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Proving that a set, which is defined via distance to a closed set, is closed

  1. Oct 20, 2011 #1
    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.

    1. The problem statement, all variables and given/known data

    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.

    3. 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?
     
  2. jcsd
  3. Oct 21, 2011 #2
    It seems that your proof is solid enough. In proving that a set is closed, I also always opt to use the approach that any limit point of the set has to be a member to the set. I never opt for the open set-complement approach.

    You have shown that for any limit point in Y, there exists an x which is a limit point of the corresponding convergent (x_nk) subsequence by the bounded sequence property. Moreover, since X is closed, then x has to be in X.

    Moreover, you have also noted that the other corresponding sequence (r_n) converges to some r, which is necessary to be have norm r, since the sequence (r_n) belongs to the closed set {z:|z|=r}(closed ball of radius r minus open ball of radius r).

    And ultimately, you've shown that y-x=r.

    I belive your proof is correct.
     
  4. Oct 21, 2011 #3
    I think you make a good point here. While for easier situations, the strategy of proving that the complement is open is often quicker and easier, the limit point strategy seems to be more fool-proof and widely applicable. So I'm learning it's often best to start using the limit-point strategy, then only revert to the complement set if its clear to see that the method will work.

    Thanks for the feedback.
     
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