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diligence
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This was a problem on a recent graduate level introductory analysis midterm. The entire class completely bombed (class average was 18%), so we have to rewrite the exam as homework. So with those considerations, i don't want any explicit help, but feedback on the other hand would be great. I'm basically trying to gauge whether I'm in the ballpark, or if I'm in the ballpark but took the long way here, or if I'm not in the correct stadium at all.
Let X be a closed subset of R^n, let r > 0 be fixed, and let Y = { y in R^n such that |x - y| = r, for some x in X}. Prove Y is closed.
This is a tricky proof (for me at least). Proving that Y^c is open is a tall task, so we will pick a limit point of Y and prove it is contained in Y.
Let {y_n} be a Cauchy sequence in Y such that (y_n) converges to y. Then for each y_n there is at least one corresponding x_n in X such that |x_n - y_n| = r. Furthermore, for each x_n, y_n pairing, there is a corresponding r_n so that y_n = x_n + r_n, where r_n is in the set of all R^n such that |r_n| = r (note that this set is closed).
So for the sequence {y_n} we have at least one corresponding sequence {x_n}, but while {y_n} is Cauchy it is not necessarily the case that {x_n} is also Cauchy (each x_n is r-distance away from y_n but it could be in a different direction than the previous term).
However, note for any e > 0 there is a natural number N so that if m,n > N, we have:
|y_n - y_m| = |(x_n + r_n) - (x_m + r_m)| = |(x_n - x_m) - (r_m - r_n)| < e.
Thus by the reverse triangle inequality, we have:
|x_n - x_m| - |r_n - r_m| < e, hence
|x_n - x_m| < |r_n - r_m| + e <= |r_n| + |r_m| + e = 2r + e.
This implies that {x_n} is bounded. Therefore there exists a convergent subsequence {x_nk}, and since X is closed, (x_nk) --> x in X. But we also have {x_nk} = {y_nk - r_nk}, so that (y_nk - r_nk) --> x. This implies {r_nk} is convergent, else {y_nk - r_nk} would diverge. Hence (y_nk - r_nk ) --> y - r = x. Furthermore |r| = r, since the sequence {r_n} is in a closed set. Thus y - x = r, and we see that y is in Y and that Y is therefore closed.
phew...thoughts?
Homework Statement
Let X be a closed subset of R^n, let r > 0 be fixed, and let Y = { y in R^n such that |x - y| = r, for some x in X}. Prove Y is closed.
The Attempt at a Solution
This is a tricky proof (for me at least). Proving that Y^c is open is a tall task, so we will pick a limit point of Y and prove it is contained in Y.
Let {y_n} be a Cauchy sequence in Y such that (y_n) converges to y. Then for each y_n there is at least one corresponding x_n in X such that |x_n - y_n| = r. Furthermore, for each x_n, y_n pairing, there is a corresponding r_n so that y_n = x_n + r_n, where r_n is in the set of all R^n such that |r_n| = r (note that this set is closed).
So for the sequence {y_n} we have at least one corresponding sequence {x_n}, but while {y_n} is Cauchy it is not necessarily the case that {x_n} is also Cauchy (each x_n is r-distance away from y_n but it could be in a different direction than the previous term).
However, note for any e > 0 there is a natural number N so that if m,n > N, we have:
|y_n - y_m| = |(x_n + r_n) - (x_m + r_m)| = |(x_n - x_m) - (r_m - r_n)| < e.
Thus by the reverse triangle inequality, we have:
|x_n - x_m| - |r_n - r_m| < e, hence
|x_n - x_m| < |r_n - r_m| + e <= |r_n| + |r_m| + e = 2r + e.
This implies that {x_n} is bounded. Therefore there exists a convergent subsequence {x_nk}, and since X is closed, (x_nk) --> x in X. But we also have {x_nk} = {y_nk - r_nk}, so that (y_nk - r_nk) --> x. This implies {r_nk} is convergent, else {y_nk - r_nk} would diverge. Hence (y_nk - r_nk ) --> y - r = x. Furthermore |r| = r, since the sequence {r_n} is in a closed set. Thus y - x = r, and we see that y is in Y and that Y is therefore closed.
phew...thoughts?