Chapter 7: Mappings in Product Spaces and Connectedness
In the previous chapters, we have established the foundations of topological spaces, continuous mappings, and product topologies. This chapter will first supplement the universal property of continuous mappings in product spaces and the concept of metrizable spaces, and then delve into the core topological invariants with strong geometric intuition in point-set topologyβConnectedness and Path-connectedness.
1. Continuous Mappings in Product Spaces and Metrizability
When studying the product space \(\prod X_i\), we naturally care about how to determine whether a mapping into this product space is continuous. The construction of the product topology (the coarsest topology making all projection mappings continuous) provides us with an extremely elegant criterion.
1.1 Continuity of Mappings into Product Spaces
Let \((X_i, \tau_i)_{i \in I}\) be a family of topological spaces, and \(\prod_{i \in I} X_i\) be the product space endowed with the product topology. Let \(p_i: \prod_{j \in I} X_j \to X_i\) be the natural projection mapping of the \(i\)-th coordinate.
Theorem 7.1 (Continuity of Mappings into Product Spaces)
Let \(Y\) be an arbitrary topological space, and \(f: Y \to \prod_{i \in I} X_i\) be a mapping. We define the \(i\)-th coordinate function of \(f\) as \(f_i = p_i \circ f: Y \to X_i\).
Then the mapping \(f\) is continuous if and only if every coordinate function \(f_i\) is continuous.
Proof of Theorem 7.1
Necessity (\(\implies\)):
Given that \(f\) is a continuous mapping.
According to the definition of the product topology, every natural projection mapping \(p_i: \prod X_j \to X_i\) is continuous.
Since the composition of continuous mappings remains a continuous mapping, every coordinate function \(f_i = p_i \circ f\) must be continuous.
Sufficiency (\(\impliedby\)):
Given that for all \(i \in I\), the coordinate functions \(f_i\) are continuous. We need to prove that \(f\) is continuous.
To prove that \(f\) is continuous, it suffices to prove that the preimage of any subbasis element in the product topology is an open set in \(Y\).
The standard form of a subbasis element in the product topology is \(p_i^{-1}(U_i)\), where \(U_i\) is an open set in \(X_i\).
Consider the preimage of this subbasis element under \(f\):
Since it is assumed that \(f_i\) is continuous and \(U_i\) is an open set, \(f_i^{-1}(U_i)\) must be an open set in \(Y\).
Since the pullbacks of all subbasis elements are open sets, and any open set can be expressed as arbitrary unions of finite intersections of subbasis elements, the mapping \(f\) is continuous on the entire product space. \(\square\)
1.2 Metrizable Spaces
Although topological spaces abandon the dependence on distance, not all topological spaces become "elusive". The open set structure of some topological spaces can actually be entirely generated by some hidden metric.
Definition 7.1 (Metrizable Space)
A topological space \((X, \tau)\) is called metrizable if there exists a metric (distance function) \(d: X \times X \to \mathbb{R}\) on \(X\) such that the metric topology \(\tau_d\) induced by the open balls of the metric \(d\) is exactly equal to the original topology \(\tau\) (i.e., \(\tau = \tau_d\)).
For example, the real space \(\mathbb{R}\) and the Euclidean space \(\mathbb{R}^n = \prod_{i=1}^n \mathbb{R}\) are metrizable spaces under the standard topology. In \(\mathbb{R}^2\), the Euclidean distance \(d_1(x, y) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\) and the maximum metric \(d_2(x, y) = \max\{|x_1 - x_2|, |y_1 - y_2|\}\) induce the same topology, because their open balls can be nested within each other.
2. Basic Concepts of Connectedness
Intuitively, if a space is a "single piece" and has not been split into several parts, we call it connected. We strictly define this concept through the opposite of "separation".
Definition 7.2 (Separation and Connectedness)
Let \((X, \tau)\) be a topological space.
