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

Step 1: Start from the equality of open balls 
We are given that the open ball centered at \((0,0)\) of radius \(\tfrac{1}{7}\) in the metric \(d_1\) is equal to the open ball centered at \((0,0)\) of radius \(\tfrac{1}{\alpha}\) in the metric \(d_2\).

This means that for any point \((x_1,x_2)\in\mathbb{R}^2\),

\[ d_1\big((x_1,x_2),(0,0)\big) < \frac{1}{7} \quad \Longleftrightarrow \quad d_2\big((x_1,x_2),(0,0)\big) < \frac{1}{\alpha}. \]

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Step 2: Express both conditions using \(d_1\)
Recall that \[ d_1\big((x_1,x_2),(0,0)\big) = |x_1| + |x_2| \] and \[ d_2 = \frac{d_1}{1+d_1}. \]

So the two conditions become:

\[ |x_1| + |x_2| < \frac{1}{7} \] and \[ \frac{|x_1| + |x_2|}{1 + |x_1| + |x_2|} < \frac{1}{\alpha}. \]

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Step 3: Convert the \(d_2\)-inequality into a \(d_1\)-inequality
Starting from \[ \frac{|x_1| + |x_2|}{1 + |x_1| + |x_2|} < \frac{1}{\alpha}, \] multiply both sides by \(1 + |x_1| + |x_2|\) (which is positive):

\[ |x_1| + |x_2| < \frac{1}{\alpha}\big(1 + |x_1| + |x_2|\big). \]

Rearranging,

\[ \left(1 - \frac{1}{\alpha}\right)(|x_1| + |x_2|) < \frac{1}{\alpha}. \]

Thus,

\[ |x_1| + |x_2| < \frac{1}{\alpha - 1}. \]

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Step 4: Match the two ball descriptions
Since the two open balls are equal, their descriptions in terms of \(|x_1|+|x_2|\) must match:

\[ \frac{1}{7} = \frac{1}{\alpha - 1}. \]

Solving,

\[ \alpha - 1 = 7 \quad \Rightarrow \quad \alpha = 8. \]

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Step 5: Conclusion
The value of \(\alpha\) for which the two open balls coincide is

\[ \boxed{\alpha = 8}. \]

Final Answer: \(\boxed{8}\)