Check the Hypotheses Before Using a Separation Theorem
Separation and extension arguments often fail not because their central idea is wrong, but because a hypothesis has been dropped or a conclusion has been reversed. A separation theorem concerns particular kinds of sets in a particular kind of space; an extension theorem has its own domain and range conditions. This tutorial checks four frequent errors: confusing normality with complete regularity, separating sets that are not closed, omitting boundedness when using the bounded Tietze theorem, and interchanging the values assigned by a Urysohn function. It also examines a related continuity error: assuming that a pointwise limit of continuous functions must be continuous.
We use the following conventions. A space \(X\) is normal if every pair of disjoint closed subsets has disjoint open neighborhoods. This definition does not include \(T_1\); some texts build \(T_1\) into their definition of normality. A space is \(T_1\) if every singleton is closed. We say here that \(X\) is completely regular if, whenever \(x\in X\) and \(F\subseteq X\) is closed with \(x\notin F\), there is a continuous \(f:X\to[0,1]\) such that \(f(x)=0\) and \(f(y)=1\) for every \(y\in F\).
These definitions ask for different things. Normality supplies open neighborhoods of two disjoint closed sets. Complete regularity asks for a continuous real-valued function separating a point from a closed set. The distinction matters especially when singletons are not closed.
For a \(T_1\) normal space, this result does imply complete regularity: \(\{x\}\) is closed, so for a closed \(F\) not containing \(x\), the sets \(\{x\}\) and \(F\) are disjoint and closed, and Urysohn's lemma applies. The \(T_1\) condition is what lets us use the singleton as one of the closed sets. Without it, normality alone does not guarantee the point-versus-closed-set function condition.
Worked Example: Normality Without Point-Closed-Set Separation
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). Its closed sets, obtained by taking complements of open sets, are \(\varnothing,\{b\},X\). The space is normal under the definition above: the only disjoint pairs of closed sets are pairs in which at least one set is empty, and \(\varnothing\) and \(X\) are disjoint open neighborhoods of such a pair. But \(X\) is not \(T_1\), since \(\{a\}\) is not closed.
Now take the point \(a\) and the closed set \(F=\{b\}\). They are disjoint. Suppose a continuous function \(f:X\to[0,1]\) had \(f(a)=0\) and \(f(b)=1\). The set \((1/2,3/2)\cap[0,1]=(1/2,1]\) is open in the subspace \([0,1]\). Its preimage under \(f\) would be \(\{b\}\), which is not open in \(X\). This contradicts continuity, so no such function exists. This example shows why normality without \(T_1\) cannot simply be substituted for complete regularity. If a text defines “normal” to include \(T_1\), this particular space is not normal under that text's convention.
Do Not Drop the Closedness Hypothesis
Urysohn's lemma requires disjoint closed sets. Disjointness alone does not ensure that a continuous function can assign one set the value \(0\) and the other the value \(1\). The obstruction is that points of one set can approach a point of the other, forcing incompatible values by continuity.
Worked Example: Disjoint Sets That Cannot Be Separated by a Continuous Function
Give \(\mathbb{R}\) its standard topology, and let \(A=(0,1)\) and \(B=\{1\}\). The sets are disjoint, but \(A\) is not closed: every open interval around \(1\) meets \(A\), while \(1\notin A\). Suppose there were a continuous \(f:\mathbb{R}\to[0,1]\) with \(f(x)=0\) for \(x\in A\) and \(f(1)=1\).
For each integer \(n\ge2\), set \(x_n=1-1/n\). Then \(0<x_n<1\), so \(x_n\in A\) and \(f(x_n)=0\). Also \(x_n\to1\). Continuity at \(1\) would imply \(f(x_n)\to f(1)=1\), but the sequence of function values is constantly \(0\), so it converges to \(0\). Since a real sequence cannot have two different limits, this is a contradiction. The error was applying a theorem for disjoint closed sets to a pair for which one set is not closed.
The same check is useful in extension arguments: establish that the domain on which the given function is defined is closed before invoking a Tietze extension theorem. A continuous function on an arbitrary subset need not have the extension guaranteed by that theorem.
Boundedness Is a Hypothesis of the Bounded Tietze Theorem
The bounded Tietze extension theorem says that if \(X\) is normal, \(A\subseteq X\) is closed, and \(f:A\to[-M,M]\) is continuous for some finite \(M\ge0\), then \(f\) has a continuous extension \(F:X\to[-M,M]\). Both the closedness of \(A\) and the bounded range condition are part of this statement. There is also an unbounded version of the Tietze extension theorem for real-valued functions, but it is a different result; the bounded theorem does not by itself justify that version.
Worked Example: Why an Unbounded Function Has No Bounded Extension
Let \(X=\mathbb{R}\), \(A=\mathbb{Z}\), and define \(f:A\to\mathbb{R}\) by \(f(n)=n\). First, \(A\) is closed: if \(x\notin\mathbb{Z}\), write \(x=m+t\) with \(m\in\mathbb{Z}\) and \(0<t<1\); the interval centered at \(x\) with radius \(\min(t,1-t)/2\) contains no integer. Each integer \(n\) is isolated in the subspace \(A\), because \((n-1/2,n+1/2)\cap A=\{n\}\). Thus every subset of \(A\) is open in \(A\), so \(f\) is continuous.
The space \(\mathbb{R}\) is normal. To verify this, for nonempty disjoint closed sets \(C,F\subseteq\mathbb{R}\), define \(d(x,C)=\inf_{c\in C}|x-c|\) and \(d(x,F)=\inf_{y\in F}|x-y|\). Each distance function is continuous, since the triangle inequality gives \(|d(x,C)-d(z,C)|\le |x-z|\), and likewise for \(F\). The open sets \(U=\{x:d(x,C)<d(x,F)/2\}\) and \(V=\{x:d(x,F)<d(x,C)/2\}\) contain \(C\) and \(F\), respectively: at a point of either closed set, its distance to that set is zero and its distance to the other closed set is positive. They are disjoint, since both displayed strict inequalities cannot hold at once. If either closed set is empty, \(X\) and \(\varnothing\) are disjoint open neighborhoods. This verifies normality.
Now any extension \(F:\mathbb{R}\to\mathbb{R}\) of \(f\) must satisfy \(F(n)=n\) for every integer \(n\). In particular, for every positive integer \(N\), \(F(N)=N\), so \(F\) cannot be bounded. The bounded Tietze theorem cannot be applied to this \(f\): its values on \(A\) are unbounded. The obstruction is not a failure of continuity or closedness; it is precisely the missing boundedness hypothesis for a bounded extension.
Track Which Set Receives Which Value
Urysohn's lemma is often stated with the convention \(f(C)=\{0\}\) and \(f(F)=\{1\}\). A proof may instead be arranged with the values reversed, but then the resulting function does not satisfy the original prescription. The reversal can be corrected by replacing \(f\) with \(1-f\), because \(1-f\) is continuous whenever \(f\) is continuous.
Worked Example: Check the Endpoint Values
On \(I=[0,1]\) with its subspace topology, let \(C=\{0\}\) and \(F=\{1\}\). These are disjoint closed sets. The function \(f(x)=x\) is continuous and has \(f(0)=0\), \(f(1)=1\), so it has the required orientation.
A reversed formula, \(g(x)=1-x\), instead gives \(g(0)=1-0=1\) and \(g(1)=1-1=0\). It is continuous, but it does not meet the stated convention: the value on \(C\) is \(1\), not \(0\), and the value on \(F\) is \(0\), not \(1\). Applying \(1-g\) repairs the orientation: \((1-g)(x)=1-(1-x)=x\), with \((1-g)(0)=0\) and \((1-g)(1)=1\). The check is to substitute points from each specified set into the final formula, not merely to confirm continuity.
Pointwise Limits Need Not Be Continuous
A sequence of functions \(f_n:X\to\mathbb{R}\) converges pointwise to \(f\) if, for each fixed \(x\in X\), the real sequence \(f_n(x)\) converges to \(f(x)\). This condition allows the rate of convergence to depend on \(x\). It does not, by itself, preserve continuity.
Worked Example: A Pointwise Limit With a Discontinuity
On \(I=[0,1]\), define \(f_n(x)=x^n\) for \(n\in\mathbb{N}\). Each \(f_n\) is continuous, since it is a polynomial restricted to \(I\). If \(0\le x<1\), then \(x^n\to0\); at \(x=1\), \(f_n(1)=1^n=1\) for every \(n\). Thus the pointwise limit is \(f(x)=0\) for \(0\le x<1\), and \(f(1)=1\).
