Solving Problems About Countable Products
Compactness of a product is useful when the constraints on its points are coordinate-by-coordinate, even if there are infinitely many coordinates. The key is to keep two facts separate: compactness comes from the compactness of the factor spaces, while neighborhoods in the product topology restrict only finitely many coordinates at a time. We will apply these facts to interval products, products of finite spaces, and a mixed product.
For a family of spaces \((X_n,\mathcal{T}_n)\), indexed by \(n\in\mathbb{N}\), write \(X=\prod_{n\in\mathbb{N}}X_n\) and give \(X\) the product topology. The projection \(\pi_n:X\to X_n\) sends a point to its \(n\)-th coordinate. A set is compact if every open cover of it has a finite subcover.
This finite-coordinate condition is often the first step in a product problem. It follows directly from the definition of the product topology: the sets \(\pi_n^{-1}(U_n)\), with \(U_n\) open in \(X_n\), form a subbasis, and finite intersections of subbasic sets form a basis.
Interval Factors and Finite-Coordinate Neighborhoods
Worked Example: A Product of Shrinking Intervals
Let \(K=\prod_{n\in\mathbb{N}}[0,1/n]\), with each interval carrying its subspace topology from \(\mathbb{R}\) and \(K\) carrying the product topology. Show that \(K\) is compact and describe a basic neighborhood of a particular point.
For each \(n\in\mathbb{N}\), the interval \([0,1/n]\) is compact by the Heine–Borel theorem, and it is nonempty because \(0\in[0,1/n]\). Tychonoff’s theorem therefore says that \(K\) is compact. To verify nonemptiness directly, set \(p_n=1/(2n)\). Since \(0\leq 1/(2n)\leq 1/n\) for every \(n\), the point \(p=(p_n)_{n\in\mathbb{N}}\) belongs to \(K\).
Now restrict the first coordinate to \(U_1=(1/4,3/4)\cap[0,1]\) and the third coordinate to \(U_3=(1/12,1/4)\cap[0,1/3]\). These sets are open in their respective interval factors. The point \(p\) meets both restrictions: \(p_1=1/2\), and \(1/4<1/2<3/4\); also \(p_3=1/6\), and \(1/12<1/6<1/4\). Thus \[ \{x\in K:x_1\in U_1,\ x_3\in U_3\} \] is a basic open neighborhood of \(p\). There is no restriction on \(x_2\) or on any coordinate \(x_n\) with \(n\geq4\).
Check: the specified coordinates really are in their factors: \(U_1\subseteq[0,1]\) and \(U_3\subseteq[0,1/3]\). The neighborhood restricts exactly two coordinates, so it has the required finite-coordinate form.
This example illustrates why product neighborhoods are manageable even when the product has infinitely many factors. A basis element does not need to control the entire sequence; it can control just the coordinates relevant to the question.
Finite Spaces and Closed Coordinate Conditions
A finite space with the discrete topology is compact: given an open cover, choose one member containing each point. There are only finitely many points, so the chosen members form a finite subcover. We will also use the following consequence of compactness.
Proof. Let \(\{V_\alpha\}_{\alpha\in A}\) be an open cover of \(F\) in the subspace topology. For each \(\alpha\), the definition of that topology gives an open set \(U_\alpha\subseteq X\) with \(V_\alpha=F\cap U_\alpha\). The sets \(U_\alpha\), together with \(X\setminus F\), cover \(X\). Compactness of \(X\) gives a finite subcover. The selected sets \(U_\alpha\) among them cover \(F\), because \(X\setminus F\) contains no point of \(F\). Consequently the corresponding finite collection of \(V_\alpha\) covers \(F\). Thus \(F\) is compact. \(\square\)
Worked Example: A Cylinder in Binary Sequence Space
Let \(D=\{0,1\}\) have the discrete topology and let \(P=\prod_{n\in\mathbb{N}}D\) have the product topology. Consider \(C=\{x\in P:x_2=1,\ x_5=0\}\). Show that \(P\) is compact and that \(C\) is a compact open subset.
The two-point discrete space \(D\) is compact by the finite-space argument. Every factor of \(P\) is therefore compact, so Tychonoff’s theorem gives compactness of \(P\). The cylinder \(C\) is a basic open set: it restricts the second coordinate to \(\{1\}\) and the fifth coordinate to \(\{0\}\), both open in \(D\).
Each singleton in \(D\) is also closed. Therefore \(\pi_2^{-1}(\{1\})\) and \(\pi_5^{-1}(\{0\})\) are closed, and \(C=\pi_2^{-1}(\{1\})\cap\pi_5^{-1}(\{0\})\) is closed. It is nonempty: the sequence \(x\) given by \(x_2=1\), \(x_5=0\), and \(x_n=0\) for every other \(n\) belongs to \(C\). The closed-subset proposition now gives that \(C\) is compact.
Check: the exhibited sequence satisfies both conditions, since its second coordinate is \(1\) and its fifth coordinate is \(0\). The cylinder imposes no condition on any other coordinate.
Mixed Products and Closed Constraints
Worked Example: Matching a Real Coordinate to a Binary Coordinate
Let \(D=\{0,1\}\) have the discrete topology, and give \(X=[0,1]\times\prod_{n\in\mathbb{N}}D\) the product topology, with \([0,1]\) carrying its usual subspace topology from \(\mathbb{R}\). Show that \(X\) is compact and that \(E=\{(t,s)\in X:t=s_1\}\) is compact.
