Constructing Countably Many Separating Functions

From a Countable Base to Function Coordinates

Continuous functions can record topological information about points. To distinguish two points, it is enough to find a continuous function that takes different values at them. A stronger goal is to choose functions that also recover every open set: around each point of an open set, one function should be positive there and vanish outside the open set. In a regular second-countable space, a countable base supplies the data for making these choices countably many times.

Definition: A space is second-countable if its topology has a countable base $\mathcal{B}$: every open set is a union of members of $\mathcal{B}$. We use the convention that regular means $T_1$ and that a point and a closed set not containing it have disjoint open neighborhoods. A space is normal if any two disjoint closed sets have disjoint open neighborhoods.

The key base pairs are not arbitrary. We need pairs $B,C\in\mathcal{B}$ for which $\overline{B}\subseteq C$. For such a pair, the closed sets $\overline{B}$ and $X\setminus C$ are disjoint. Urysohn's lemma can then give a continuous function equal to $1$ on $\overline{B}$ and $0$ on $X\setminus C$. We first justify that the required normality is available.

Why Regularity and a Countable Base Give Normality

Proposition: Every regular second-countable space is normal.

Proof. Let $A$ and $D$ be disjoint closed subsets of $X$. If either is empty, the open sets $\varnothing$ and $X$ separate them. Assume both are nonempty. A second-countable space, and each of its subspaces, is Lindelöf: every open cover has a countable subcover. Indeed, the subspace has a countable base. For each basis element that is contained in a member of the cover, choose one such cover member. The chosen members form a countable subcover, since each point lies in a basis element contained in a cover member.

For each $a\in A$, regularity gives an open set $U_a$ containing $a$ whose closure misses $D$. To see the closure condition, take disjoint open neighborhoods of $a$ and $D$; the closure of the first misses the second. The sets $U_a$ cover $A$, so the Lindelöf property of the subspace $A$ gives a countable subcover $(U_i)$. Similarly, choose a countable open cover $(V_j)$ of $D$ such that each $\overline{V_j}$ misses $A$. Define \[ U=\bigcup_i\left(U_i\setminus\bigcup_{j\leq i}\overline{V_j}\right), \qquad V=\bigcup_j\left(V_j\setminus\bigcup_{i\leq j}\overline{U_i}\right). \] Each term in these unions is open, because it removes only a finite union of closed sets from an open set. Each $a\in A$ lies in some $U_i$ and in none of the $\overline{V_j}$, so $a\in U$; likewise $D\subseteq V$. If a point belonged to the $i$-th term of $U$ and the $j$-th term of $V$, then $i\leq j$ would make the latter term exclude $\overline{U_i}$, or $j\leq i$ would make the former exclude $\overline{V_j}$. Both are impossible. Thus $U$ and $V$ are disjoint open neighborhoods of $A$ and $D$, proving normality. $\square$

We now use Urysohn's lemma, the standard result that in a normal space, for disjoint closed sets $F,G$, there is a continuous function $f:X\to[0,1]$ with $f=1$ on $F$ and $f=0$ on $G$. The interval $[0,1]$ has its usual subspace topology from $\mathbb{R}$. The lemma is used here in this precise form.

The Countable Family from Nested Base Pairs

Theorem: If $X$ is regular and second-countable, there is a countable family of continuous functions $f:X\to[0,1]$ such that (1) any two distinct points have different values under at least one function, and (2) for every open $O\subseteq X$, $$ O=\bigcup_{f\in\mathcal{F}}\{x\in X:f(x)>0\}. $$ Here $\mathcal{F}$ is the subfamily of functions selected for base pairs whose outer member is contained in $O$.

