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kingwinner
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1) Prove rigorously that S={(x,y) | 1< x^2 + y^2 <4} in R^n is open using the following definition of an open set:
A set S C R^n is "open" if for all x E S, there exists some r>0 s.t. all y E R^n satisfying |y-x|<r also belongs to S.
[My attempt:
Let x E S, r1 = 2 - |x|, r2= |x| - 1, r = min {r1,r2}
Let y E R^n s.t. |y-x|<r
=>|y-x|<r1 and |y-x|<r2
|y|<=|y-x|+|x| (triangle inequality) [<= means less than or equal to]
<r1+|x|=2-|x|+|x|=2
So |y|<2
Now if I can prove that |y|>1, then I am done (y belongs to S). However, I tried many different ways, but still unable to prove that |y|>1, what should I do? ]
2) Let Li denote the line segment in R^2 from the origin (0,0) to the point (1/i, sqrt(1/i) ) on the curve f(x)=sqrt x. Prove that the infinite union S=U(i=1 to infinity) Li is compact using the Bolzano-Weierstrass Theorem.
Bolzano-Weierstrass Theorem: Let S C R^n. Then S is compact (bounded and closed) iff every sequence of points in S has a convergent subsequence whose limit lies in S.
Any help, explanation, or hints? I am feeling totally blank on this question...
My textbook basically has no examples, so I don't know how to use the theorems and I am really frustrated when doing the exercises...
I hope that someone would be kind enough to help me out! Thank you!
A set S C R^n is "open" if for all x E S, there exists some r>0 s.t. all y E R^n satisfying |y-x|<r also belongs to S.
[My attempt:
Let x E S, r1 = 2 - |x|, r2= |x| - 1, r = min {r1,r2}
Let y E R^n s.t. |y-x|<r
=>|y-x|<r1 and |y-x|<r2
|y|<=|y-x|+|x| (triangle inequality) [<= means less than or equal to]
<r1+|x|=2-|x|+|x|=2
So |y|<2
Now if I can prove that |y|>1, then I am done (y belongs to S). However, I tried many different ways, but still unable to prove that |y|>1, what should I do? ]
2) Let Li denote the line segment in R^2 from the origin (0,0) to the point (1/i, sqrt(1/i) ) on the curve f(x)=sqrt x. Prove that the infinite union S=U(i=1 to infinity) Li is compact using the Bolzano-Weierstrass Theorem.
Bolzano-Weierstrass Theorem: Let S C R^n. Then S is compact (bounded and closed) iff every sequence of points in S has a convergent subsequence whose limit lies in S.
Any help, explanation, or hints? I am feeling totally blank on this question...
My textbook basically has no examples, so I don't know how to use the theorems and I am really frustrated when doing the exercises...
I hope that someone would be kind enough to help me out! Thank you!
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