The interval \([0,1]\) is compact by Heine–Borel, and each factor \(D\) is compact because it is finite. Tychonoff’s theorem gives compactness of \(X\). The constraint set can be written \[ E=\bigl(\{0\}\times\{s\in D^{\mathbb{N}}:s_1=0\}\bigr) \ \cup\ \bigl(\{1\}\times\{s\in D^{\mathbb{N}}:s_1=1\}\bigr). \] Indeed, \(s_1\) is either \(0\) or \(1\), and the equation \(t=s_1\) then forces \(t\) to equal that same value.
The sets \(\{0\}\) and \(\{1\}\) are closed in \([0,1]\). The coordinate sets \(\{s:s_1=0\}\) and \(\{s:s_1=1\}\) are closed because they are preimages of closed singletons under the first-coordinate projection. Each product appearing in the union is closed in \(X\), and a finite union of closed sets is closed. Thus \(E\) is closed in the compact space \(X\), so it is compact by the proposition.
For a neighborhood calculation, take \(s_1=1\), \(s_4=0\), and consider the point \((1,s)\in E\). The set \((3/4,1]\) is open relative to \([0,1]\), and the restrictions \(s_1=1,\ s_4=0\) form a basic open cylinder in \(D^{\mathbb{N}}\). Their product is a basic neighborhood in \(X\). Intersecting this neighborhood with \(E\) forces \(t=1\), while leaving all sequence coordinates other than the first and fourth unrestricted.
Check: the point \((1,s)\) satisfies \(t=s_1\), because both values are \(1\). The neighborhood’s real-coordinate restriction contains \(1\), and its sequence-coordinate restrictions match \(s_1=1\) and \(s_4=0\).
Nested Closed Constraints
Proof. Suppose instead that \(\bigcap_{m\in\mathbb{N}}F_m=\varnothing\). Then the open sets \(X\setminus F_m\) cover \(X\). Compactness gives a finite subcover, say \(X\setminus F_{m_1},\ldots,X\setminus F_{m_k}\). Let \(M\) be the largest of the indices \(m_1,\ldots,m_k\). Since the sets \(F_m\) decrease as \(m\) increases, \(F_M\subseteq F_{m_i}\) for each \(i\). A point in the nonempty set \(F_M\) therefore belongs to none of \(X\setminus F_{m_1},\ldots,X\setminus F_{m_k}\), contradicting that these sets cover \(X\). Hence the intersection is nonempty. \(\square\)
Worked Example: Binary Sequences with No Adjacent Ones
In the compact product \(P=D^{\mathbb{N}}\), let \(F_m\) be the set of sequences for which no two adjacent entries among the first \(m\) entries are both \(1\). Prove that there is a sequence with no adjacent \(1\)s anywhere.
For each \(m\), the condition defining \(F_m\) is a finite list of restrictions: for every \(j\) with \(1\leq j<m\), the pair \((x_j,x_{j+1})\) is not \((1,1)\). Each restriction depends on two coordinates and defines a closed set, since \(D\times D\) is discrete and \(\{(1,1)\}\) is closed. Thus \(F_m\), a finite intersection of closed sets, is closed. It is nonempty because the all-zero sequence belongs to it. Also \(F_{m+1}\subseteq F_m\), since satisfying all the restrictions among the first \(m+1\) entries includes satisfying those among the first \(m\).
The nested-closed-sets theorem applies because \(P\) is compact and every \(F_m\) is closed and nonempty. Hence \(\bigcap_m F_m\ne\varnothing\). By the definitions, a sequence belongs to this intersection exactly when no adjacent pair at any position is \((1,1)\). For a direct check of the conclusion, the sequence \(x_n=1\) for odd \(n\) and \(x_n=0\) for even \(n\) belongs to every \(F_m\): each adjacent pair has one odd index and one even index, so its entries are \(1\) and \(0\), not \(1\) and \(1\).
Check: the first four entries of this sequence are \(1,0,1,0\); the adjacent pairs are \((1,0)\), \((0,1)\), and \((1,0)\). The same alternating rule verifies every adjacent pair, including those beyond the first four.
A Common Pitfall: Restricting Every Coordinate at Once
In \(P=\{0,1\}^{\mathbb{N}}\), consider \(V=\{x:x_n=0\text{ for every }n\in\mathbb{N}\}\), the singleton consisting of the all-zero sequence. A common error is to call \(V\) a basic open neighborhood because it specifies an open condition in every coordinate. That calculation is wrong: a basic open set in the product topology restricts only finitely many coordinates.
In fact, \(V\) is not open. If it were open, some basic open set containing the all-zero sequence would be contained in \(V\). Such a basic set restricts only a finite set \(F\) of coordinates. Choose \(k\in\mathbb{N}\setminus F\), and take the sequence that is \(1\) in coordinate \(k\) and \(0\) in every other coordinate. It satisfies every restriction imposed by the basic set, but it is not in \(V\). This contradicts containment in \(V\). The distinction is essential: the product topology allows finitely many coordinate restrictions in a basic neighborhood, not infinitely many.
Check Your Understanding
For each question, identify the factor spaces and the finite-coordinate restrictions involved.
- Why does Tychonoff’s theorem apply to \(\prod_{n\in\mathbb{N}}[0,2/n]\) with the product topology?
- In \(\{0,1\}^{\mathbb{N}}\), describe a basic open neighborhood of a sequence that specifies coordinates \(1\), \(3\), and \(8\).
- Show that \(\{x\in\{0,1\}^{\mathbb{N}}:x_1=0,\ x_3=1\}\) is both open and closed.
- In \([0,1]\times\{0,1\}^{\mathbb{N}}\), explain why the set defined by \(t=x_1\) is closed.
- Give a sequence in \(\{0,1\}^{\mathbb{N}}\) with no adjacent \(1\)s, and check the first five adjacent pairs.