From Unit Values to Prescribed Values
Urysohn’s lemma gives a continuous function whose values are \(0\) and \(1\) on two disjoint closed sets. Those particular values are often only an intermediate step: we may want different endpoint values, values in a smaller interval, or the same value on both sets. Affine transformations and the complement \(1-u\) make these changes while preserving continuity.
Throughout, \(X\) is a normal topological space, and all real-valued functions have codomain \(\mathbb{R}\) with its standard topology unless a smaller codomain is specified. We use the convention that the Urysohn function is \(0\) on the first closed set and \(1\) on the second.
Here \([0,1]\) has its subspace topology from \(\mathbb{R}\). The rescaling argument below uses the lemma’s function, but does not depend on how that function was constructed.
The Rescaling Theorem
Proof. If \(c=d\), take the constant function \(f(x)=c\) for every \(x\in X\). It is continuous because the preimage of any open set is either \(X\) or \(\varnothing\); it has values in \(J\) because \(c\in J\), and it has the required value on both sets.
Now suppose \(c\ne d\). If both \(E\) and \(F\) are nonempty, take \(u:X\to[0,1]\) from Urysohn’s lemma and define \(f(x)=c+(d-c)u(x)\). This is continuous: the affine function \(h(t)=c+(d-c)t\) satisfies \(|h(s)-h(t)|=|d-c||s-t|\), so it is continuous, and \(f=h\circ u\) is continuous by the composition rule. On \(E\), \(u=0\), hence \(f=c\); on \(F\), \(u=1\), hence \(f=c+(d-c)=d\).
For every \(x\in X\), write \(t=u(x)\), so \(0\le t\le1\). Then \(f(x)=(1-t)c+td\). This is a weighted average of \(c\) and \(d\), with nonnegative weights \(1-t,t\) whose sum is \(1\). Therefore \(f(x)\) lies between \(c\) and \(d\). Since \(J\) is an interval containing both \(c\) and \(d\), it contains every value between them; thus \(f(X)\subseteq J\).
If one of \(E,F\) is empty and the other is nonempty, the constant function with the prescribed value on the nonempty set works. If both are empty, any constant function with value in \(J\) works; such a value exists because \(c\in J\). This handles all cases. \(\square\)
The same formula works whether \(d\) is larger or smaller than \(c\). When \(d<c\), the coefficient \(d-c\) is negative, so the function decreases as \(u\) increases. Equivalently, one may first replace \(u\) by its complement \(1-u\), which is \(1\) on \(E\) and \(0\) on \(F\). The complement is continuous because \(t\mapsto1-t\) is continuous.
Worked Example: Rescaling to a Larger Interval
In \(\mathbb{R}\) with its standard topology, let \(E=(-\infty,0]\) and \(F=[2,\infty)\). These sets are closed and disjoint. Define
On each of the three intervals the formula is continuous. At \(x=0\), the middle formula gives \(0/2=0\), agreeing with the left formula. At \(x=2\), it gives \(2/2=1\), agreeing with the right formula. Thus the pieces join continuously, and \(u=0\) on \(E\), \(u=1\) on \(F\), with \(0\le u(x)\le1\).
We want values \(-2\) on \(E\) and \(4\) on \(F\), with range in \(J=[-3,5]\). The affine rescaling is \(f(x)=-2+(4-(-2))u(x)=-2+6u(x)\). For \(x\in E\), \(u(x)=0\), so \(f(x)=-2+6(0)=-2\). For \(x\in F\), \(u(x)=1\), so \(f(x)=-2+6(1)=4\). For every \(x\), \(0\le u(x)\le1\), hence \(-2\le -2+6u(x)\le4\). Since \([-2,4]\subseteq[-3,5]\), the range lies in the requested interval.
Check: at \(x=1\), the middle formula gives \(u(1)=1/2\), and then \(f(1)=-2+6(1/2)=-2+3=1\), which lies between \(-2\) and \(4\).
Worked Example: Reversing the Assigned Values
Use the same space, sets, and function \(u\), but now assign \(3/4\) on \(E\) and \(1/4\) on \(F\), keeping the range in \(J=[0,1]\). Because the values are reversed relative to \(u\), first take \(v=1-u\). Then \(v=1\) on \(E\), \(v=0\) on \(F\), and \(v\) is continuous. Define \(f(x)=1/4+(1/2)v(x)\).
On \(E\), \(v=1\), so \(f=1/4+1/2=3/4\). On \(F\), \(v=0\), so \(f=1/4\). Since \(0\le v(x)\le1\), the formula gives \(1/4\le f(x)\le3/4\), which is contained in \([0,1]\).
