Paracompact spaces. A topological space $X$ is paracompact if
- $X$ is Hausdorff and
- for every open cover $X = \bigcup_{i \in I}U_{i},$ there is a locally finite open cover $X = \bigcup_{j \in I}V_{j},$ refining the given cover.
Personal remark. The second condition is mouthful. Saying that $\{V_{j}\}_{j \in J}$ is locally finite in $X$ means that for any $x \in X,$ there are at most finitely many $V_{j}$ that contain $x.$ Saying that $\{V_{j}\}_{j \in J}$ refines $\{U_{i}\}_{i \in I}$ means that for each $V_{j}$ there is $U_{i} \supset V_{j}.$ This seems to me counter-intuitive at first, but I suppose I can think like "every small guy should refine some big guy" because if some $V_{j}$ does not lie in any $U_{i},$ it does not look like we are "refining" when we draw a picture.
Example. If $X$ is Hausdorff and compact, then $X$ is paracompact, because for any open cover of $X,$ we can find a finite open subcover, which is certainly a locally finite refinement that covers $X.$
Example. If $X$ is a CW complex, then $X$ is paracompact. To see $X$ is Hausdorff, note that if any two distinct points lie in a cell with dimension $> 0,$ then we can take disjoint open neighborhoods of those points in the cell. Otherwise, both are $0$-cells, which can be separated by open sets in $X.$
Of course, a finite CW complex (a CW complex with finitely many cells in total) is already compact, so it is paracompact.
For a general CW complex, we may have infinitely many cells, so it cannot be compact. For instance, the real line $\mathbb{R}$ with the Euclidean topology is a CW complex because we can take $\mathbb{Z}$ to be $0$-cells, and the segments $$\cdots, [-2, -1], [-1, 0], [0, 1], [1, 2], \dots$$ $1$-cells. However, we know that $\mathbb{R}$ is not compact because it is not bounded.
The proof that a general CW complex is paracompact seems to be considered "elementary" by the community, but Fritsch and Piccinini (Theorem 1.3.5) spends about two full pages, which I decided to skip for now.
Example. Any topological manifold is paracompact.
We shall spend the rest of this posting to prove this statement.
Remark. Topological manifolds are somewhat more general notion of spaces incorporating smooth manifolds. A topological space $M$ is a topological manifold if
The second-countablity axiom will be used as the following form:
Lemma. If $X$ is a second-countable topological space, any basis of $X$ has a countable subbasis.
Proof. Let $\mathcal{U} = \{U_{i}\}_{i \in I}$ be any basis and $\mathcal{B} = \{B_{j}\}_{j=1}^{\infty}$ be a countable basis of $X.$ We follow a clever observation from these notes: any collection of open subsets with which we can write each $B_{j}$ as a countable union is a countable basis. Since $U_{i}$ form a basis, we can write $$B_{j} = \bigcup_{i \in I_{j}}U_{i},$$ and we would have been done if $I_{j} \subset I$ was countable. If not, for each $x \in B_{j},$ choose any $j_{x} \in J$ such that $x \in B_{j_{x}} \subset U_{i_{x}}$ for some $i_{x} \in I_{j},$ which we choose and fix. There can be some repetitions for $B_{j_{x}}$ and $U_{i_{x}},$ but the size of the collection of all $B_{j}$ where $j$ can appear as $j = j_{x}$ is greater than equal to those $U_{i}$ where $i$ can appear as $i = i_{x}.$ Such $i$ form an at most countable subcollection of $I_{j},$ which finishes the proof. $\Box$
Of course, a finite CW complex (a CW complex with finitely many cells in total) is already compact, so it is paracompact.
For a general CW complex, we may have infinitely many cells, so it cannot be compact. For instance, the real line $\mathbb{R}$ with the Euclidean topology is a CW complex because we can take $\mathbb{Z}$ to be $0$-cells, and the segments $$\cdots, [-2, -1], [-1, 0], [0, 1], [1, 2], \dots$$ $1$-cells. However, we know that $\mathbb{R}$ is not compact because it is not bounded.
The proof that a general CW complex is paracompact seems to be considered "elementary" by the community, but Fritsch and Piccinini (Theorem 1.3.5) spends about two full pages, which I decided to skip for now.
Example. Any topological manifold is paracompact.
We shall spend the rest of this posting to prove this statement.
