1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Analysis: Closed sets and extrema

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data
    a) Let U be a closed subset of the reals with an upper bound. You know that U has a supremum, say z. Prove that z is an element of U.

    b) Suppose U is a closed subset of the real numbers with an upper and lower bound. Prove that U has a maximum and minimum.

    3. The attempt at a solution

    I was able to come up with a proof for b) by simply finding the maximum and minimum of U however I feel like the proof is flawed.

    b) Since U is a closed and bounded subset of R, we can write U as a finite collection of closed intervals of R.

    [tex] U = \cup^{n}_{1} I_{i}[/tex] where [tex] I_{i} = [a_{i},b_{i}]. [/tex]

    Hence for each [tex]I_{i}[/tex],

    [tex]max(I_{i}) = b_{i}[/tex]


    [tex]min(I_{i}) = a_{i}[/tex].

    Then consider the set

    [tex]S = (a_{i})\cup(b_{i}) [/tex] for i = 1,...,n.

    Then [tex]max(s) = b_{max} [/tex] and [tex] min(S) = a_{min}[/tex].

    So max(U) and min(U) both exist and are equal to bmax and amax respectively.

    a) I wasn't sure how to do this one. I tried using some inequalities and defining the maximum of U but couldn't come up with anything useful.

    Any help is greatly appreciated.
  2. jcsd
  3. Oct 25, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    b) is deeply flawed. You CANNOT write every closed and bounded subset of R as a finite union of closed intervals, take e.g. {1/n for n in N} union {0}. Go back to the first one. You have a supremum z. Suppose z is NOT in U. Can you show every neighborhood of z contains an element of U? Can you use that to construct a sequence in U converging to z? Can you see where to go from here?
  4. Oct 25, 2008 #3


    User Avatar
    Science Advisor

    Or: for (a), since U is closed, its complement is open. If is in complement of U, it is an interior point, there exist some [itex]\delta[/itex] so that the [itex]\delta[/itex] neighborhood of z is in U, ....

    (b) If a set of real numbers has an upper bound, then it has a sup. What does (a) tell you about that sup? If a set of real numbers has a lower bound, then it has an inf. You should be able to prove a theorem similar to (a) for infimums.
  5. Oct 26, 2008 #4
    Thanks for your replies. I can see where I should be going with a) but I'm still working on it.

    a) Consider the metric space (R,d) where d is the euclidean metric d(x,y)=|x-y|. Note that if (X,d) is a metric space then a subset Y of X is called closed if for every sequence (y(n)) of elements of Y, if (y(n)) has a limit L in X, then L is in Y. U is a closed and bounded above subset of R with supremum z. Then if there is a sequence (u(n)) of elements of U that converges to z, then z is in U.

    Now all I need to do is find such a sequence, any ideas?

    For now assume a) is true.

    b) Let U be a closed and bounded subset of R. Since U is bounded there are real numbers
    x=inf(U) and y=sup(U). It follows from a) that y is in U and via a similar proof that x is also in U. Since x,y are inf,sup respectively, x<=u<=y for all u in U. Thus min(U)=x and max(U) = y.
    Last edited: Oct 26, 2008
  6. Oct 26, 2008 #5


    User Avatar
    Science Advisor

    That's not the proof I would give because I prefer to use the equivalent definition of "closed": A set is closed if and only if its complement is open. However, it certainly can be used:
    If [itex]\alpha[/itex] is an upper bound for X, then there exist a member, x, of X such that [itex]d(x,\alpha)< 1/n[/itex] for every integer n. Let {xn} be a member of x in [itex]d(x,\alpha)< 1/n[/itex].

    And if you talk about a "similar" proof, you had better be ready to give that "similar" proof explicitely.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook