This question asks about the properties of closed maps between metric spaces. A map $f: X \to Y$ between metric spaces $(X, d_X)$ and $(Y, d_Y)$ is defined as a closed map if for every closed set $F$ in $X$, the image $f(F)$ is a closed set in $Y$. We need to evaluate each given statement.

Metric Space Definition Review

A metric space $(X, d_X)$ is a set $X$ equipped with a distance function $d_X$ that satisfies certain properties. A subset $B$ of a metric space is given the induced metric, meaning distances within $B$ are measured using the metric of the larger space.

The product metric on $X \times Y$ is given by $d((x, y), (x', y')) = \max\{d_X(x, x'), d_Y(y, y')\}$. This is one of the common ways to define a metric on a product space.

Closed Map Properties Analysis

Let's analyze each statement:

Option 1: For any subset $A \subseteq X$ the inclusion map $i: A \to X$ is closed.

• The inclusion map $i: A \to X$ is defined by $i(a) = a$ for all $a \in A$.
• $A$ is considered as a metric space with the induced metric. A set $F \subseteq A$ is closed in $A$ if and only if $F$ is the intersection of $A$ with a closed set in $X$. That is, $F = C \cap A$ where $C$ is closed in $X$.
• The image of $F$ under the inclusion map is $i(F) = F$.
• For $i: A \to X$ to be closed, for any $F$ closed in $A$, $i(F) = F$ must be closed in $X$.
• However, consider $X = \mathbb{R}$ with the standard metric, and $A = (0, 1)$. $A$ is a subset of $X$. Let $F = (0, 1/2]$. This set $F$ is closed in $A$ (since $F = [0, 1/2] \cap (0, 1)$, and $[0, 1/2]$ is closed in $\mathbb{R}$).
• The image $i(F)$ is $(0, 1/2]$ in $X = \mathbb{R}$. This set $(0, 1/2]$ is not closed in $\mathbb{R}$.
• Therefore, the inclusion map is not always a closed map for any subset $A \subseteq X$.

Option 2: The projection map $p_1: X \times Y \to X$ given by $p_1(x, y) = x$ is closed.

• For $p_1$ to be closed, for any closed set $F$ in $X \times Y$, the image $p_1(F) = \{x \in X \mid \exists y \in Y, (x, y) \in F\}$ must be closed in $X$.
• Consider $X = \mathbb{R}$ and $Y = \mathbb{R}$ with the standard metric. The product space $\mathbb{R} \times \mathbb{R} = \mathbb{R}^2$ has the metric $d((x, y), (x', y')) = \max\{|x-x'|, |y-y'|\}$.
• Consider the set $F = \{(x, y) \in \mathbb{R}^2 \mid xy = 1\}$. This set is the graph of the function $y=1/x$. This set is closed in $\mathbb{R}^2$ because it is the inverse image of the closed set $\{1\}$ under the continuous multiplication map $(x,y) \mapsto xy$.
• The projection of $F$ onto the first coordinate is $p_1(F) = \{x \in \mathbb{R} \mid \exists y \in \mathbb{R}, xy=1\}$. If $x \ne 0$, we can find $y=1/x$. If $x=0$, there is no such $y$. So $p_1(F) = \{x \in \mathbb{R} \mid x \ne 0\} = \mathbb{R} \setminus \{0\}$.
• The set $\mathbb{R} \setminus \{0\}$ is not closed in $\mathbb{R}$ (its complement $\{0\}$ is not open).
• Therefore, the projection map is not always a closed map. (Note: Projection maps are usually open maps, not closed maps).

Option 3: Suppose that $f: X \to Y$, $g: Y \to Z$ are continuous maps. If $g \circ f: X \to Z$ is a closed map then $g|_{f(X)}: f(X) \to Z$ is closed. Here $g|_{f(X)}$ means the map $g$ restricted to $f(X)$.

