Question

Let \( (\mathbb{R}^2, d_1) \) and \( (\mathbb{R}^2, d_2) \) be two metric spaces with \[ d_1\left( (x_1, x_2), (y_1, y_2) \right) = |x_1 - y_1| + |x_2 - y_2| \] \[ {and} \quad d_2\left( (x_1, x_2), (y_1, y_2) \right) = \frac{d_1\left( (x_1, x_2), (y_1, y_2) \right)}{1 + d_1\left( (x_1, x_2), (y_1, y_2) \right)}. \] If the open ball centered at \( (0,0) \) with radius \( \frac{1}{7} \) in \( (\mathbb{R}^2, d_1) \) is equal to the open ball centered at \( (0,0) \) with radius \( \frac{1}{\alpha} \) in \( (\mathbb{R}^2, d_2) \), then the value of \( \alpha \) is (in integer).

Hint:

In metric spaces, the relationship between the radii of open balls can be used to compute the corresponding values in different metrics. Here, we used the formula for \( d_2 \) to find the relation between the radii.

Answer 1

Correct Answer: 8

Step 1: Understanding the Metrics The metric \( d_1 \) is the Manhattan distance (also known as the taxicab distance) on \( \mathbb{R}^2 \), which is defined as: \[ d_1\left( (x_1, x_2), (y_1, y_2) \right) = |x_1 - y_1| + |x_2 - y_2|. \] The metric \( d_2 \) is defined as: \[ d_2\left( (x_1, x_2), (y_1, y_2) \right) = \frac{d_1\left( (x_1, x_2), (y_1, y_2) \right)}{1 + d_1\left( (x_1, x_2), (y_1, y_2) \right)}. \] Step 2: The Open Balls An open ball in \( (\mathbb{R}^2, d_1) \) centered at \( (0, 0) \) with radius \( r \) is given by: \[ B_1(0, r) = \left\{ (x_1, x_2) \in \mathbb{R}^2 : |x_1| + |x_2|<r \right\}. \] Similarly, an open ball in \( (\mathbb{R}^2, d_2) \) centered at \( (0, 0) \) with radius \( r \) is given by: \[ B_2(0, r) = \left\{ (x_1, x_2) \in \mathbb{R}^2 : \frac{|x_1| + |x_2|}{1 + |x_1| + |x_2|}<r \right\}. \] Step 3: Relating the Radii of the Open Balls We are given that the open ball with radius \( \frac{1}{7} \) in \( (\mathbb{R}^2, d_1) \) is equal to the open ball with radius \( \frac{1}{\alpha} \) in \( (\mathbb{R}^2, d_2) \). From the condition that these balls are equal, we equate their radii in terms of \( d_1 \) and \( d_2 \). Thus, the relationship between \( r_1 \) and \( r_2 \) can be established as: \[ r_2 = \frac{r_1}{1 + r_1}. \] Step 4: Solving for \( \alpha \) Substituting \( r_1 = \frac{1}{7} \) and solving for \( r_2 \), we get: \[ r_2 = \frac{\frac{1}{7}}{1 + \frac{1}{7}} = \frac{\frac{1}{7}}{\frac{8}{7}} = \frac{1}{8}. \] Now, since \( r_2 = \frac{1}{\alpha} \), we have \( \frac{1}{\alpha} = \frac{1}{8} \), so \( \alpha = 8 \). Step 5: Conclusion Thus, the value of \( \alpha \) is \( \boxed{8} \). \[ \boxed{8} \quad \alpha = 8 \]