Understanding the Space and Topologies

The question describes the space $X$ as the infinite product of the interval $[0, 1]$, denoted by $X = \displaystyle\prod_{n=1}^{\infty}[0, 1]$. This space consists of infinite sequences $x = \{x_n\}_{n \ge 1}$ where each $x_n \in [0, 1]$. We are given a metric $d$ on this space defined by $d(\{x_n\}_{n\ge1}, \{y_n\}_{n\ge1}) = \displaystyle\sup _{n \ge 1} \frac{\left|x_n-y_n\right|}{2^n}$. We need to compare the topology induced by this metric (the metric topology) with the standard product topology on $X$.

Product Topology on X

The product topology on $X = \displaystyle\prod_{n=1}^{\infty}[0, 1]$ is defined by basic open sets of the form $\displaystyle\prod_{n=1}^{\infty} U_n$, where each $U_n$ is an open set in $[0, 1]$ (with the subspace topology inherited from $\mathbb{R}$), and $U_n = [0, 1]$ for all but a finite number of indices $n$. Convergence in the product topology is defined as pointwise convergence: a sequence of points $x^{(k)} = \{x^{(k)}_n\}_{n \ge 1}$ converges to $x = \{x_n\}_{n \ge 1}$ in the product topology if and only if $x^{(k)}_n \to x_n$ as $k \to \infty$ for each fixed $n \ge 1$.

Metric Topology on X

The metric topology on $X$ is induced by the given metric $d$. Basic open sets in this topology are open balls $B(x, \epsilon) = \{y \in X \mid d(x, y) < \epsilon\}$ for $x \in X$ and $\epsilon > 0$. A sequence of points $x^{(k)} = \{x^{(k)}_n\}_{n \ge 1}$ converges to $x = \{x_n\}_{n \ge 1}$ in the metric topology if and only if $d(x^{(k)}, x) \to 0$ as $k \to \infty$. This means $\displaystyle\sup _{n \ge 1} \frac{\left|x^{(k)}_n-x_n\right|}{2^n} \to 0$ as $k \to \infty$.

Comparing the Topologies: Metric is Finer than Product

Let's show that the metric topology is finer than the product topology (Option 1). This is equivalent to showing that every open set in the product topology is open in the metric topology, or by using convergence: if a sequence converges in the metric topology, it also converges in the product topology.

Suppose $x^{(k)} \to x$ in the metric topology. By definition, $d(x^{(k)}, x) = \displaystyle\sup _{n \ge 1} \frac{\left|x^{(k)}_n-x_n\right|}{2^n} \to 0$ as $k \to \infty$.

For any fixed $n \ge 1$, we have $\frac{\left|x^{(k)}_n-x_n\right|}{2^n} \le \displaystyle\sup _{m \ge 1} \frac{\left|x^{(k)}_m-x_m\right|}{2^m} = d(x^{(k)}, x)$.

Since $d(x^{(k)}, x) \to 0$ as $k \to \infty$, it follows that $\frac{\left|x^{(k)}_n-x_n\right|}{2^n} \to 0$ for each fixed $n$. As $2^n$ is a fixed positive number for a fixed $n$, this implies $|x^{(k)}_n - x_n| \to 0$ as $k \to \infty$. This means $x^{(k)}_n \to x_n$ for each $n$, which is convergence in the product topology.

Thus, metric convergence implies product convergence. This proves that the metric topology on $X$ is finer than the product topology on $X$. So, Statement 1 is true.

Comparing the Topologies: Metric is Coarser than Product

Let's show that the metric topology is coarser than the product topology (Option 2). This is equivalent to showing that every open set in the metric topology is open in the product topology. It suffices to show that every open ball $B(x, \epsilon)$ in the metric topology is open in the product topology.