The direct affine formula gives the same answer: \(f(x)=3/4+(1/4-3/4)u(x)=3/4-(1/2)u(x)\). At \(x=1\), \(u(1)=1/2\), so this gives \(f(1)=3/4-1/4=1/2\). The complement formula gives \(v(1)=1-1/2=1/2\), and then \(1/4+(1/2)(1/2)=1/2\) as well. This checks both the reversed assignment and the equivalence of the formulas.
Prescribing Values Inside Any Bounded Interval
The theorem applies to any interval \(J\) as long as the prescribed values belong to it. In particular, if \(J=[L,U]\) with \(L<U\), then the rescaling \(f=L+(U-L)u\) maps the whole unit interval onto \([L,U]\): \(u=0\) gives \(L\), and \(u=1\) gives \(U\). If the desired values are interior points \(c,d\), use \(f=c+(d-c)u\) instead. Its range is between \(c\) and \(d\), so it need not fill all of \(J\).
There is no need for the interval to be closed. For example, if \(J=(0,1)\) and \(c,d\in(0,1)\), the segment between \(c\) and \(d\) stays inside \((0,1)\), so the same formula takes values in \(J\). The endpoint-rescaling formula \(L+(U-L)u\), however, takes the values \(L\) and \(U\); it does not map into an open interval \((L,U)\). To use an open interval as codomain, choose prescribed values strictly inside it.
Worked Example: A Finite Discrete Space
Let \(X=\{p,q,r\}\) with the discrete topology, in which every subset is open. Set \(E=\{p\}\) and \(F=\{q,r\}\). Both are closed because their complements are open, and they are disjoint. Define \(u(p)=0\), \(u(q)=1\), and \(u(r)=1\). This function is continuous: for every open subset \(V\) of \(\mathbb{R}\), its preimage \(u^{-1}(V)\) is a subset of \(X\), hence open in the discrete topology.
Take \(J=[-4,2]\), and prescribe \(-3\) on \(E\) and \(1/2\) on \(F\). The affine formula is \(f=-3+(1/2-(-3))u=-3+(7/2)u\). Thus \(f(p)=-3+(7/2)(0)=-3\), while \(f(q)=-3+(7/2)(1)=1/2\) and \(f(r)=-3+(7/2)(1)=1/2\). Its only values are \(-3\) and \(1/2\), both in \([-4,2]\).
Check: the required assignments hold on every point of each set: \(E\) contains only \(p\), and both points \(q,r\) of \(F\) receive \(1/2\). The numerical formula also gives \(-3+7/2=1/2\) at each point where \(u=1\).
A Normalized Formulation and a Common Pitfall
Proof. Given such a function \(u\), set \(f=c+(d-c)u\). The affine map is continuous, and the endpoint values follow by substituting \(u=0\) on \(E\) and \(u=1\) on \(F\). Its values lie between \(c\) and \(d\) because \(0\le u\le1\).
Conversely, given \(f\) with values between \(c\) and \(d\), define \(u=(f-c)/(d-c)\). The denominator is nonzero because \(c\ne d\), and this is a composition of continuous affine maps with \(f\), so \(u\) is continuous. If \(d>c\), the inequalities \(c\le f(x)\le d\) imply \(0\le(f(x)-c)/(d-c)\le1\). If \(d<c\), then \(d\le f(x)\le c\); both numerator and denominator are nonpositive, and division by the negative denominator gives \(0\le(f(x)-c)/(d-c)\le1\) again. Substitution gives \(u=0\) on \(E\) and \(u=1\) on \(F\). Thus the two functions determine each other. \(\square\)
Common pitfall. The scale factor must be the difference of the target values, not the second target value itself. For example, if \(u=0\) on \(E\), \(u=1\) on \(F\), and the desired values are \(2\) and \(5\), the incorrect formula \(2+5u\) gives \(2\) on \(E\) but \(2+5(1)=7\) on \(F\). The correct formula is \(2+(5-2)u=2+3u\). Substitution checks it: on \(E\) the value is \(2+3(0)=2\), and on \(F\) it is \(2+3(1)=5\).
Rescaling changes the numerical values, not the separation mechanism. Normality supplies the original continuous separation; an affine map adjusts its range and endpoint values, and complementation reverses which closed set receives the larger value.
Check Your Understanding
In each question, state the assigned values and verify that the resulting function stays in the claimed range.
- If \(u=0\) on \(E\) and \(u=1\) on \(F\), what affine formula assigns values \(a\) on \(E\) and \(b\) on \(F\)?
- Why does \(1-u\) reverse the endpoint assignments while preserving continuity and the range \([0,1]\)?
- Suppose \(a=b\). Give a continuous function with that value on both closed sets, including when one or both sets are empty.
- If \(J=(0,1)\) and \(a,b\in J\), explain why \(a+(b-a)u\) still takes values in \(J\).
- For targets \(4\) on \(E\) and \(-2\) on \(F\), write the affine rescaling and check its two endpoint values.