Remark. Topological manifolds are somewhat more general notion of spaces incorporating smooth manifolds. A topological space $M$ is a topological manifold if
- $M$ is Hausdorff,
- $M$ is second-countable (meaning it has a countable basis),
- $M$ is locally Euclidean of dimension $n$ for some $n \in \mathbb{Z}_{\geq 0}.$
The second-countablity axiom will be used as the following form:
Lemma. If $X$ is a second-countable topological space, any basis of $X$ has a countable subbasis.
Proof. Let $\mathcal{U} = \{U_{i}\}_{i \in I}$ be any basis and $\mathcal{B} = \{B_{j}\}_{j=1}^{\infty}$ be a countable basis of $X.$ We follow a clever observation from these notes: any collection of open subsets with which we can write each $B_{j}$ as a countable union is a countable basis. Since $U_{i}$ form a basis, we can write $$B_{j} = \bigcup_{i \in I_{j}}U_{i},$$ and we would have been done if $I_{j} \subset I$ was countable. If not, for each $x \in B_{j},$ choose any $j_{x} \in J$ such that $x \in B_{j_{x}} \subset U_{i_{x}}$ for some $i_{x} \in I_{j},$ which we choose and fix. There can be some repetitions for $B_{j_{x}}$ and $U_{i_{x}},$ but the size of the collection of all $B_{j}$ where $j$ can appear as $j = j_{x}$ is greater than equal to those $U_{i}$ where $i$ can appear as $i = i_{x}.$ Such $i$ form an at most countable subcollection of $I_{j},$ which finishes the proof. $\Box$
The last axiom for topological manifolds means that for any $x \in M,$ there is an open neighborhood $U \ni x$ in $M$ and a homeomorphic open embedding $U \hookrightarrow \mathbb{R}^{n}.$ The number $n$ appearing here is called the dimension of $M.$ Note that for a smooth manifold, the homeomorphic open embeddings are also need to be "smoothly compatible" (see $\mathscr{C}^{\infty}$-compatbility on Section 5.2 in Tu's book). Note that any topological manifold $M$ is locally compact, meaning that every $x \in M$ of it has an open neighborhood contained in a compact subset of $M.$
Lemma. Let $X$ be a topological space. If $X$ is Hausdorff, then the following are equivalent:
- $X$ is locally compact;
- Every $x \in X$ has a precompact (compact closure) open neighborhood in $X$;
- $X$ has a basis consisting of precompact open subsets of $X.$
Proof. Assume Statement 1. For $x \in X,$ take $x \in U \subset K,$ where $U$ is open in $X$ and $K$ is compact. Since $X$ is Hausdorff, we see that $K \subset X$ is closed. Thus, we have $\bar{U} \subset K,$ so $U$ is precompact. This shows Statement 2.
Assume Statement 2. For any given basis $\mathcal{B}$ of $X,$ let $\mathcal{B}'$ be the subcollection of $\mathcal{B}$ consisting of only precompact open subsets. We show that $\mathcal{B}'$ is still a basis of $X.$ For any $x \in X,$ using the assumption, pick any precompact open $U \ni x.$ Then we may find $B \in \mathcal{B}$ such that $x \in B \subset U.$ This implies that $\bar{B} \subset \bar{U},$ and since $\bar{U}$ is compact, its closed subset $\bar{B}$ must be compact as well. This implies that $B \in \mathcal{B}',$ showing our claim that it is a basis of $X.$ This shows Statement 3.
Statement 3 evidently implies Statement 1. $\Box$
Lemma 1. Let $X$ be a topological space that is
Then $X$ admits an exhaustion by compact subsets, meaning an sequence $K_{1} \subset K_{2} \subset \cdots$ of compact subsets whose union covers $X$ such that $K_{i} \subset K_{i+1}^{\mathrm{o}}$ for all $i \geq 1.$
- second-countable,
- locally compact, and
- Hausdorff.
