Proposition

Let \((X_{\alpha})_{\alpha \in A}\) be an indexed family of topological spaces.

1. A subset of \(\coprod_{\alpha \in A} X_{\alpha}\) is closed if and only if its intersection with each $X_{\alpha}$ is closed.
2. Each canonical injection \(i_{\alpha}: X_{\alpha} \rightarrow \coprod_{\alpha \in A} X_{\alpha}\) is a topological embedding and an open and closed map.
3. If each $X_{\alpha}$ is Hausdorff, then so is $\coprod_{\alpha \in A} X_{\alpha}$.
4. If each $X_{\alpha}$ is first countable, then so is $\coprod_{\alpha \in A} X_{\alpha}$.
5. If each $X_{\alpha}$ is second countable and the index set $A$ is countable, then $\coprod_{\alpha \in A} X_{\alpha}$ is second countable.

Solution

First, we will show some properties of disjoint union spaces. Let \(C \subset \coprod_{\alpha \in A} X_{\alpha}\). For each $\alpha$, let $C_{\alpha} = \{ x \mid (x, \beta) \in C, \beta = \alpha \}$. Then $C_{\alpha}$ denotes the intersection of $C$ with each $X_{\alpha}$. Moreover, \(C = \coprod_{\alpha \in A} C_{\alpha}\) because

We claim that $(\coprod X_{\alpha}) \setminus (\coprod C_{\alpha}) = \coprod (X_{\alpha} \setminus C_{\alpha})$.

These properties are true for any subset $C$ of \(\coprod_{\alpha \in A} X_{\alpha}\).

1

Let $C \subset \coprod X_{\alpha}$ be a closed subset. Then $(\coprod X_{\alpha}) \setminus C$ is open. Let $C_{\alpha}$ defined as above. $(\coprod X_{\alpha}) \setminus C = (\coprod X_{\alpha}) \setminus (\coprod C_{\alpha}) = \coprod (X_{\alpha} \setminus C_{\alpha})$.

Since $(\coprod X_{\alpha}) \setminus C$ is open, the intersection with each $\alpha$ is open. The intersection with each $\alpha$ is $X_{\alpha} \setminus C_{\alpha}$. Thus each $C_{\alpha}$ is closed. Moreover, the intersection of $C = \coprod C_{\alpha}$ with each $X_{\alpha}$ is $C_{\alpha}$. Therefore, if $C$ is a closed subset, the intersection of $C$ with each $X_{\alpha}$ is closed.

On the other hand, suppose that the intersection of a subset $C$ with each $X_{\alpha}$ is closed. Then $C_{\alpha}$ is closed in $X_{\alpha}$ for each $\alpha$, so $X_{\alpha} \setminus C_{\alpha}$ is open for each $\alpha$. This implies that $\coprod (X_{\alpha} \setminus C_{\alpha})$ is open. This is equivalent to saying $(\coprod X_{\alpha}) \setminus C$ is open, so $C$ is closed.

2

Let $\alpha_0$ be given.

• Injective?
  • For any $x, y \in X_{\alpha_0}$, $i_{\alpha_0}(x) = i_{\alpha_0}(y) \implies (x, \alpha_0) = (y, \alpha_0) \implies x = y$.
• Continuous?

  • Let $U \subset \coprod_{\alpha \in A} X_{\alpha}$ be an open subset. Then the intersection $U_{\alpha}$ of $U$ with each $X_{\alpha}$ is open. In particular, $U_{\alpha_0}$ is open in $X_{\alpha_0}$.

    \[\begin{align*} (i_{\alpha_0})^{-1}(U) &= (i_{\alpha_0})^{-1}(\coprod_{\alpha \in A} U_{\alpha}) \\ &= U_{\alpha_0}. \end{align*}\]

    Therefore, $i_{\alpha_0}$ is continuous.

Hence, $i_{\alpha_0}$ is a topological embedding.

Let $U \subset X_{\alpha_0}$ be an open subset. Then $i_{\alpha_0}(U) = \coprod_{\alpha \in A} U_{\alpha}$ where

Since each $U_{\alpha}$ is open in their space, $i_{\alpha_0}$ is an open map.

Let $C \subset X_{\alpha_0}$ be a closed subset. Then $i_{\alpha_0}(C) = \coprod_{\alpha \in A} C_{\alpha}$ where

Since each $C_{\alpha}$ is closed in their space, $i_{\alpha_0}(C)$ is closed by (1). Therefore, $i_{\alpha_0}$ is a closed map.

3

Suppose that each $X_{\alpha}$ is Hausdorff. Let $(x_1, \alpha_1) \ne (x_2, \alpha_2) \in \coprod X_{\alpha}$ be given.

