January 2nd, 2017, 11:08 AM  #1 
Newbie Joined: Mar 2014 Posts: 11 Thanks: 2  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.

January 2nd, 2017, 01:08 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,488 Thanks: 749 
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 01:11 PM. 
January 3rd, 2017, 01:05 PM  #3 
Newbie Joined: Mar 2014 Posts: 11 Thanks: 2 
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. 
January 3rd, 2017, 09:25 PM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 165 Thanks: 74 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.

January 4th, 2017, 10:47 AM  #5 
Newbie Joined: Mar 2014 Posts: 11 Thanks: 2 
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. 
January 4th, 2017, 11:12 AM  #6  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,488 Thanks: 749  Quote:
 

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