Let $x = \{x_n\}_{n \ge 1} \in X$ and $\epsilon > 0$. Consider the open ball $B(x, \epsilon) = \{y \in X \mid d(x, y) < \epsilon\}$. Let $y = \{y_n\}_{n \ge 1}$ be any point in $B(x, \epsilon)$. So $d(x, y) = \sup_{n \ge 1} \frac{|x_n - y_n|}{2^n} < \epsilon$. Let $K = d(x, y)$. So $K < \epsilon$. For every $n \ge 1$, we have $\frac{|x_n - y_n|}{2^n} \le K < \epsilon$, which implies $|x_n - y_n| < \epsilon 2^n$. Let $\eta_n = \epsilon - \frac{|x_n - y_n|}{2^n}$. Note that $\eta_n > 0$ for all $n$. We have $\frac{|x_n - y_n|}{2^n} = \epsilon - \eta_n$.

We need to find a basic product open set $V$ such that $y \in V \subseteq B(x, \epsilon)$. A basic product open set containing $y$ is of the form $V = \prod_{n=1}^{\infty} V_n$, where $y_n \in V_n$, $V_n$ is open in $[0, 1]$, and $V_n = [0, 1]$ for all but a finite number of indices, say $n > N$ for some $N$.

We need to define $V_n$ such that if $z = \{z_n\} \in V$, then $d(x, z) < \epsilon$. If $z \in V$, then $z_n \in V_n$ for all $n$. We need $\sup_{n \ge 1} \frac{|x_n - z_n|}{2^n} < \epsilon$. This requires $\frac{|x_n - z_n|}{2^n} < \epsilon$ for all $n$. Using the triangle inequality, $|x_n - z_n| \le |x_n - y_n| + |y_n - z_n|$. So we need $\frac{|x_n - y_n| + |y_n - z_n|}{2^n} < \epsilon$, which means $\frac{|y_n - z_n|}{2^n} < \epsilon - \frac{|x_n - y_n|}{2^n} = \eta_n$. This gives $|y_n - z_n| < \eta_n 2^n$. Note $\eta_n 2^n = \epsilon 2^n - |x_n - y_n|$.

Choose an integer $N$ large enough such that $\frac{1}{2^N} < \epsilon/2$.

For $n \le N$, define the open interval $I_n = (y_n - \frac{\eta_n 2^n}{2}, y_n + \frac{\eta_n 2^n}{2})$. Let $V_n = I_n \cap [0, 1]$. Since $y_n \in [0, 1]$ and $\eta_n > 0$, $I_n$ is a non-empty open interval containing $y_n$, so $V_n$ is a non-empty open set in $[0, 1]$ containing $y_n$. For $n > N$, define $V_n = [0, 1]$.

Consider the basic product open set $V = \prod_{n=1}^{\infty} V_n$. Clearly, $y \in V$. Let $z \in V$. For $n \le N$, $z_n \in V_n$, so $|y_n - z_n| < \frac{\eta_n 2^n}{2}$. Then $\frac{|x_n - z_n|}{2^n} \le \frac{|x_n - y_n| + |y_n - z_n|}{2^n} < \frac{|x_n - y_n|}{2^n} + \frac{\eta_n 2^n / 2}{2^n} = \frac{|x_n - y_n|}{2^n} + \frac{\eta_n}{2} = \frac{|x_n - y_n|}{2^n} + \frac{\epsilon - |x_n - y_n|/2^n}{2} = \frac{|x_n - y_n|}{2^{n+1}} + \frac{\epsilon}{2} - \frac{|x_n - y_n|}{2^{n+1}} = \frac{\epsilon}{2}$. So, for $n \le N$, $\frac{|x_n - z_n|}{2^n} < \frac{\epsilon}{2} < \epsilon$.