- Separation: If there exist two non-empty open sets \(U\) and \(V\) satisfying:
Then the pair \((U, V)\) is called a separation of the space \(X\).
- Connected: If there does not exist any separation for the topological space \(X\), then \(X\) is said to be connected.
For a subset \(A \subseteq X\) of \(X\), if \(A\) endowed with the relative topology (subspace topology) is connected as a topological space, then the set \(A\) is called a connected subset in \(X\).
We have the following equivalent criterion for connectedness, which is very useful in actual proofs:
- Equivalent Characterization: A topological space \(X\) is connected if and only if the subsets in \(X\) that are both open and closed (called Clopen sets) are only the empty set \(\emptyset\) and the entire space \(X\) itself.
Examples Analysis:
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\(X = [0, 1) \cup (1, 2]\) (endowed with the relative topology induced from \(\mathbb{R}\)) is disconnected. We can take the relative open sets \(U = [0, 1)\) and \(V = (1, 2]\). They are non-empty, disjoint, and their union is \(X\).
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The set of rational numbers \(\mathbb{Q}\) is extremely disconnected (totally disconnected space). Because for any irrational number \(\alpha\) (e.g., \(\sqrt{2}\)), we can always use it as a "knife" to cut \(\mathbb{Q}\). Let \(U = (-\infty, \alpha) \cap \mathbb{Q}\) and \(V = (\alpha, +\infty) \cap \mathbb{Q}\). Both sets are relative open sets (and also relative closed sets) in \(\mathbb{Q}\), non-empty, disjoint, and constitute a separation of \(\mathbb{Q}\).
3. Core Properties of Connected Sets
Connected sets exhibit excellent preservation properties under common topological operations (such as taking unions, taking closures, and continuous mappings).
3.1 Union of Connected Sets
Theorem 7.2 (The Union of a Collection of Connected Sets with a Common Point is Connected)
Let \(\{A_i\}_{i \in I}\) be a collection of connected subsets in a topological space \(X\).
If this collection of sets has a common point, i.e., \(\bigcap_{i \in I} A_i \neq \emptyset\), then their union \(A = \bigcup_{i \in I} A_i\) is also connected.
Proof of Theorem 7.2
Using proof by contradiction.
Assume that \(A\) is disconnected, meaning there exists a separation of \(A\). This implies there exist non-empty relative open sets \(C\) and \(D\) in \(A\) satisfying \(C \cap D = \emptyset\) and \(C \cup D = A\).
Since all \(A_i\) have a common intersection point, let us take \(p \in \bigcap_{i \in I} A_i\).
Since \(p\) belongs to \(A = C \cup D\), then \(p\) must belong to either \(C\) or \(D\). Without loss of generality, assume \(p \in C\).
Now let's examine each individual connected set \(A_i\).
Since \(C\) and \(D\) are open sets in \(A\), \(A_i \cap C\) and \(A_i \cap D\) form relative open sets on \(A_i\), and they are disjoint with their union being \(A_i\). This looks like a separation of \(A_i\).
But it is known that \(A_i\) is connected, and a connected set cannot be separated. Therefore, one of the two sets \(A_i \cap C\) and \(A_i \cap D\) must be empty, and the other must be equal to the entire \(A_i\).
Since the common point \(p \in A_i\), and we have assumed \(p \in C\), there must be \(p \in A_i \cap C\). This implies \(A_i \cap C \neq \emptyset\).
This forces \(A_i \cap D = \emptyset\), which further implies \(A_i \subseteq C\).
This deduction holds for all \(i \in I\). Therefore, the entire union \(A = \bigcup_{i \in I} A_i\) is contained within \(C\).
This means that \(D\) must be an empty set (since \(C \cup D = A \subseteq C\) and \(C \cap D = \emptyset\)). But this contradicts the definition that \(C, D\) must both be "non-empty sets" as a separation!
