> [!definitionb] Definition > > Let $X^0$ be a discrete [[Topological Space|topological space]], then $X^0$ is a **$0$-skeleton**. > > Let $n \in \nat$ and $I_n$ be an index set. For each $i \in I_n$, let $D^n_i = e^n_i \sqcup \partial e^n_i$ be [[Homeomorphism|homeomorphic]] to the closed unit ball in $\real^n$, with $e^n_i$ corresponding to the open unit ball and $\partial e^n_i$ corresponding to the boundary, and $\varphi_i^n: \partial e_i^n \to X^{n - 1}$ be a [[Continuity|continuous]] map. Let $\td X^n = X^{n - 1} \cup \bigsqcup_{i \in I}D_i^n$ be equipped with the disjoint union topology, and take $\sim$ to be the smallest [[Equivalence Relation|equivalence relation]] on $\td X^n$ generated by the relation $x \sim y$ if $y = \varphi_i^n(x)$ for any $x \in \partial e_i^n$ and $y \in X^{n - 1}$. > > For each $x \in \td X$, let $[x]_\sim$ denote its equivalence class by $\sim$, then > > 1. For each $y \in X^{n - 1}$, $[y]_\sim = \bracs{y} \sqcup \bigsqcup_{i \in I_n}(\varphi_i^n)^{-1}(y)$. > 2. For each $x \in e_i^n$, $[x]_\sim = \bracs{x}$. > > The [[Quotient Topology|quotient]] $X^n = \td X^n / \sim$ is an **$n$-skeleton**, and the inclusion $\Phi_i^n = \pi \circ \varphi_i^n: D_i^n \to X^n$ is the **characteristic map** of the **cell** $e_i^n$. > > Let $\pi: \td X^n \to X^n$ be the canonical projection, then > > 3. The inclusion $\pi \circ \iota: X^{n - 1} \to X^n$ is a [[Topological Embedding|topological embedding]]. > 4. For each $i \in I_n$, the restriction $\Phi_i^n|_{e_i^n}: e_i^n \to X^n$ is a topological embedding. > > Hence we identify $X^{n - 1}$ and $e_i^n$ as subspaces of $X^n$. > > Let $X$ be the [[Direct Limit of Topological Spaces|direct limit]] of $X^n$ as $n \to \infty$, then $X$ is a **CW complex**, and $\bracs{D_i^n: i \in I_n, n \in \nat}$ is a **cellulation** of $X$. > > *Proof of Equivalence Relation*. First observe that for any $x \in \partial e_i^n$, $x \sim \varphi_i^n(x)$. Thus when considering equivalence classes, it is sufficient to consider the representative $X^{n-1} \sqcup \bigcup_{i \in I_n}e_i^n$. > > For any $y \in X^{n - 1}$, let $[y]_\approx = \bracs{y} \sqcup \bigsqcup_{i \in I_n}(\varphi_i^n)^{-1}(y)$. For each $x \in e^n_i$, let $[x]_\approx = \bracs{x}$, then since > $ > \td X = X^{n - 1} \sqcup \bigsqcup_{i \in I_n}\partial e_i^n\sqcup \bigsqcup_{i \in I_n}e_i^n = \paren{\bigsqcup_{y \in X^{n - 1}}[y]_\approx} \sqcup \paren{\bigsqcup_{i \in I_n}\bigsqcup_{x \in e_i^n}[x]_\approx} > $ > the sets $\bracsn{[x]_\approx: x \in \td X^n}$ partitions $\td X^n$, and defines an equivalence relation $\approx$. > > For each $i \in I_n$, $y \in X^{n - 1}$, and $x \in \partial e_i^n$, if $x \sim y$, then $x \in (\varphi_i^n)^{-1}(y)$. Thus $[y]_\approx \subset [y]_\sim$. On the other hand, observe that $\sim$ is contained in $(X^{n - 1} \cup \bigsqcup_{i \in I_n}\partial e_i^n)^2$, so $[x]_\approx = [x]_\sim = \bracs{x}$ for all $x \in \bigsqcup_{i \in I_n}e_i^n$. Therefore $\approx$ is a smaller equivalence relation than $\sim$. By minimality of $\sim$, the two relations coincide, so $(1)$ and $(2)$ hold. > > *Proof of Embeddings.