Consider the metric spaces \( X = (\mathbb{R}, d_1) \) and \( Y = ([0, 1], d_2) \) with the metrics defined by \[ d_1(x, y) = |x - y|, \quad x, y \in \mathbb{R}, \quad {and} \quad d_2(x, y) = |x - y|, \quad x, y \in [0, 1], \] respectively. Then, which one of the following is TRUE?

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).

Answer 1

To determine which of the given statements about the set \( [0, \frac{1}{4}] \) is true, we need to understand the concepts of open and closed sets in metric spaces.

In the context of a metric space, a set is open if, for every point in the set, there is an epsilon-neighborhood entirely contained within the set.
Firstly, consider the set \( [0, \frac{1}{4}] \) in the metric space \( X = (\mathbb{R}, d_1) \).
- The metric \( d_1(x, y) = |x - y| \) is the standard Euclidean metric on the real line.
- The set \( [0, \frac{1}{4}] \) is not open in \(\mathbb{R}\) because, for instance, at \( x = 0 \), any epsilon-neighborhood \( (0 - \epsilon, 0 + \epsilon) \) will have some part that lies outside \( [0, \frac{1}{4}] \).
Now, consider the set \( [0, \frac{1}{4}] \) in the metric space \( Y = ([0, 1], d_2) \).
- The metric \( d_2(x, y) = |x - y| \) is also the standard Euclidean metric but restricted to the interval \([0, 1]\).
- In the subspace topology induced on \([0, 1]\), \( [0, \frac{1}{4}] \) can be considered an open set because it does not include any extrinsic boundary points within the interval \([0, 1]\). The intervals inside are included perfectly.

Therefore, the correct answer is: \( [0, \frac{1}{4}] \) is open in \( Y \) but not in \( X \).

Answer 2

To determine which of the given statements about the set \( [0, \frac{1}{4}] \) is true, we need to understand the concepts of open and closed sets in metric spaces.

In the context of a metric space, a set is open if, for every point in the set, there is an epsilon-neighborhood entirely contained within the set.
Firstly, consider the set \( [0, \frac{1}{4}] \) in the metric space \( X = (\mathbb{R}, d_1) \).
- The metric \( d_1(x, y) = |x - y| \) is the standard Euclidean metric on the real line.
- The set \( [0, \frac{1}{4}] \) is not open in \(\mathbb{R}\) because, for instance, at \( x = 0 \), any epsilon-neighborhood \( (0 - \epsilon, 0 + \epsilon) \) will have some part that lies outside \( [0, \frac{1}{4}] \).
Now, consider the set \( [0, \frac{1}{4}] \) in the metric space \( Y = ([0, 1], d_2) \).
- The metric \( d_2(x, y) = |x - y| \) is also the standard Euclidean metric but restricted to the interval \([0, 1]\).
- In the subspace topology induced on \([0, 1]\), \( [0, \frac{1}{4}] \) can be considered an open set because it does not include any extrinsic boundary points within the interval \([0, 1]\). The intervals inside are included perfectly.

Therefore, the correct answer is: \( [0, \frac{1}{4}] \) is open in \( Y \) but not in \( X \).

Consider the metric spaces \( X = (\mathbb{R}, d_1) \) and \( Y = ([0, 1], d_2) \) with the metrics defined by \[ d_1(x, y) = |x - y|, \quad x, y \in \mathbb{R}, \quad {and} \quad d_2(x, y) = |x - y|, \quad x, y \in [0, 1], \] respectively. Then, which one of the following is TRUE?

Show Hint

The Correct Option is B

Solution and Explanation

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