The assumption is wrong, so the union \(A\) must be connected. \(\square\)
3.2 Closure of Connected Sets
Theorem 7.3 (The Closure of a Connected Set Remains Connected)
Let \(A\) be a connected subset in a topological space \(X\).
If \(B\) is a subset satisfying \(A \subseteq B \subseteq \overline{A}\), then \(B\) must also be connected. In particular, when \(B = \overline{A}\), the closure \(\overline{A}\) of \(A\) is also connected.
This theorem tells us that if you add some (or all) of its limit points to a connected set, it will never become "fractured" because of these newly added points.
3.3 Continuous Mappings Preserve Connectedness
This is the most important theorem for connectedness as a topological invariant.
Theorem 7.4 (Continuous Mappings Preserve Connectedness)
Let \(f: X \to Y\) be a continuous mapping. If the domain \(X\) is connected, then its image set \(f(X)\) is also connected in \(Y\).
Proof of Theorem 7.4
Using proof by contradiction.
Assume the image set \(f(X)\) is disconnected. Then there exists a separation of \(f(X)\), i.e., there exist relative open sets \(U\) and \(V\) in \(f(X)\) satisfying \(U \neq \emptyset\), \(V \neq \emptyset\), \(U \cap V = \emptyset\), and \(U \cup V = f(X)\).
Since \(U, V\) are relative open sets of \(f(X)\), their preimages \(f^{-1}(U)\) and \(f^{-1}(V)\) under the continuous mapping \(f\) must be open sets in \(X\).
Observe these two preimages:
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Since \(U\) and \(V\) are non-empty and both within the image set, the preimages \(f^{-1}(U)\) and \(f^{-1}(V)\) are both non-empty.
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Since \(U \cap V = \emptyset\), after pulling back, it must also hold that \(f^{-1}(U) \cap f^{-1}(V) = f^{-1}(U \cap V) = \emptyset\).
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Since \(U \cup V = f(X)\), every point in \(X\) is mapped into \(U \cup V\), therefore \(f^{-1}(U) \cup f^{-1}(V) = X\).
Thus, we have found two non-empty, disjoint open sets \(f^{-1}(U)\) and \(f^{-1}(V)\) in \(X\) whose union is the entire space.
This indicates that \(X\) is disconnected, but this completely contradicts the given condition "\(X\) is connected"!
Therefore, the assumption does not hold, and the image set \(f(X)\) must be connected. \(\square\)
Corollary (The broadest form of the Intermediate Value Theorem): If \(X\) is a connected space and there exists a continuous mapping \(f: X \to \mathbb{R}\), then \(f(X)\) must be a connected subset on the real line (i.e., an interval). This means that for any two points \(a, b \in f(X)\) in the image set, any value \(c\) between \(a\) and \(b\) must belong to the image set \(f(X)\) (i.e., a preimage exists).
3.4 Connectedness of Product Spaces
Theorem 7.5
If both \(X_1\) and \(X_2\) are connected spaces, then their product space \(X_1 \times X_2\) is also connected. By mathematical induction, the product of any finite number of connected spaces remains connected.
Brief Outline of Proof: Fix a point \(y \in X_2\), the slice \(X_1 \times \{y\}\) is homeomorphic to \(X_1\) and thus is connected. Similarly, for any \(x \in X_1\), the slice \(\{x\} \times X_2\) is also connected. Construct a cross-shaped set \(C_x = (X_1 \times \{y\}) \cup (\{x\} \times X_2)\). Since these two parts intersect at the point \((x, y)\), by Theorem 7.2, the union \(C_x\) is connected. Then we write the entire space as \(X_1 \times X_2 = \bigcup_{x \in X_1} C_x\), and all \(C_x\) have the common part \(X_1 \times \{y\}\). Applying Theorem 7.2 again proves the entire space is connected.
4. Path Connectedness
Connectedness is a topological concept purely dependent on open sets, whereas in intuitive geometry, we usually think "connected" means being able to connect any two points with a curve, which leads to path-connectedness.