Then $X$ admits an exhaustion by compact subsets, meaning an sequence $K_{1} \subset K_{2} \subset \cdots$ of compact subsets whose union covers $X$ such that $K_{i} \subset K_{i+1}^{\mathrm{o}}$ for all $i \geq 1.$
Proof. Since $X$ is locally compact Hausdorff, we may find a basis consisting of precompact open subsets of $X.$ As we have shown above, the hypothesis that $X$ is second countable allows us to assume that this basis is countable. In particular, we have a countable precompact open cover $X = \bigcup_{i=1}^{\infty}U_{i}.$ We define $K_{1} := \overline{U_{1}}$ in $X,$ which is compact because $U_{1}$ is precompact. Write $j_{1} := 1.$ Since $K_{1}$ is compact, there must be $j_{2} \in \mathbb{Z}_{\geq 1}$ such that $$K_{1} \subset U_{1} \cup U_{2}, \cdots \cup U_{j_{2}}.$$ We define $K_{2} := \overline{U_{1}} \cup \cdots \cup \overline{U_{j_{2}}},$ which is compact in $X$ due to precompactness of $U_{i}.$ This way, whenever we have can construct $$K_{i} := \overline{U_{1}} \cup \cdots \cup \overline{U_{j_{i}}}$$ and pick $j_{i+1} > j_{i}$ such that $$K_{i} \subset U_{1} \cup \cdots \cup U_{j_{i+1}}.$$ This way, we construct $K_{1} \subset K_{2} \subset K_{3} \subset \cdots$ recursively such that $K_{i} \subset K_{i+1}^{\mathrm{o}}$ for all $i \geq 1.$ This finishes the proof $\Box$
Lemma 2. Let $X$ be a topological space that admits an exhaustion by compact subsets. Then for any open cover $\mathcal{U}$ and any basis $\mathcal{B}$ of $M,$ there is a refinement $\mathcal{U}'$ of $\mathcal{U}$ such that
Lemma 2. Let $X$ be a topological space that admits an exhaustion by compact subsets. Then for any open cover $\mathcal{U}$ and any basis $\mathcal{B}$ of $M,$ there is a refinement $\mathcal{U}'$ of $\mathcal{U}$ such that
- $\mathcal{U}'$ is still an open cover of $X,$
- $\mathcal{U}'$ is countable,
- $\mathcal{U}'$ is locally finite, and
- $\mathcal{U}' \subset \mathcal{B}.$
Proof. Let $K_{1} \subset K_{2} \subset K_{3} \subset \cdots$ be a compact exhaustion of $X.$ Define $$R_{i} := K_{i+1} \smallsetminus K_{i}^{\mathrm{o}}$$ and $$T_{i} := K_{i+2}^{\mathrm{o}} \smallsetminus K_{i-1} \supset R_{i},$$ where $K_{0} := \emptyset.$ Note that $R_{i}$ is compact because it is a closed subset of a compact set $K_{i+1}.$ For each $x \in R_{i},$ we may take $U_{x} \in \mathcal{U}$ and $B_{x} \in \mathcal{B}$ such that $x \in B_{x} \subset U_{x} \cap T_{i}.$ Since $R_{i}$ is compact, we can find finitely many $B_{x_{i,1}}, \dots, B_{x_{i,n_{i}}}$ such that $$R_{i} \subset B_{x_{i,1}} \cup B_{x_{i,2}} \cup \cdots \cup B_{x_{i,n_{i}}} \subset (U_{i,x_{1}} \cup U_{i,x_{2}} \cup \cdots \cup U_{i,x_{n_{i}}}) \cap T_{i}.$$ Now, we may take $$\mathcal{U}' := \{B_{i,j} : i \in \mathbb{Z}_{\geq 1}, 1 \leq j \leq n_{i}\},$$ and check that it is locally finite and covers $X.$ This finishes the proof $\Box$
Corollary. Every locally compact, Hausdorff, and second-countable topoloigical space is paracompact. In particular, topoloigcal manifolds are paracompact.
Paracompactness implies normality. Recall that a topological space $X$ is said to be normal if for any disjoint closed subsets $A, B$ of $X,$ we can find disjoint open subsets $V \supset A$ and $W \supset B$ in $X.$ Note that this condition is equivalent to say that for every closed $A \subset X$ and open $U \supset A$ in $X,$ we can find another open subset $V$ of $X$ such that $A \subset \bar{V} \subset U,$ where the closure is taken in $X.$
Proof. Let $X$ be normal. Given any closed subset $A \subset X$ and an open subset $U \supset A$ in $X,$ the complement $B := X \smallsetminus U$ is a closed subset of $X$ disjoint to $A.$ Hence, we may find disjoint open subsets $V \supset A$ and $W \supset B$ in $X$ using normality. But then $$A \subset V \subset X \smallsetminus W \subset X \smallsetminus B = U,$$ and since $X \smallsetminus W$ is closed in $X,$ it contains $\bar{V},$ which implies what we want.