• Let $F$ be a closed set in $f(X)$. $f(X)$ is considered as a metric space with the induced metric from $Y$.
• Since $F$ is closed in the subspace $f(X)$, there exists a closed set $C$ in $Y$ such that $F = C \cap f(X)$.
• We need to show that the image of $F$ under $g|_{f(X)}$ is closed in $Z$. The map $g|_{f(X)}: f(X) \to Z$ is defined by $(g|_{f(X)})(y) = g(y)$ for all $y \in f(X)$.
• So we need to show that $g(F)$ is closed in $Z$.
• Since $F \subseteq f(X)$, $g(F) = g(C \cap f(X))$. Since $F = C \cap f(X)$, we have $g(F) = g(C \cap f(X))$.
• Let's consider a different approach. Let $F$ be closed in $f(X)$. Let $y \in F$. Then $y \in f(X)$, so there exists $x \in X$ such that $f(x)=y$. Since $F$ is closed in $f(X)$, we can write $F = C \cap f(X)$ for some closed set $C \subseteq Y$.
• Consider the set $A = f^{-1}(F) = \{x \in X \mid f(x) \in F\}$. Since $f$ is continuous and $F$ is closed in $f(X)$, and $f(X)$ has the induced topology from $Y$, $F$ is the intersection of a $Y$-closed set with $f(X)$. However, continuity of $f: X \to Y$ means that the preimage of a closed set in $Y$ is closed in $X$. Is $F$ necessarily closed in $Y$? No, only closed in the subspace $f(X)$.
• Let's use the property that $F$ is closed in $f(X)$. This means that if $(y_n)$ is a sequence in $F$ converging to $y \in f(X)$ (with respect to the metric of $Y$), then $y \in F$.
• Consider the set $A = f^{-1}(F) = \{x \in X \mid f(x) \in F\}$. Since $F \subseteq f(X)$, $f(A) = F$.
• Since $F$ is closed in the subspace $f(X)$, $F = C \cap f(X)$ for some closed set $C$ in $Y$. Since $f$ is continuous, $f^{-1}(C)$ is closed in $X$. Also, $A = f^{-1}(F) = f^{-1}(C \cap f(X)) = f^{-1}(C) \cap f^{-1}(f(X))$. $f^{-1}(f(X))$ might not be $X$.
• Let's rethink. $F$ is closed in the subspace $f(X)$. Let $y \in F$. Then $y=f(x)$ for some $x \in X$.
• Let $y_0 \in Z$ be a limit point of $g(F)$. We want to show $y_0 \in g(F)$.
• There exists a sequence $(z_n)$ in $g(F)$ such that $z_n \to y_0$ in $Z$.
• Since $z_n \in g(F)$, $z_n = g(y_n)$ for some $y_n \in F$. Since $y_n \in F$ and $F \subseteq f(X)$, $y_n \in f(X)$.
• Since $y_n \in f(X)$, $y_n = f(x_n)$ for some $x_n \in X$.
• So $z_n = g(f(x_n)) = (g \circ f)(x_n)$. We have a sequence $(x_n)$ in $X$ such that $(g \circ f)(x_n) \to y_0$ in $Z$.
• Since $g \circ f$ is a closed map and $y_0$ is in the closure of the image of the sequence $(x_n)$ under $g \circ f$, the set $\{(g \circ f)(x_n)\}$ has $y_0$ as a limit point.
• Consider the set $A = \{x_n \mid n \in \mathbb{N}\}$. The image $(g \circ f)(A) = \{z_n\}$. The closure of this set contains $y_0$.
• Let $F'$ be the closure of $\{x_n\}$ in $X$. $(g \circ f)(F')$ is not necessarily closed.
• Let's use the definition directly. Let $F$ be closed in $f(X)$. We need to show $g(F)$ is closed in $Z$.
• Let $y_0 \in Z$ be a limit point of $g(F)$. There exists a sequence $(z_n)$ in $g(F)$ converging to $y_0$.
• $z_n = g(y_n)$ for some $y_n \in F$. Since $y_n \in F \subseteq f(X)$, $y_n = f(x_n)$ for some $x_n \in X$.
• $z_n = g(f(x_n))$. So $(g \circ f)(x_n) \to y_0$.
• Consider the set $A = f^{-1}(F) = \{x \in X \mid f(x) \in F\}$. $F = f(A)$. $g(F) = g(f(A)) = (g \circ f)(A)$.
• We need to show that if $F$ is closed in $f(X)$, then $(g \circ f)(f^{-1}(F))$ is closed in $Z$.
• Let $F$ be closed in $f(X)$. Let $y \in f(X)$. $y \in \text{cl}_{f(X)}(F)$ implies $y \in F$.
• Let $z \in Z$ be a limit point of $g(F)$. There is a sequence $z_n \in g(F)$ such that $z_n \to z$.
• $z_n = g(y_n)$ for $y_n \in F$. Since $y_n \in F \subseteq f(X)$, $y_n = f(x_n)$ for $x_n \in X$.
• $z_n = (g \circ f)(x_n)$. So $(g \circ f)(x_n) \to z$.
• Consider the set $A = \{x_n \mid n \in \mathbb{N}\}$. The closure $\text{cl}_X(A)$ contains points.
• Since $g \circ f$ is a closed map, for any closed set $K \subseteq X$, $(g \circ f)(K)$ is closed in $Z$.
• Let $F$ be a closed set in $f(X)$. $F = C \cap f(X)$ for some closed $C \subseteq Y$.
• Consider the set $A = f^{-1}(F) = \{x \in X \mid f(x) \in F\}$. Since $F \subseteq f(X)$, this is well-defined.
• Let $z \in Z$ be a limit point of $g(F) = g(f(A)) = (g \circ f)(A)$. There exists a sequence $(a_n)$ in $A$ such that $(g \circ f)(a_n) \to z$.
• Let $A' = \text{cl}_X(A)$. Since $g \circ f$ is closed, $(g \circ f)(A')$ is closed in $Z$.