* $(3)$: Let $U \subset X^{n - 1}$ be open, then $U$ is also open in $\td X^{n}$. For each $i \in I_n$, by continuity of $\varphi_i^n$, $(\varphi_i^n)^{-1}(U)$ is open in $D^i_n$, and hence in $\td X^n$ as well. Let $V = U \cup \bigcup_{i \in I}(\varphi_i^n)^{-1}(U)$, then $V$ is also open in $\td X^n$ with $V \cap X^{n - 1} = U$. In addition, for each $x \in X^{n - 1} \cap V$, $V \supset [x]_\sim = \bracs{x} \cup (\varphi_i^n)^{-1}(x)$. For each $i \in I_n$ and $x \in e_i^n \cap V$, $[x]_\sim = \bracs{x} \subset V$. Thus $V$ is saturated with respect to $\sim$, and $\pi(V)$ is open in $X^n$. From here, observe that > $ > \pi(V) \cap \pi(X^{n - 1}) = \underbrace{\pi(V \cap X^{n-1})}_{\pi(U)} \cup \underbrace{\braks{\pi\paren{V \cap \bigcup_{i \in I_n}e_n^i} \cap \pi(X^{n - 1})}}_{\emptyset} > $ > Therefore $\pi(U)$ is relatively open in $\pi(X^{n - 1})$ with respect to $\td X^n$, and $\pi \circ \iota: X^{n - 1} \to X^n$ is an embedding. > > $(4)$: Let $i \in I_n$ and $U \subset e_i^n$ be open, then $U$ is also open in $\td X^n$. Since $[x]_\sim = \bracs{x}$ for all $x \in e_i^n$, $U$ is saturated with respect to $\sim$, and $\pi(U)$ is open in $X^n$. Therefore $\pi \circ \iota: e_i^n \to X^n$ is an embedding. > [!theorem] > > Let $X$ be a CW-complex with cellulation $\bracs{(D_\alpha^n, e_\alpha^n, \varphi_\alpha^n, \Phi_\alpha^n)}$ and $A \subset X$, then > 1. $A$ is open if and only if $(\Phi_\alpha^n)^{-1}(A)$ is open for all $\alpha$. > 2. $A$ is closed if and only if $(\Phi_\alpha^n)^{-1}(A)$ is closed for all $\alpha$. > > *Proof*. $(1, 2, \Rightarrow)$: By continuity of $\Phi_\alpha^n$. > > $(1, \Leftarrow)$: Since $X^0$ is discrete, $A \cap X^0$ is both open and closed. > > Suppose that $A$ is open in $X^n$. Let $\td X^n, \td X^{n+1}$ and $\pi_n: \td X^n \to X^n$ and $\pi_{n+1}: \td X^{n+1} \to X^{n+1}$ be as in the construction of the CW-complex, then there is a decomposition > $ > \begin{align*} > A \cap X^{n+1} &= (A \cap X^n) \cup \braks{A \cap (X^{n+1} \setminus X^n)} \\ > \pi_{n+1}^{-1}(A \cap X^{n+1}) &= (A \cap X^n) \cup \bigcup_{\alpha \in I_n}\pi_{n+1}^{-1}(A \cap \Phi_\alpha^{n+1}(D_\alpha^{n+1})) \\ > \pi_{n+1}^{-1}(A \cap X^{n+1}) &= (A \cap X^n) \cup\bigcup_{\alpha \in I_n}(\Phi_{\alpha}^{n+1})^{-1}(A) > \end{align*} > $ > as a saturated open set in $\td X^{n+1}$. Thus $A$ is open in $X^{n+1}$. > [!definition] > > Let $X$ be a CW-complex with cellulation $\bracs{(D_\alpha^n, e_\alpha^n, \varphi_\alpha^n, \Phi_\alpha^n)}$ and $A \subset X$ be [[Compactness|compact]], then $A$ intersects at most finitely many cells. > > *Proof*. For each $n \in \nat$, let $J_n \subset I_n$ be the family of indices such that $A \cap e_\alpha^n$ for all $\alpha \in J_n$. For each $\alpha \in J_n$, let $x_\alpha \in e_\alpha^n$, and $S = \bracs{x_\alpha: n \in \nat, \alpha \in J_n}$. > > For each $\alpha \in J_0$, $(\Phi_\alpha^0)^{-1}(S)$ is a singleton, and is closed in $D_\alpha^0$. Suppose that $(\Phi_\alpha^{n})^{-1}(S)$ is closed, then $S$ is closed in $X^n$. Thus for any $\alpha \in J_{n+1}$, > $ > (\Phi_\alpha^{n+1})^{-1}(S) = (\varphi_\alpha^{n+1})^{-1}(S \cap X^{n}) \cup \bracs{x_\alpha} > $ > By continuity of $\varphi_\alpha^{n+1}$, $(\Phi_\alpha^{n+1})^{-1}(S)$ is closed. > > In addition, observe that $A \cap X^n$ has the discrete topology for all $n \in \nat$. Thus $A$ has the discrete topology. > > Since $A \subset S$ is a closed subset of a compact set, it is compact and thus finite. > [!definition] > > Let $X$ be a CW-complex with cellulation $\bracs{(D_\alpha^n, e_\alpha^n, \varphi_\alpha^n, \Phi_\alpha^n)}$, $A \subset X$, and $\eps: \bigcup_{n \in \nat}I_n \to \real_{> 0}$. Let $N_\eps^0(A) = A \cap X^0$. For each $n \in \nat$ and $\alpha \in I_n$, let > $ > N_{\eps, \alpha}^{n+1}(A) = B(A \cap e_\alpha^{n+1}, \eps(\alpha)) \cup \braks{(1 - \eps_\alpha, 1] \times N_\eps^{n}(A) \cap \partial D_\alpha^{n+1}} > $ > where $B(E, \eps(\alpha))$ is the $\eps(\alpha)$-fattening of $E$ in $e_\alpha^{n+1}$, and the product $(1 - \eps_\alpha, 1] \times N_\eps^{n}(A) \cap \partial e_\alpha^{n+1}$ is taken over spherical coordinates. Now define $N_\eps^{n+1}(A) = \bigcup_{\alpha \in I_{n+1}}N_{\eps, \alpha}^{n+1}(A)$, then $N_\eps(A) = \bigcup_{n \in \nat_0}N_\eps^n(A)$ is open in $X$, and is known as the **$\eps$-neighbourhood** of $A$ in $X$. > > *Proof*. Since $X^0$ is discrete, $N_\eps^0(A)$ is open in $X^0$. Suppose that $N_\eps^n(A) \subset X^n$ is open, then for each $\alpha \in I_{n+1}$, > $ > (\Phi_\alpha^{n+1})^{-1}(N_\eps^{n+1}(A)) = B(A \cap e_{\alpha}^{n+1}, \eps(\alpha)) \cup (\varphi_{\alpha}^{n+1})^{-1}(N_\eps^n(A)) > $ > is a union of open sets by continuity of $\varphi_\alpha^{n+1}$. Thus $N_\eps^{n+1}(A) \subset X^{n+1}$ is open, and $N_\eps(A)$ is open in $X$. > [!theorem] > > Let $X$ be a CW-complex, then $X$ is normal. > > *Proof*. Let $\bracs{(D_\alpha^n, e_\alpha^n, \varphi_\alpha^n, \Phi_\alpha^n)}$ be a cellulation of $X$, and $A, B \subset X$ be closed sets. > > Let $n \in \nat$ and suppose that $N_\eps^{n}(A)$ and $N_\eps^n(B)$ have disjoint closures. By compactness of each disk, for each $\alpha \in I_{n+1}$, there exists $\eps(\alpha) > 0$ such that $N_{\eps, \alpha}^{n+1}(A)$ and $N_{\eps, \alpha}^{n+1}(B)$ are disjoint. > > Proceeding by induction, $N_\eps(A)$ and $N_\eps(B)$ are desired disjoint neighbourhoods.