This limit is not continuous at \(1\). For integers \(k\ge2\), let \(y_k=1-1/k\). Then \(0\le y_k<1\), \(y_k\to1\), and \(f(y_k)=0\) for every \(k\), whereas \(f(1)=1\). If \(f\) were continuous at \(1\), continuity would require \(f(y_k)\to f(1)\), but the left side is constantly \(0\). This example disproves the claim that a pointwise limit of continuous functions is necessarily continuous.
One useful sufficient condition is uniform convergence: \(f_n\) converges uniformly to \(f\) if, for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(|f_n(x)-f(x)|<\varepsilon\) for every \(x\in X\) and every \(n\ge N\). Unlike pointwise convergence, this controls the error at all points using the same \(N\).
Proof. Let \(x_0\in X\), and let \(\varepsilon>0\). By uniform convergence, choose \(N\in\mathbb{N}\) such that \(|f_N(x)-f(x)|<\varepsilon/3\) for every \(x\in X\). Since \(f_N\) is continuous at \(x_0\), there is an open neighborhood \(U\) of \(x_0\) such that \(|f_N(x)-f_N(x_0)|<\varepsilon/3\) for all \(x\in U\). For any \(x\in U\), the triangle inequality gives \[ |f(x)-f(x_0)|\le |f(x)-f_N(x)|+|f_N(x)-f_N(x_0)|+|f_N(x_0)-f(x_0)|<\varepsilon. \] The first and third terms are each less than \(\varepsilon/3\) by the same uniform bound, and the middle term is less than \(\varepsilon/3\) by continuity of \(f_N\). This is the \(\varepsilon\)-neighborhood condition for continuity of \(f\) at \(x_0\). Since \(x_0\) was arbitrary, \(f\) is continuous. \(\square\)
A Short Diagnostic Before Concluding
When checking a separation or extension proof, identify the exact theorem being used and match every hypothesis to the objects in the argument. In particular, ask whether the space has the separation axiom required, whether the relevant sets are closed, whether the function is bounded when the bounded Tietze theorem is invoked, and whether the prescribed values are assigned to the correct sets. For a continuity claim about a limit, ask whether convergence is only pointwise or is uniform.
Common pitfall. A wrong argument might say, “The sets \(A=(0,1)\) and \(B=\{1\}\) are disjoint, so Urysohn's lemma gives a continuous function equal to \(0\) on \(A\) and \(1\) on \(B\).” Disjointness alone is not the hypothesis: \(A\) is not closed. The sequence \(1-1/n\in A\) for \(n\ge2\) converges to \(1\in B\), so the demanded values would violate continuity. The correction is to check closedness before invoking Urysohn's lemma, and not to treat an unverified hypothesis as automatic.
| Claim being considered | Condition to check |
|---|---|
| Separate a point from a closed set by a continuous function | Complete regularity; if deducing it from normality, check \(T_1\) so the singleton is closed. |
| Apply Urysohn's lemma to two sets | The sets must be disjoint and closed, and the space must be normal. |
| Use the bounded Tietze extension theorem | The domain subset must be closed, the ambient space normal, and the given function bounded. |
| Claim a limit function is continuous | Pointwise convergence alone is insufficient; uniform convergence is a sufficient condition. |
Check Your Understanding
For each question, state which hypothesis or value check matters, and justify your answer.
- In the two-point topology \(\{\varnothing,\{a\},X\}\), which singleton is not closed, and why does that matter when deducing complete regularity from normality?
- Can Urysohn's lemma be applied to \(A=(0,1)\) and \(B=\{1\}\) in \(\mathbb{R}\)? Identify the failed hypothesis.
- If a continuous function on a closed subset is unbounded, what prevents the bounded Tietze theorem from guaranteeing a bounded extension?
- For \(C=\{0\}\), \(F=\{1\}\), and \(g(x)=1-x\) on \([0,1]\), compute both endpoint values and state which prescription \(g\) satisfies.
- What is the pointwise limit of \(f_n(x)=x^n\) on \([0,1]\), and at which point is it discontinuous?