• Since $(g \circ f)(a_n) \in (g \circ f)(A)$ for all $n$, and $(g \circ f)(a_n) \to z$, $z$ must be in the closure of $(g \circ f)(A)$.
• If $A$ were closed in $X$, then $(g \circ f)(A)$ would be closed in $Z$, and thus $z \in (g \circ f)(A) = g(f(A)) = g(F)$, which is what we want.
• But $A = f^{-1}(F)$ is not necessarily closed in $X$. $A = f^{-1}(C \cap f(X))$. Since $f$ is continuous, $f^{-1}(C)$ is closed in $X$.
• Let $y \in F$. Then $y \in f(X)$. Consider a sequence $(x_n)$ in $X$ such that $f(x_n) \to y$. Since $f$ is continuous, $f(\text{cl}_X(\{x_n\})) \subseteq \text{cl}_Y(f(\{x_n\}))$.
• Let's use the definition of closed set in a subspace. $F$ is closed in $f(X)$ means that if a sequence $(y_n)$ in $F$ converges to a point $y \in f(X)$, then $y \in F$.
• Let $z \in Z$ be a limit point of $g(F)$. There exists a sequence $(z_n)$ in $g(F)$ such that $z_n \to z$.
• $z_n = g(y_n)$ for some $y_n \in F$. Since $y_n \in F \subseteq f(X)$, $y_n = f(x_n)$ for some $x_n \in X$.
• So $z_n = (g \circ f)(x_n)$. We have $(g \circ f)(x_n) \to z$ with $(x_n)$ in $X$.
• Let $A = \{x_n \mid n \in \mathbb{N}\}$. Let $A' = \text{cl}_X(A)$.
• Since $(g \circ f)(x_n) \to z$, $z$ is in the closure of $(g \circ f)(A)$.
• Since $g \circ f$ is a closed map, $(g \circ f)(A')$ is closed in $Z$.
• Since $(g \circ f)(A) \subseteq (g \circ f)(A')$, $\text{cl}_Z((g \circ f)(A)) \subseteq (g \circ f)(A')$.
• So $z \in (g \circ f)(A')$. This means $z = (g \circ f)(x')$ for some $x' \in A'$.
• $z = g(f(x'))$. We need to show $z \in g(F)$, which means we need to show $f(x') \in F$.
• We know $y_n = f(x_n) \in F$ for all $n$.
• Consider the sequence $(y_n) = (f(x_n))$ in $Y$.
• We have $(g(y_n)) = (z_n) \to z = g(f(x'))$. Since $g$ is continuous, if $y_n \to y$, then $g(y_n) \to g(y)$.
• However, we don't know if $(y_n)$ converges. But $(g(y_n))$ converges.
• Let's consider the closure of $F$ in $Y$. $\text{cl}_Y(F)$. Since $F$ is closed in $f(X)$, $\text{cl}_{f(X)}(F) = F$.
• Let $y \in \text{cl}_Y(F)$. There exists a sequence $(y_n)$ in $F$ such that $y_n \to y$ in $Y$.
• Since $y_n \in F \subseteq f(X)$, $y_n \in f(X)$ for all $n$. Since $f(X)$ is a metric subspace, its closure in $Y$ is $\text{cl}_Y(f(X))$. So $y \in \text{cl}_Y(f(X))$.
• Since $g$ is continuous, $g(\text{cl}_Y(F)) \subseteq \text{cl}_Z(g(F))$.
• We have $z$ is a limit point of $g(F)$, so $z \in \text{cl}_Z(g(F))$.
• Let $F$ be closed in $f(X)$. Let $y \in \text{cl}_Y(F)$. There is sequence $y_n \in F$, $y_n \to y$. $y_n \in f(X)$ for all $n$.
• $g(y_n) \in g(F)$. $g(y_n) \to g(y)$ since $g$ is continuous.
• Since $z$ is a limit point of $g(F)$, $z$ is in the closure of $g(F)$.
• Let's assume $z \notin g(F)$. Then $z \in \text{cl}_Z(g(F)) \setminus g(F)$.
• Consider the set $A = f^{-1}(F) = \{x \in X \mid f(x) \in F\}$.
• $g(F) = g(f(A)) = (g \circ f)(A)$.
• Let $z \in \text{cl}_Z((g \circ f)(A))$. We want to show $z \in (g \circ f)(A)$.
• There is a sequence $(w_n)$ in $(g \circ f)(A)$ such that $w_n \to z$. $w_n = (g \circ f)(a_n)$ for $a_n \in A$.
• Let $A' = \{a_n \mid n \in \mathbb{N}\} \subseteq A$. Let $A'' = \text{cl}_X(A')$.
• $(g \circ f)(A'')$ is closed in $Z$ because $g \circ f$ is closed.
• Since $(g \circ f)(A') \subseteq (g \circ f)(A'')$, $\text{cl}_Z((g \circ f)(A')) \subseteq (g \circ f)(A'')$.
• $z \in \text{cl}_Z((g \circ f)(A')) \subseteq (g \circ f)(A'')$. So $z = (g \circ f)(x')$ for some $x' \in A'' = \text{cl}_X(\{a_n\})$.
• We have $a_n \in A$ for all $n$, so $f(a_n) \in F$ for all $n$.
• $x' \in \text{cl}_X(\{a_n\})$. Since $f$ is continuous, $f(x') \in \text{cl}_Y(\{f(a_n)\})$.
• The sequence $(f(a_n))$ is in $F$. Since $F$ is closed in $f(X)$, any limit point of a sequence in $F$ that lies in $f(X)$ must be in $F$.
• Let $y' = f(x')$. $y' \in \text{cl}_Y(\{f(a_n)\})$.
• The sequence $(f(a_n))$ is in $F$. If $y' \in f(X)$, then since $y' \in \text{cl}_Y(\{f(a_n)\}) \subseteq \text{cl}_Y(F)$, and $F$ is closed in $f(X)$, if $y' \in f(X)$, then $y' \in F$.
• Is $f(x')$ necessarily in $f(X)$? Yes, since $x' \in X$, $f(x')$ is in the image of $f$, which is $f(X)$.
• So $f(x') \in F$.
• We have $z = g(f(x'))$. Since $f(x') \in F$, $z \in g(F)$.
• Thus, $g(F)$ is closed in $Z$. The statement is true.