Definition 7.3 (Path and Path-connectedness)
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Path: Let \(X\) be a topological space, and \(x, y \in X\). A path from point \(x\) to point \(y\) is a continuous mapping \(\gamma: [0, 1] \to X\) from the closed interval \([0, 1]\) to \(X\), satisfying the initial condition \(\gamma(0) = x\) and the terminal condition \(\gamma(1) = y\).
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Path-connected: If for any two points \(x\) and \(y\) in space \(X\), there exists a path \(\gamma\) connecting them, then the space \(X\) is said to be path-connected.
Theorem 7.6
If a space is path-connected, then it must be connected.
Proof Highlights: For any given point \(x_0 \in X\), and for any other point \(y \in X\), there exists a path \(\gamma_y: [0, 1] \to X\). The interval \([0, 1]\) as a sub-interval of real numbers is connected, hence the image set \(C_y = \gamma_y([0,1])\) is a connected subset. And all \(C_y\) contain the common point \(x_0 = \gamma_y(0)\). Therefore, the entire space \(X = \bigcup_{y \in X} C_y\) is the union of a collection of connected sets, so it must be connected.
Note! The converse is not true. A connected space is not necessarily path-connected.
- Classic Counterexample: Topologist's Sine Curve. Define the set \(A = \{(x, \sin(1/x)) \mid x \in (0, 1]\} \subset \mathbb{R}^2\). Since \(A\) is the continuous image of the interval \((0,1]\), \(A\) is connected. Take the closure of \(A\) in the plane, \(\overline{A} = A \cup (\{0\} \times [-1, 1])\). By Theorem 7.3, \(\overline{A}\) is connected. However, \(\overline{A}\) is not path-connected. Because the vertical line segment at the origin and the wildly oscillating curve part are "torn" apart, you cannot use a continuous path to cross infinite vertical oscillations in finite time \(t\) to walk from the origin to the curve on the right.
5. Connected Components
When a space is not connected as a whole, we can study the "largest" connected fragments within it.
Definition 7.4 (Connected Components)
In a topological space \(X\), define an equivalence relation \(\sim\):
This equivalence relation partitions the entire space \(X\) into several disjoint equivalence classes. These equivalence classes are called the Connected Components of space \(X\).
Connected components have an extremely beautiful closure property: Any connected component must be a closed set. (This is because if a connected component were not a closed set, according to Theorem 7.3, taking its closure would yield a strictly larger connected subset, which contradicts the nature of the equivalence class as a "maximal connected subset").
6. Distinguishing Topological Spaces using Connectedness
Homeomorphisms not only preserve connectedness, but also preserve local connected structures (such as connectedness after removing a point). This provides us with a powerful "topological invariant" tool to prove that spaces are not homeomorphic.
Cut-point Method: If a space \(X\) becomes disconnected after removing a point \(p\) (i.e., \(X \setminus \{p\}\)), we call \(p\) a cut point.
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Example Analysis: Intervals and Circles are not Homeomorphic
Consider the open interval \((0, 1)\), the half-open half-closed interval \([0, 1)\), and the unit circle \(S^1\).
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In the open interval \((0, 1)\), any point is a cut point. After removing any point \(c \in (0, 1)\), the space breaks into two disjoint open sets \((0, c) \cup (c, 1)\).
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In the half-closed interval \([0, 1)\), after removing the endpoint \(0\), the remaining part \((0, 1)\) is still connected. So \([0, 1)\) has exactly one non-cut point.
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In the circle \(S^1\), after removing any point, the remaining part (degenerating into an open line segment) is still connected. So \(S^1\) does not have any cut points.
Since the number of cut points (or the existence of a cut point) must remain invariant under a homeomorphism, and the cut point properties of these three spaces are completely different, these three spaces are pairwise not homeomorphic.
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