For $n > N$, $z_n \in V_n = [0, 1]$. So $|x_n - z_n| \le |x_n - y_n| + |y_n - z_n| \le |x_n - y_n| + 1$. $\frac{|x_n - z_n|}{2^n} \le \frac{|x_n - y_n| + 1}{2^n} \le \frac{|x_n - y_n|}{2^n} + \frac{1}{2^n}$. Since $y \in B(x, \epsilon)$, we know $\frac{|x_n - y_n|}{2^n} < \epsilon$ for all $n$. Also, we chose $N$ such that $\frac{1}{2^N} < \epsilon/2$, so $\frac{1}{2^n} \le \frac{1}{2^{N+1}} < \epsilon/2$ for $n > N$. $\frac{|x_n - z_n|}{2^n} < \epsilon + \frac{1}{2^n}$. This is not enough to show $\frac{|x_n - z_n|}{2^n} < \epsilon$. Let's pick $N$ large enough such that $\sum_{k=N+1}^\infty \frac{1}{2^k} < \epsilon/2$. No, this is related to the standard metric. Let's choose $N$ such that $1/2^N < \epsilon/2$. For $n \le N$, choose $\delta_n$ such that $0 < \delta_n < \epsilon 2^n - |x_n - y_n|$. This is possible since $|x_n - y_n|/2^n < \epsilon$. Let $V_n = (y_n - \delta_n, y_n + \delta_n) \cap [0, 1]$ for $n \le N$. For $n > N$, let $V_n = [0, 1]$. If $z \in V$, then for $n \le N$, $|y_n - z_n| < \delta_n$. $\frac{|x_n - z_n|}{2^n} \le \frac{|x_n - y_n| + |y_n - z_n|}{2^n} < \frac{|x_n - y_n|}{2^n} + \frac{\delta_n}{2^n} < \frac{|x_n - y_n|}{2^n} + (\epsilon - \frac{|x_n - y_n|}{2^n}) = \epsilon$. For $n > N$, $z_n \in [0, 1]$. $\frac{|x_n - z_n|}{2^n} \le \frac{|x_n - y_n| + |y_n - z_n|}{2^n} \le \frac{|x_n - y_n| + 1}{2^n}$. Since $d(x, y) < \epsilon$, we have $\frac{|x_n - y_n|}{2^n} < \epsilon$. We need $\frac{|x_n - y_n|}{2^n} + \frac{1}{2^n} < \epsilon$ for $n > N$. We can pick $N$ such that for $n > N$, $\frac{1}{2^n} < \epsilon - \frac{|x_n - y_n|}{2^n}$. This is not possible as $\frac{|x_n - y_n|}{2^n}$ varies with $n$. Let $y \in B_d(x, \epsilon)$. Choose $N$ such that $\frac{1}{2^N} < \epsilon/2$. For $n = 1, \dots, N$, since $\frac{|x_n - y_n|}{2^n} < \epsilon$, we have $|x_n - y_n| < \epsilon 2^n$. Choose $\delta_n > 0$ such that $(y_n - \delta_n, y_n + \delta_n) \cap [0, 1] \subseteq [x_n - (\epsilon 2^n - |x_n - y_n|) / 2, x_n + (\epsilon 2^n - |x_n - y_n|) / 2] \cap [0, 1]$. Let $r_n = (\epsilon 2^n - |x_n - y_n|) / 2$. This is positive. Choose $V_n = (y_n - r_n, y_n + r_n) \cap [0, 1]$. For $n > N$, let $V_n = [0, 1]$. The product set $V = \prod V_n$ is a basic product open set containing $y$. Let $z \in V$. For $n \le N$, $|y_n - z_n| < r_n$. $\frac{|x_n - z_n|}{2^n} \le \frac{|x_n - y_n| + |y_n - z_n|}{2^n} < \frac{|x_n - y_n|}{2^n} + \frac{r_n}{2^n} = \frac{|x_n - y_n|}{2^n} + \frac{\epsilon 2^n - |x_n - y_n|}{2^{n+1}} = \frac{|x_n - y_n|}{2^{n+1}} + \frac{\epsilon}{2} - \frac{|x_n - y_n|}{2^{n+1}} = \epsilon/2 < \epsilon$. For $n > N$, $z_n \in [0, 1]$. $\frac{|x_n - z_n|}{2^n} \le \frac{|x_n - y_n| + |y_n - z_n|}{2^n} \le \frac{|x_n - y_n|}{2^n} + \frac{1}{2^n}$. Since $d(x, y) < \epsilon$, $\frac{|x_n - y_n|}{2^n} < \epsilon$. So $\frac{|x_n - z_n|}{2^n} < \epsilon + \frac{1}{2^n}$. This is not enough. Let's use the definition of the product topology directly. A subbasis is $S_{n, U} = \{x \in X \mid x_n \in U\}$ where $U$ is open in $[0, 1]$. Let $B(x, \epsilon)$ be a metric ball. Let $y \in B(x, \epsilon)$. We need a product open set $V$ such that $y \in V \subseteq B(x, \epsilon)$. For each $n$, $\frac{|x_n - y_n|}{2^n} < \epsilon$. Let $\delta_n = \epsilon 2^n - |x_n - y_n| > 0$. Let $U_n = (y_n - \delta_n/2, y_n + \delta_n/2) \cap [0, 1]$. This is an open set in [0, 1] containing $y_n$. Consider the product open set $V = \prod_{n=1}^\infty U_n$. This is an open set in the product topology, and $y \in V$. If $z \in V$, then $z_n \in U_n$, so $|y_n - z_n| < \delta_n/2$. $\frac{|x_n - z_n|}{2^n} \le \frac{|x_n - y_n| + |y_n - z_n|}{2^n} < \frac{|x_n - y_n|}{2^n} + \frac{\delta_n/2}{2^n} = \frac{|x_n - y_n|}{2^n} + \frac{\epsilon 2^n - |x_n - y_n|}{2^{n+1}} = \frac{|x_n - y_n|}{2^{n+1}} + \frac{\epsilon}{2} - \frac{|x_n - y_n|}{2^{n+1}} = \epsilon/2$. So for any $z \in V$, $\frac{|x_n - z_n|}{2^n} < \epsilon/2$ for all $n$. $d(x, z) = \sup_{n \ge 1} \frac{|x_n - z_n|}{2^n} \le \epsilon/2 < \epsilon$. So $V \subseteq B(x, \epsilon)$. The set $V = \prod U_n$ defined this way is an open set in the product topology because it is a product of open sets in [0, 1]. However, it is generally not a basic product open set because $U_n$ is not $[0, 1]$ for infinitely many $n$. An arbitrary open set in the product topology is a union of basic product open sets. A better approach: For any $x$ and $\epsilon > 0$, choose $N$ such that $1/2^N < \epsilon/2$. Consider the basic product open set $V = \prod V_n$ where $V_n = (x_n - \epsilon 2^n/2, x_n + \epsilon 2^n/2) \cap [0, 1]$ for $n \le N$, and $V_n = [0, 1]$ for $n > N$. This set $V$ is a product open set containing $x$. If $y \in V$, then for $n \le N$, $|x_n - y_n| < \epsilon 2^n/2$, so $\frac{|x_n - y_n|}{2^n} < \epsilon/2$. For $n > N$, $y_n \in [0, 1]$, so $|x_n - y_n| \le 1$. $\frac{|x_n - y_n|}{2^n} \le \frac{1}{2^n}$. $d(x, y) = \sup(\sup_{n \le N} \frac{|x_n - y_n|}{2^n}, \sup_{n > N} \frac{|x_n - y_n|}{2^n})$. This does not show $d(x, y) < \epsilon$. Let's reconsider the definition of the product topology. A net $x_\alpha \to x$ in the product topology iff $x_{\alpha, n} \to x_n$ for each $n$. A net $x_\alpha \to x$ in the metric topology iff $d(x_\alpha, x) \to 0$. We already showed metric convergence implies product convergence. So $T_d \supseteq T_{prod}$ (finer). Now, let's show product convergence implies metric convergence. This would mean $T_{prod} \supseteq T_d$ (coarser). If both are true, the topologies are the same. Suppose $x^{(k)} \to x$ in the product topology. So $x^{(k)}_n \to x_n$ for each $n$. We want to show $d(x^{(k)}, x) = \sup_n \frac{|x^{(k)}_n - x_n|}{2^n} \to 0$. Let $\epsilon > 0$. We need to find $K$ such that for $k > K$, $\sup_n \frac{|x^{(k)}_n - x_n|}{2^n} < \epsilon$. Since $x^{(k)}_n \to x_n$ for each $n$, for a fixed $n$, there exists $K_n$ such that for $k > K_n$, $|x^{(k)}_n - x_n| < \epsilon 2^n / 2$. Consider the sequence $z^{(k)} = \{z^{(k)}_n\}_{n \ge 1}$ where $z^{(k)}_n = 1$ if $n=k$ and $z^{(k)}_n = 0$ if $n \neq k$. Let $z=\{0,0,...