Chapter 6: Closure, Continuity, and Homeomorphism
After establishing the foundations of topological spaces and open sets in the previous chapter, this chapter will further explore the connections between points and sets in topological spaces (such as closures and limit points), and deeply study the bridge connecting different topological spacesβcontinuous mappings. We will also introduce the most core equivalent classification standard in topology: Homeomorphism. Finally, we will generalize the concept of product spaces, contrasting the differences between the box topology and the product topology.
1. Closure and Limit Points
In Chapter 5, we defined the closure via the smallest closed set containing a set. Here, we provide a more intuitive point-set criterion for closure through "neighborhood intersection."
1.1 Neighborhood Characterization of Closure
Proposition 6.1 (Point-set Characterization of Closure)
Let \(A \subseteq X\) be a subset of a topological space \(X\).
- (1) A point \(x \in \overline{A}\) if and only if every open set \(U\) containing \(x\) intersects \(A\), i.e., \(U \cap A \neq \emptyset\).
- (2) If \(\mathcal{B}\) is a basis for the topology \(\tau\), then \(x \in \overline{A}\) if and only if for every basis element \(B \in \mathcal{B}\) containing \(x\), we have \(B \cap A \neq \emptyset\).
Proof of Proposition 6.1
Sufficiency (If): Suppose there exists an open set \(U\) containing \(x\) such that \(U \cap A = \emptyset\). This means \(A \subseteq X \setminus U\). Since \(U\) is open, its complement \(X \setminus U\) is a closed set containing \(A\). According to the definition that the closure is the "smallest closed set containing \(A\)," it must follow that \(\overline{A} \subseteq X \setminus U\). Since \(x \in U\), then \(x \notin X \setminus U\), which implies \(x \notin \overline{A}\). This proves the contrapositive of the original proposition.
Necessity (Only if): Suppose \(x \notin \overline{A}\). Let \(U = X \setminus \overline{A}\). Since \(\overline{A}\) is a closed set, \(U\) is an open set containing \(x\). Clearly, \(U \cap \overline{A} = \emptyset\), therefore \(U \cap A = \emptyset\). This also proves the reverse contrapositive. \(\square\)
1.2 Limit Points (Cluster Points)
In calculus, limits reflect the trend of a sequence approaching a point. In general topology, this approach is abstracted as a "limit point."
Definition 6.1 (Limit Point / Cluster Point)
Let \(A \subseteq X\). A point \(x\) is called a limit point of set \(A\) if for every open set \(U\) containing \(x\):
In other words, every open set containing \(x\) must contain at least one point of \(A\) other than \(x\) itself.
We denote the set of all limit points of \(A\) in \(X\) as \(LP(A)\) (commonly also denoted as the derived set \(A'\)).
There is a direct connection between the closure and limit points: \(\overline{A} = A \cup LP(A)\).
Examples:
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In the real space \(\mathbb{R}\) (endowed with the standard topology), let \(A = \{\frac{1}{n} \mid n \in \mathbb{N}_{\ge 1}\}\). Although \(0 \notin A\), \(0\) is the unique limit point of \(A\), i.e., \(LP(A) = \{0\}\), and \(\overline{A} = A \cup \{0\}\).
-
Let \(X = \{a, b, c\}\) with the topology \(\tau = \{\emptyset, X, \{a\}\}\). For the subset \(A = \{a\}\), its closure is \(\overline{A} = X\), and the set of limit points is \(LP(A) = \{b, c\}\).
2. Continuity
The continuity of mappings is the most central object of study in topology. Topology discards the dependence on specific numerical differences (\(\epsilon-\delta\)) found in metric spaces and establishes continuity purely on the pullback (preimage) of open sets.
Definition 6.2 (Continuous Mapping)
Let \(f: X \to Y\) be a mapping between topological spaces.
- (1) Local Continuity: The mapping \(f\) is continuous at point \(x_0 \in X\) if for any open set \(V \subseteq Y\) containing the image point \(f(x_0)\), there exists an open set \(U\) in \(X\) containing \(x_0\) such that \(U \subseteq f^{-1}(V)\) (i.e., \(f(U) \subseteq V\)).
