# Analysis: Closed sets and extrema

1. Oct 25, 2008

### iopmar06

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.

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

Hence for each $$I_{i}$$,

$$max(I_{i}) = b_{i}$$

and

$$min(I_{i}) = a_{i}$$.

Then consider the set

$$S = (a_{i})\cup(b_{i})$$ for i = 1,...,n.

Then $$max(s) = b_{max}$$ and $$min(S) = a_{min}$$.

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. Oct 25, 2008

### Dick

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?

3. Oct 25, 2008

### HallsofIvy

Staff Emeritus
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 $\delta$ so that the $\delta$ 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.

4. Oct 26, 2008

### iopmar06

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
5. Oct 26, 2008

### HallsofIvy

Staff Emeritus
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 $\alpha$ is an upper bound for X, then there exist a member, x, of X such that $d(x,\alpha)< 1/n$ for every integer n. Let {xn} be a member of x in $d(x,\alpha)< 1/n$.

And if you talk about a "similar" proof, you had better be ready to give that "similar" proof explicitely.