Proof. Fix a countable base $\mathcal{B}$, and consider the set of ordered pairs \[ \mathcal{P}=\{(B,C)\in\mathcal{B}\times\mathcal{B}:\overline{B}\subseteq C\}. \] It is countable because it is a subset of the countable set $\mathcal{B}\times\mathcal{B}$. For every $(B,C)\in\mathcal{P}$, the sets $\overline{B}$ and $X\setminus C$ are disjoint and closed. Normality and Urysohn's lemma give a continuous $f_{B,C}:X\to[0,1]$ such that $f_{B,C}=1$ on $\overline{B}$ and $f_{B,C}=0$ on $X\setminus C$. Thus $\{x:f_{B,C}(x)>0\}\subseteq C$.

We verify that there are enough such pairs. Given $x\in O$ with $O$ open, choose $C\in\mathcal{B}$ with $x\in C\subseteq O$. Regularity gives an open $W$ containing $x$ with $\overline{W}\subseteq C$: separate $x$ from the closed set $X\setminus C$, and take the closure of the neighborhood of $x$. Choose $B\in\mathcal{B}$ with $x\in B\subseteq W$. Then $\overline{B}\subseteq\overline{W}\subseteq C$, so $(B,C)\in\mathcal{P}$, and $f_{B,C}(x)=1$. Because $C\subseteq O$, this function is zero outside $O$. We have shown that every $x\in O$ belongs to the positive set of a function from a pair with $C\subseteq O$, while every such positive set is contained in $O$. This proves the stated union formula.

Finally, if $x\ne y$, the $T_1$ property makes $O=X\setminus\{y\}$ open and it contains $x$. The construction gives a function with $f(x)=1$ and $f(y)=0$. It therefore separates $x$ and $y$. If $X$ is empty, the empty family satisfies both conclusions, so that case is included as well. $\square$

The union formula also says that the family detects the topology. Each $f$ is continuous, so its inverse images of open subsets of $[0,1]$ are open in $X$. Conversely, $(0,1]$ is open in $[0,1]$, and the formula writes every open subset of $X$ as a union of sets $f^{-1}((0,1])$. Thus the original open sets can be recovered from the functions.

Worked Constructions

Worked Example: Nested Rational Intervals in the Real Line

In $\mathbb{R}$ with its standard topology, rational-endpoint open intervals form a countable base. Consider $B=(0,2)$ and $C=(-1,3)$. Their closures satisfy $\overline{B}=[0,2]\subset C$. A function for this pair is $$ f(x)= \begin{cases} 0,&x\leq -1,\\ x+1,&-1\leq x\leq 0,\\ 1,&0\leq x\leq 2,\\ 3-x,&2\leq x\leq 3,\\ 0,&x\geq 3. \end{cases} $$

The formulas agree at every shared endpoint: the values at $-1,0,2,3$ are respectively $0,1,1,0$ from either adjoining piece. Each piece is continuous, so the matching endpoint values make $f$ continuous on $\mathbb{R}$. For $x=1$, the middle piece gives $f(1)=1$; for $x=4$, the last piece gives $f(4)=0$. Thus these two points are separated. Also, $f(x)=0$ whenever $x\notin C$, as required for the pair.

For a function positive near $1$ but zero outside $O=(0,2)$, take $B=(1/2,3/2)$ and $C=(1/4,7/4)$. Then $\overline{B}=[1/2,3/2]\subset C\subset O$. The function from this pair equals $1$ at $1$, and equals $0$ at every point outside $C$, hence outside $O$. This illustrates how the pair chosen for open-set detection can be made to lie inside the specified open set.

Worked Example: A Finite Discrete Space

Let $X=\{a,b,c\}$ have the discrete topology, so every subset is open. Use the base of singletons $\mathcal{B}=\{\{a\},\{b\},\{c\}\}$. Each singleton is closed as well as open, so $(\{a\},\{a\})$, $(\{b\},\{b\})$, and $(\{c\},\{c\})$ are suitable pairs. Their functions take value $1$ on the named singleton and $0$ elsewhere. In the order $a,b,c$, their values are $$ \begin{array}{c|ccc} & a&b&c\\ \hline f_{\{a\},\{a\}}&1&0&0\\ f_{\{b\},\{b\}}&0&1&0\\ f_{\{c\},\{c\}}&0&0&1 \end{array} $$

For example, the first and second functions distinguish $a$ from $b$, since their values at those points are $1,0$ and $0,1$, respectively. For $O=\{a,c\}$, the functions associated with $\{a\}$ and $\{c\}$ have positive sets $\{a\}$ and $\{c\}$; their union is exactly $O$. For $O=\varnothing$, the union of no positive sets is $\varnothing$. Thus the formula works for these sets, including the empty open set.