• Case 1: $\alpha_1 \ne \alpha_2$. We claim that $i_{\alpha_1}(X_{\alpha_1}), i_{\alpha_2}(X_{\alpha_2})$ separate $(x_1, \alpha_1), (x_2, \alpha_2)$.
  • By (2), the inclusion maps are open. Therefore, $i_{\alpha_1}(X_{\alpha_1}), i_{\alpha_2}(X_{\alpha_2})$ are open.
  • Since $\alpha_1 \ne \alpha_2$, $i_{\alpha_1}(X_{\alpha_1}), i_{\alpha_2}(X_{\alpha_2})$ are disjoint.
• Case 2: $\alpha_1 = \alpha_2$. Then $x_1 \ne x_2$. Since $X_{\alpha_1}$ is Hausdorff, there exist disjoint neighborhoods $U_1, U_2$ of $x_1, x_2$. By (2), $i_{\alpha_1}(U_1), i_{\alpha_1}(U_2)$ are open. Since $U_1, U_2$ are disjoint, $i_{\alpha_1}(U_1), i_{\alpha_1}(U_2)$ are disjoint. Thus we found disjoint neighborhoods of $(x_1, \alpha_1), (x_2, \alpha_1)$.

Therefore, $\coprod X_{\alpha}$ is Hausdorff.

4

Suppose that each $X_{\alpha}$ is first countable. Let $(x_0, \alpha_0) \in \coprod X_{\alpha}$. Then $x_0 \in X_{\alpha_0}$. Since $X_{\alpha_0}$ is first countable, there exists a sequence $\{ U_i \}$ of neighborhoods of $x_0$ in $X_{\alpha_0}$ such that any neighborhood of $x_0$ contains $U_i$ for some $i \in \mathbb{N}$. We claim that $i_{\alpha_0}(U_1), i_{\alpha_1}(U_2), \cdots$ is a local basis of $(x_0, \alpha_0)$.

• $i_{\alpha_0}$ is an open map as shown in (2).
• For each $i \in \mathbb{N}$, $U_i$ is a neighborhood of $x_0$. Therefore, $(x_0, \alpha_0) \in i_{\alpha_0}(U_i)$.

Let $U \subset \coprod X_{\alpha}$ be a neighborhood of $(x_0, \alpha_0)$. Since $U$ is open, the intersection of $U$ with $X_{\alpha_0}$, $U_{\alpha_0} = \{ x \mid (x, \alpha) \in U, \alpha = \alpha_0 \}$, is open in $X_{\alpha}$. Moreover, $(x_0, \alpha_0) \in U$, so $x_0 \in U_{\alpha_0}$. Therefore, $U_{\alpha_0}$ is a neighborhood of $x_0$.

Since $U_1, \cdots$ is a local basis, $U_k \subset U_{\alpha_0}$ for some $k \in \mathbb{N}$. Then $i_{\alpha_0}(U_k) \subset i_{\alpha_0}(U_{\alpha_0}) \subset U$.

Therefore, $i_{\alpha_0}(U_1), \cdots$ is a local basis of $(x_0, \alpha_0)$.

5

Since $A$ is countable, we will regard $A$ as $\mathbb{N}$. Since each $X_1, X_2, \cdots$ is second countable, for each $i \in \mathbb{N}$, $X_i$ has a countable basis $U_{i, 1}, U_{i, 2}, \cdots$. Consider $B = \bigcup_{i \in \mathbb{N}} \{ i_{i}(U_{i, j}) \mid j \in \mathbb{N} \}$. We claim that $B$ is a countable basis of $\coprod X_{i}$.

• $B$ is countable because it is a countable union of countable sets.
• Each set in $B$ is open in $\coprod X_{i}$ because the inclusion maps are open maps by (2).

Let $U \subset \coprod X_{i}$ be open. For each $i$, the intersection $U_{i}$ of $U$ with $X_{i}$ is open. Then for each $i$, there exist $a_{i, 1}, a_{i, 2}, \cdots$ such that $U_i = U_{i, a_{i, 1}} \cup U_{i, a_{i, 2}} \cup \cdots$. We claim that $U = \bigcup_{i, j \in \mathbb{N}} i_i(U_{i, a_{i, j}})$.

• For each $i, j$, $U_{i, a_{i, j}} \subset U_i$. Therefore, $i_i(U_{i, a_{i, j}}) \subset i_i(U_i) \subset U$. Thus $\bigcup_{i, j \in \mathbb{N}} i_i(U_{i, a_{i, j}}) \subset U$.
• Let $(x, i) \in U$. Then $x \in U_i$, so $x \in U_{i, a_{i, j}}$ for some $j \in \mathbb{N}$. Therefore, $(x, i) \in i_i(U_{i, a_{i, j}}) \subset U$.

Therefore, any open set in $\coprod X_i$ can be expressed as the union of elements in $B$, so $B$ is a basis of $\coprod X_i$.

Hence, $\coprod X_i$ has a countable basis, so it is second countable.