- (2) Global Continuity: The mapping \(f\) is continuous on the entire \(X\) if and only if it is continuous at every point in \(X\).
The following lemma gives equivalent characterizations for determining global continuity, which are used extremely frequently in subsequent proofs.
Lemma 6.1 (Equivalent Characterizations of Continuity)
For a mapping \(f: X \to Y\), the following four statements are equivalent (TFAE):
- (i) \(f\) is a continuous mapping.
- (ii) For any open set \(V\) in \(Y\), its preimage \(f^{-1}(V)\) is also an open set in \(X\) ("The preimage of an open set is open").
- (iii) For any subset \(A \subseteq X\), it always holds that \(f(\overline{A}) \subseteq \overline{f(A)}\).
- (iv) For any closed set \(B\) in \(Y\), its preimage \(f^{-1}(B)\) is also a closed set in \(X\) ("The preimage of a closed set is closed").
Proof of Lemma 6.1
Proof (i) \(\implies\) (ii): Let \(V\) be an open set in \(Y\). Take any \(x \in f^{-1}(V)\), so \(f(x) \in V\). Since \(f\) is continuous, there exists an open set \(U_x\) containing \(x\) such that \(U_x \subseteq f^{-1}(V)\). Therefore, \(f^{-1}(V)\) can be expressed as the union of all such \(U_x\):
Since the union of any collection of open sets is still an open set, \(f^{-1}(V)\) is open.
Proof (ii) \(\implies\) (iv): Let \(B \subseteq Y\) be a closed set. Then \(Y \setminus B\) is an open set. Since the preimage operation of a mapping commutes with the complement operation, i.e., \(f^{-1}(Y \setminus B) = X \setminus f^{-1}(B)\). According to (ii), this is an open set in \(X\). Therefore, \(f^{-1}(B)\) must be a closed set.
Proof (iv) \(\implies\) (iii): Given any \(A \subseteq X\). Since the closure \(\overline{f(A)}\) is a closed set in \(Y\), according to (iv), \(f^{-1}(\overline{f(A)})\) is a closed set in \(X\). Clearly, \(A \subseteq f^{-1}(f(A)) \subseteq f^{-1}(\overline{f(A)})\). Because \(f^{-1}(\overline{f(A)})\) is a closed set containing \(A\), by the property that the closure is the smallest closed set, we must have \(\overline{A} \subseteq f^{-1}(\overline{f(A)})\). Applying \(f\) to both sides yields \(f(\overline{A}) \subseteq \overline{f(A)}\). (The proofs for (iii) \(\implies\) (i) and (ii) \(\iff\) (i) are similar, completing the logical loop). \(\square\)
Notes: 1. The identity mapping \(id: X \to X\) is always continuous. 2. For an inclusion mapping, if the subset \(A \subseteq X\) is endowed with the subspace topology \(\tau_A\) (the weakest topology making the inclusion mapping continuous), then it must be continuous.
3. Homeomorphism
In the eyes of topology, if two spaces can be transformed into each other through continuous deformation (without tearing or gluing), they are considered identical in topological structure. This "topological equivalence" is precisely defined through Homeomorphism.
Definition 6.3 (Homeomorphism)
Let \(f: X \to Y\) be a mapping. If it satisfies the following three conditions, then \(f\) is called a Homeomorphism:
- (i) \(f\) is a bijection (i.e., injective and surjective, ensuring a one-to-one correspondence between points in the two spaces).
- (ii) \(f\) is continuous.
- (iii) The inverse mapping \(f^{-1}\) is also continuous.
If there exists a homeomorphism from \(X\) to \(Y\), we say that the topological spaces \(X\) and \(Y\) are homeomorphic, denoted by \(X \cong Y\).
Examples and Counterexamples:
- The open interval \((-\pi/2, \pi/2)\) is homeomorphic to the entire real line \(\mathbb{R}\) (e.g., via the tangent function \(f(x) = \tan(x)\)).