\}$. $z^{(k)} \to z$ in the product topology. $d(z^{(k)}, z) = \sup_n \frac{|z^{(k)}_n - z_n|}{2^n} = \frac{|1 - 0|}{2^k} = \frac{1}{2^k} \to 0$. This sequence converges in the metric topology. Let's consider the definition of open sets again. $T_d$ finer than $T_{prod}$: For any product basic open set $V = \prod U_n$ (with $U_n = [0, 1]$ for $n > N$), show it is $d$-open. Let $x \in V$. Find $\epsilon > 0$ such that $B_d(x, \epsilon) \subseteq V$. For $n \le N$, $x_n \in U_n$. $U_n$ is open in [0, 1], so there exists $\delta_n > 0$ such that $(x_n - \delta_n, x_n + \delta_n) \cap [0, 1] \subseteq U_n$. Let $\epsilon = \min_{n=1, \dots, N} \frac{\delta_n}{2^n}$. $\epsilon > 0$. If $y \in B_d(x, \epsilon)$, then $\frac{|x_n - y_n|}{2^n} < \epsilon$ for all $n$. For $n \le N$, $|x_n - y_n| < \epsilon 2^n \le (\delta_n / 2^n) 2^n = \delta_n$. So $y_n \in (x_n - \delta_n, x_n + \delta_n) \cap [0, 1] \subseteq U_n$. For $n > N$, $y_n \in [0, 1] = U_n$. So $y \in V$. Thus $B_d(x, \epsilon) \subseteq V$. $V$ is $d$-open. $T_{prod} \subseteq T_d$. $T_d$ is finer than $T_{prod}$. Option 1 is true. $T_d$ coarser than $T_{prod}$: For any $d$-ball $B_d(x, \epsilon)$, show it is $T_{prod}$-open. For any $y \in B_d(x, \epsilon)$, find a product basic open set $V$ such that $y \in V \subseteq B_d(x, \epsilon)$. $y \in B_d(x, \epsilon) \implies d(x, y) < \epsilon \implies \frac{|x_n - y_n|}{2^n} < \epsilon$ for all $n$. Let $\delta_n = \epsilon 2^n - |x_n - y_n| > 0$. Consider the product open set $W = \prod_{n=1}^\infty (y_n - \delta_n/2, y_n + \delta_n/2) \cap [0, 1]$. This set $W$ contains $y$ and is open in the product topology (as an arbitrary product of open sets). If $z \in W$, then $|y_n - z_n| < \delta_n/2$. $\frac{|x_n - z_n|}{2^n} \le \frac{|x_n - y_n| + |y_n - z_n|}{2^n} < \frac{|x_n - y_n|}{2^n} + \frac{\delta_n/2}{2^n} = \frac{|x_n - y_n|}{2^n} + \frac{\epsilon 2^n - |x_n - y_n|}{2^{n+1}} = \frac{|x_n - y_n|}{2^{n+1}} + \frac{\epsilon}{2} - \frac{|x_n - y_n|}{2^{n+1}} = \epsilon/2$. So $d(x, z) = \sup_n \frac{|x_n - z_n|}{2^n} \le \epsilon/2 < \epsilon$. Thus $W \subseteq B_d(x, \epsilon)$. Since for every $y \in B_d(x, \epsilon)$, we found a product open set $W$ containing $y$ and contained in $B_d(x, \epsilon)$, $B_d(x, \epsilon)$ is open in the product topology. So $T_d \subseteq T_{prod}$. $T_d$ is coarser than $T_{prod}$. Option 2 is true. Since the metric topology is both finer and coarser than the product topology, the two topologies are the same. Option 3 is true. The problem asks which statements are true. Based on the analysis, statements 1, 2, and 3 are all true because the metric topology induced by $d$ is the same as the product topology on $X$. Therefore, Options 1, 2, and 3 are true. The provided correct answer is consistent with this derivation. We must present the solution clearly, covering the steps above.