Worked Example: A Convergent Sequence with Its Limit

Let $X=\{0\}\cup\{1/n:n\in\mathbb{N}\}$ with the subspace topology inherited from $\mathbb{R}$. Consider the relative open sets $B=X\cap(-1/8,1/4)$ and $C=X\cap(-1/4,1/2)$. We have $B=\{0\}\cup\{1/n:n\geq5\}$ and $C=\{0\}\cup\{1/n:n\geq3\}$. The closure of $B$ in $X$ is $B$: its complement in $X$ is the finite set $\{1,1/2,1/3,1/4\}$, whose points are isolated and therefore form an open set. Hence $\overline{B}=B\subset C$.

Define a continuous function on $\mathbb{R}$ by $$ F(t)= \begin{cases} 0,&t\leq -1/4,\\ 8(t+1/4),&-1/4\leq t\leq -1/8,\\ 1,&-1/8\leq t\leq 1/4,\\ 2-4t,&1/4\leq t\leq 1/2,\\ 0,&t\geq 1/2. \end{cases} $$ The formulas agree at $-1/4,-1/8,1/4,1/2$, with values $0,1,1,0$, respectively. Restrict $F$ to $X$ to obtain a continuous function $f:X\to[0,1]$. We have $f(0)=1$ and $f(1)=0$. Also, $f(1/2)=2-4(1/2)=0$, $f(1/3)=2-4/3=2/3$, and $f(1/4)=1$. For $n\geq5$, $1/n<1/4$, so $f(1/n)=1$. The values confirm that $f$ is positive on points of $C$ and zero outside $C$, including at $1$ and $1/2$. In particular, this function detects the neighborhood $X\setminus\{1\}$ of $0$: its positive set contains $0$ and is contained in that neighborhood.

Check: the decreasing-piece calculation at $1/3$ gives $2-4/3=2/3>0$, while at $1/2$ it gives $2-2=0$. These values match the claimed inclusion of the positive set in $C$.

A Common Pitfall

Common mistake: choosing any two base elements that contain the desired point and assuming Urysohn's lemma applies. For instance, in $\mathbb{R}$, take $B=(0,2)$ and $C=(1,3)$. Although both contain $1.5$, the required closed sets intersect: $$ \overline{B}\cap(\mathbb{R}\setminus C)=[0,2]\cap\bigl((-\infty,1]\cup[3,\infty)\bigr)=[0,1]\ne\varnothing. $$ They are not disjoint, so Urysohn's lemma does not give a function equal to $1$ on $\overline{B}$ and $0$ outside $C$. The correction is to choose a genuinely nested pair, such as $B=(0,2)$ and $C=(-1,3)$, for which $[0,2]\subset(-1,3)$. The closure containment is the condition that makes the construction work.

Check Your Understanding

Use the nested-pair construction and verify the closure containment whenever you select a pair.

In $\mathbb{R}$, choose rational-endpoint intervals $B,C$ with $1\in B$, $\overline{B}\subset C$, and $C\subset(0,2)$. What does the associated function do at $1$ and outside $(0,2)$?
Explain why the family of ordered pairs from a countable base is countable.
In the theorem, why is $X\setminus\{y\}$ open when $X$ is $T_1$? How does this give a function separating $x\ne y$?
For $X=\{0\}\cup\{1/n:n\in\mathbb{N}\}$, find a relative open neighborhood of $0$ that excludes $1$, and describe what the detecting function must satisfy.
Explain why the set $(0,1]$ is open in the subspace $[0,1]$, and how that fact recovers open sets from the functions.