Proving Sum of Open Balls B_r(α)+B_s(β) = B_{r+s}(α+β)

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SUMMARY

The discussion focuses on proving the mathematical statement that the sum of open balls B_r(α) and B_s(β) equals the open ball B_{r+s}(α+β). The open ball B_r(α) is defined as the set of points ε such that the distance from α is less than r, specifically expressed as {ε: |α - ε| < r}. The primary challenge presented is demonstrating that B_{r+s}(α+β) is a subset of the sum B_r(α) + B_s(β), with the assumption that α and β are points in three-dimensional space.

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Homework Statement


Prove that the sum [tex]B_r(\alpha)+B_s(\beta)[/tex] is exactly the ball [tex]B_{r+s}(\alpha+\beta)[/tex]

Homework Equations


open ball [tex]B_r(\alpha)[/tex] of radius [tex]r[/tex] about the center [tex]\alpha[/tex] is [tex]\left \{\epsilon: |\alpha-\epsilon|<r\right\}[/tex]

The Attempt at a Solution


I have difficulty of proving [tex]B_{r+s}(\alpha+\beta) \subset B_r(\alpha)+B_s(\beta)[/tex]
any hints?

Thanks
 
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Are [tex]\alpha[/tex] and [tex]\beta[/tex] arbitrary complex numbers?

Edit: I read it again. I'll assume that alpha and beta are points in 3-space since you're talking about balls.
 

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