Conversely, assume the condition on $X$ we hoped to be equivalent to nomarlity of $X.$ Fix any disjotint closed subsets $A, B \subset X.$ Then we have $A \subset U := X \smallsetminus B,$ so we may find an open subset $V \subset X$ such that $A \subset V \subset \bar{V} \subset U.$ This implies that $$W := X \smallsetminus \bar{V} \supset B$$ and $W$ is an open subset of $X$ disjoint to $W.$ This shows that $X$ is normal. $\Box$
Theorem. Any paracompact topological space is nomal.
Proof. Let $A, B \subset X$ be disjoint closed subsets. We want to find disjoint open $V, W \subset X$ such that $A \subset V$ and $B \subset W.$
Case 1. Assume that $A = \{a\},$ a singleton.
For each $x \in B,$ using Hausdorffness, choose open $V_{x} \ni a$ and $W_{x} \ni x$ such that $V_{x} \cap W_{x} = \emptyset.$ In particular, we have $$\overline{W_{x}} \subset X \smallsetminus V_{x},$$ so $a \notin \overline{W_{x}}.$
This way, we construct an open cover $\{W_{x}\}_{x \in B}$ for $B$ in $X.$ Then $\{W_{x}\}_{x \in B} \cup \{X \smallsetminus B\}$ is an open cover of $X,$ so by paracompactness of $X,$ there is a locally finite refinement that is an open cover of $X.$ We may assume that it is of the form $\{W'_{x}\}_{x \in B} \cup \{U_{B}\},$ where $W'_{x} \subset W_{x}$ and $U_{B} \subset X \smallsetminus B.$ Note that $\{W'_{x}\}_{x \in B}$ is an open cover of $B$ in $X$ that is locally finite and $a \notin \overline{W'_{x}}.$
The key observation is that due to local finiteness, we have $$\overline{\bigcup_{x \in B} W'_{x}} = \bigcup_{x \in B} \overline{W'_{x}}$$ because any limit point of $\bigcup_{x \in B} W'_{x}$ is necessarily a limit point of the union of some finitely many $W'_{x}.$ In particular, this shows that the left-hand side does not contain $a.$ Then taking $$V := X \smallsetminus \overline{\bigcup_{x \in B} W'_{x}} \ni a$$ and $$W := \bigcup_{x \in B} W'_{x} \supset B,$$ we finish the proof for this case.
Case 2. Now we consider the general case.
For each $a \in A,$ take an open subset $V_{a} \ni a$ and $W_{a} \supset B$ in $X$ such that $V_{a} \cap W_{a} = \emptyset,$ using Case 1. Note that $\overline{V_{a}} \subset X \smallsetminus B,$ so $\overline{V_{a}} \cap B = \emptyset.$
Then $\{V_{a}\}_{a \in A}$ gives an open cover of $A,$ so $\{V_{a}\}_{a \in A} \cup \{X \smallsetminus B\}$ is an open cover of $X.$ Using paracompactness, we may find a locally finite refinement of this cover that is also an open cover of $X.$ We can write it in the form $\{V'_{a}\}_{a \in A} \cup \{U_{B}\}$ where $V'_{a} \subset V_{a}$ and $U_{B} \subset X \setminus B.$ Again, the upshot is that we have $$\overline{\bigcup_{a \in A} V'_{a}} = \bigcup_{a \in A} \overline{V'_{a}}$$ due to local finiteness. In particular, the left-hand side is disjoint to $B,$ so can obtain an open subset $$W := X \smallsetminus \overline{\bigcup_{a \in A} V'_{a}} \supset B$$ in $X.$ Since $$V := \bigcup_{a \in A} V'_{a} \supset A$$ is open and disjoint to $V,$ this finishes the proof $\Box$
Paracompactness implies normality. Recall that a topological space $X$ is said to be normal if for any disjoint closed subsets $A, B$ of $X,$ we can find disjoint open subsets $V \supset A$ and $W \supset B$ in $X.$ Note that this condition is equivalent to say that for every closed $A \subset X$ and open $U \supset A$ in $X,$ we can find another open subset $V$ of $X$ such that $A \subset \bar{V} \subset U,$ where the closure is taken in $X.$
Proof. Let $X$ be normal. Given any closed subset $A \subset X$ and an open subset $U \supset A$ in $X,$ the complement $B := X \smallsetminus U$ is a closed subset of $X$ disjoint to $A.$ Hence, we may find disjoint open subsets $V \supset A$ and $W \supset B$ in $X$ using normality. But then $$A \subset V \subset X \smallsetminus W \subset X \smallsetminus B = U,$$ and since $X \smallsetminus W$ is closed in $X,$ it contains $\bar{V},$ which implies what we want.