- Consider the half-open interval \([0, 1)\) and the unit circle \(S^1\). We can construct a bijective mapping \(f: [0, 1) \to S^1\), such as \(f(t) = (\cos(2\pi t), \sin(2\pi t))\). This mapping is clearly continuous and bijective. However, its inverse \(f^{-1}: S^1 \to [0, 1)\) is discontinuous at the point \((1, 0)\) (as points on the circle approach \((1, 0)\), they approach \(0\) and \(1\) respectively in \([0,1)\), causing a tear). Therefore, this mapping is not a homeomorphism! In fact, the compact space \(S^1\) can never be homeomorphic to the non-compact space \([0, 1)\).
4. Infinite Products: Box vs Product Topology
In Chapter 5, we discussed the finite product of two spaces. Now we generalize this to the Cartesian product of an arbitrary infinite family of topological spaces. For infinite products, there are two very different natural ways to construct a topology.
Let \((X_i, \tau_i)_{i \in I}\) be a family of topological spaces (where the index set \(I\) can be infinite). We consider their Cartesian product \(\prod_{i \in I} X_i\).
4.1 Box Topology
Definition 6.4 (Box Topology)
Box Topology \(\tau_b\) is the topology generated by the following basis \(\mathcal{B}_b\):
That is, the basis elements of the box topology are restricted by the corresponding open set \(U_i\) in every component.
4.2 Product Topology
Definition 6.5 (Product Topology / Tychonoff Topology)
Product Topology \(\tau_p\) is the topology generated by the following basis \(\mathcal{B}_p\):
Here, "for almost all \(i\)" means that, except for a finite number of exceptions, all other components must be the entire space \(X_i\) itself.
Comparison between the two: Clearly, the basis \(\mathcal{B}_p\) is entirely contained within \(\mathcal{B}_b\). Therefore, the box topology is finer than the product topology (contains more open sets), i.e., \(\tau_b \ge \tau_p\). When the index set \(I\) is a finite set, these two are equivalent (\(\tau_b = \tau_p\)). However, in infinite products, the box topology often causes many topological properties (such as connectedness and compactness) to fail due to having too many open sets, therefore, in modern mathematics, the default topology used for product spaces is the product topology \(\tau_p\).
4.3 Universal Property of Product Topology
Why is the product topology \(\tau_p\) more important than the box topology? Because the product topology is exactly the "minimum cost" to ensure that all natural projection mappings are continuous.
Define the projection mapping for the \(i\)-th coordinate \(p_i: \prod_{j \in I} X_j \to X_i\), whose action is to extract the element at the \(i\)-th position in the sequence.
Proposition 6.2
The product topology \(\tau_p\) is the smallest (coarsest) topology such that all projection mappings \(p_i\) are continuous.
Proof of Proposition 6.2
1. Prove that \(p_i\) is continuous under \(\tau_p\): Let \(U_i \in \tau_i\) be any open set in \(X_i\). Consider its preimage under the projection mapping \(p_i\):
According to the definition of the basis for the product topology, this set is the entire space \(X_j\) in all components except for \(i\) (restricted in only a finite number, i.e., 1 component). Therefore, it is itself a basis element in \(\mathcal{B}_p\), naturally an open set in \(\tau_p\). This proves all \(p_i\) are continuous.
2. Prove that \(\tau_p\) is the coarsest topology making \(p_i\) continuous: Assume \(\mathcal{T}\) is another topology endowed on the Cartesian product, and it is required that all projection mappings \(p_i\) are continuous under \(\mathcal{T}\). Since \(p_i\) is continuous, for any \(U_i \in \tau_i\), its preimage \(p_i^{-1}(U_i)\) must belong to the topology \(\mathcal{T}\). We examine any basis element \(B \in \mathcal{B}_p\) of the product topology \(\tau_p\). By definition, \(B\) is an open set \(U_{i_m}\) in a finite number of components \(i_1, i_2, \dots, i_k\), and the entire space in the rest. Thus, \(B\) can be written as the intersection of these finite preimages:
Since the topology \(\mathcal{T}\) must contain all these preimage sets, and the topological structure is closed under finite intersections, it must hold that \(B \in \mathcal{T}\). Since every basis element of \(\tau_p\) is contained in \(\mathcal{T}\), the entire product topology generated by the basis must also be contained in \(\mathcal{T}\), i.e., \(\tau_p \subseteq \mathcal{T}\). This proves the minimality of \(\tau_p\). \(\square\)