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January 2nd, 2017, 12:08 PM   #1
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Distance from a Point to Two Sets

Show that if A and B are open sets in R^n, then {x in R^n| d(x, A) < d(x, B)} is open.
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January 2nd, 2017, 02:08 PM   #2
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have you made any attempts you can show us?

one definition of an open set $S$ is

$x \in S \Rightarrow \exists \delta >0 \ni d(x,y) < \delta \Rightarrow y \in S$

Last edited by romsek; January 2nd, 2017 at 02:11 PM.
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January 3rd, 2017, 02:05 PM   #3
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I certainly have, but still am learning how to use the symbols here.

Let x ∈ U = {x ∈ R^n| d(x, A) < d(x, B)}. Must find an ε > 0 such that B(x, ε) ⊆ U.

As is often the case with these problems, the trick is to find the right ε. I have tried ε = d(x, B) - d(x, A) and ε = d(x, A) and the Triangle Inequality on both without success.

By the way, I am not a student. I am a Community College Professor who doesn't want to forget what he learned in college. Am a little rusty.
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January 3rd, 2017, 10:25 PM   #4
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Math Focus: Dynamical systems, analytic function theory, numerics
Fix $x \in U$ so $d(x,B) - d(x,A) = \eta$ is positive. Suppose $\epsilon < \frac{\eta}{3}$ and fix $y \in B_{\epsilon}(x)$. By applying the triangle inequality, we see that $d(x,B) - \epsilon$ is a lower bound on $d(y,B)$. Another application implies that $d(x,A) + \epsilon$ is an upper bound on $d(y,A)$. Therefore, $d(y,B) - d(y,A) > \eta - 2\epsilon = \frac{\eta}{3} > 0$ implying that $y \in U$. Thus, $B_{\epsilon}(x) \subset U$ proving that $U$ is open.
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January 4th, 2017, 11:47 AM   #5
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By the way, how you you type the symbols? I see no way here to do it.

It won't let me paste from MathType.
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January 4th, 2017, 12:12 PM   #6
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Quote:
Originally Posted by dpsmith View Post
By the way, how you you type the symbols? I see no way here to do it.

It won't let me paste from MathType.
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