Option 4: If $f: X \to Y$ takes closed balls into closed sets then $f$ is closed.

• A closed ball in $X$ centered at $x$ with radius $r > 0$ is $B_X[x, r] = \{x' \in X \mid d_X(x, x') \le r\}$. These are closed sets in $X$.
• The statement says $f(B_X[x, r])$ is closed in $Y$ for all $x \in X, r > 0$.
• We need to check if this implies that for any closed set $F$ in $X$, $f(F)$ is closed in $Y$.
• Consider $X = \mathbb{R}$ and $Y = \mathbb{R}$ with the standard metric. Let $f(x) = \arctan(x)$.
• $f: \mathbb{R} \to (-\pi/2, \pi/2)$. The range of $f$ is $(-\pi/2, \pi/2)$, which is an open set in $\mathbb{R}$.
• Any closed ball in $\mathbb{R}$ is a closed interval of the form $[a, b]$. For example, $B[0, r] = [-r, r]$.
• $f([-r, r]) = [\arctan(-r), \arctan(r)]$. Since $\arctan$ is continuous, this is a closed interval, hence closed in $\mathbb{R}$.
• So $f$ takes closed balls to closed sets.
• However, $f(\mathbb{R}) = (-\pi/2, \pi/2)$, which is not closed in $\mathbb{R}$. $\mathbb{R}$ itself is a closed set in $\mathbb{R}$.
• Since $f(\mathbb{R})$ is not closed, $f$ is not a closed map.
• Therefore, taking closed balls to closed sets does not imply the map is closed.

Based on the analysis, only Option 3 is true.