Conversely, assume the condition on $X$ we hoped to be equivalent to nomarlity of $X.$ Fix any disjotint closed subsets $A, B \subset X.$ Then we have $A \subset U := X \smallsetminus B,$ so we may find an open subset $V \subset X$ such that $A \subset V \subset \bar{V} \subset U.$ This implies that $$W := X \smallsetminus \bar{V} \supset B$$ and $W$ is an open subset of $X$ disjoint to $W.$ This shows that $X$ is normal. $\Box$
Theorem. Any paracompact topological space is nomal.
Proof. Let $A, B \subset X$ be disjoint closed subsets. We want to find disjoint open $V, W \subset X$ such that $A \subset V$ and $B \subset W.$
Case 1. Assume that $A = \{a\},$ a singleton.
For each $x \in B,$ using Hausdorffness, choose open $V_{x} \ni a$ and $W_{x} \ni x$ such that $V_{x} \cap W_{x} = \emptyset.$ In particular, we have $$\overline{W_{x}} \subset X \smallsetminus V_{x},$$ so $a \notin \overline{W_{x}}.$
This way, we construct an open cover $\{W_{x}\}_{x \in B}$ for $B$ in $X.$ Then $\{W_{x}\}_{x \in B} \cup \{X \smallsetminus B\}$ is an open cover of $X,$ so by paracompactness of $X,$ there is a locally finite refinement that is an open cover of $X.$ We may assume that it is of the form $\{W'_{x}\}_{x \in B} \cup \{U_{B}\},$ where $W'_{x} \subset W_{x}$ and $U_{B} \subset X \smallsetminus B.$ Note that $\{W'_{x}\}_{x \in B}$ is an open cover of $B$ in $X$ that is locally finite and $a \notin \overline{W'_{x}}.$
The key observation is that due to local finiteness, we have $$\overline{\bigcup_{x \in B} W'_{x}} = \bigcup_{x \in B} \overline{W'_{x}}$$ because any limit point of $\bigcup_{x \in B} W'_{x}$ is necessarily a limit point of the union of some finitely many $W'_{x}.$ In particular, this shows that the left-hand side does not contain $a.$ Then taking $$V := X \smallsetminus \overline{\bigcup_{x \in B} W'_{x}} \ni a$$ and $$W := \bigcup_{x \in B} W'_{x} \supset B,$$ we finish the proof for this case.
Case 2. Now we consider the general case.
For each $a \in A,$ take an open subset $V_{a} \ni a$ and $W_{a} \supset B$ in $X$ such that $V_{a} \cap W_{a} = \emptyset,$ using Case 1. Note that $\overline{V_{a}} \subset X \smallsetminus B,$ so $\overline{V_{a}} \cap B = \emptyset.$
Then $\{V_{a}\}_{a \in A}$ gives an open cover of $A,$ so $\{V_{a}\}_{a \in A} \cup \{X \smallsetminus B\}$ is an open cover of $X.$ Using paracompactness, we may find a locally finite refinement of this cover that is also an open cover of $X.$ We can write it in the form $\{V'_{a}\}_{a \in A} \cup \{U_{B}\}$ where $V'_{a} \subset V_{a}$ and $U_{B} \subset X \setminus B.$ Again, the upshot is that we have $$\overline{\bigcup_{a \in A} V'_{a}} = \bigcup_{a \in A} \overline{V'_{a}}$$ due to local finiteness. In particular, the left-hand side is disjoint to $B,$ so can obtain an open subset $$W := X \smallsetminus \overline{\bigcup_{a \in A} V'_{a}} \supset B$$ in $X.$ Since $$V := \bigcup_{a \in A} V'_{a} \supset A$$ is open and disjoint to $V,$ this finishes the proof $\Box$
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