Proving a metric is continuous

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SUMMARY

The discussion focuses on proving the continuity of a metric \( d: X \times X \rightarrow \mathbb{R} \). The approach involves defining neighborhoods \( U \) around points \( x \) and \( y \) in \( X \times X \) and demonstrating that the image of these neighborhoods under \( d \) is contained within a specified interval \( (a, b) \). The proof utilizes open balls \( B_1(x, r_1) \) and \( B_2(y, r_2) \) to establish that the diameter of their union adheres to the condition \( 2(r_1 + r_2) = b - a \), confirming the continuity of the metric.

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Homework Statement



So, given a metric d : X x X --> R, prove that d is continuous.

The Attempt at a Solution



Let (x, y) be a point in X x X, V = <a, b> a neighborhood of d(x, y). One needs to find a neighborhood of U of (x, y) such that d(U) is contained in V. U is of the form U1 x U2, where U1 is a neighborhood of x, and U2 a neighborhood of y. I claim that every union U of two open balls B1(x, r1) and B2(y, r2), where 2(r1 + r2) = b - a, must satisfy d(U) \subseteq <a, b>.

The diameter of B1 is 2r1, and the diameter of B2 is 2r2. The diameter of their union is b = a + 2r1 + 2r2, where a is the distance between B1 and B2.

Does this work?
 

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