separation together with a unital algebra’s sums, scalar multiples, and products.	Assuming separation alone gives finite interpolation for an arbitrary family.
Does separation survive uniform approximation?	Compare the perturbation size with the original value gap.	Assuming an arbitrary large perturbation preserves a particular separator.

Takeaway: A family separates points when each distinct pair can be distinguished by some member. In a unital algebra, pairwise separators can be combined to interpolate arbitrary data on a finite set, and a positive value gap makes separation stable under sufficiently small uniform perturbations.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

Why does every family of functions separate points on a set with exactly one point?
For \(g(x)=x^2\), explain why the algebra \(\mathbb{R}[g]\) separates points on \([0,1]\) but not on \([-1,1]\).
In the finite interpolation proof, why is the denominator in \(q_{ij}\) nonzero?
How does the product defining \(r_i\) ensure that it vanishes at every prescribed point other than \(x_i\)?
If \(h(x)\ne h(y)\) and \(d=|h(x)-h(y)|\), why does \(\|g-h\|_\infty<d/3\) guarantee that \(g(x)\ne g(y)\)?

What You'll Learn