Take Home Exam 1 Key
MA 366
1. (30 points) Let’s kick things off with a few fun True/False questions.
In each case clearly state whether the assertion is true or false. If true,
provide a proof. If false, provide a counterexample.
(a) If {xk } is a nondecreasing (xk+1 ≥ xk ) sequence that’s bounded
above then all subsequences of {xk } converge to the same limit as
{xk }.
Solution: This is true. First, since xk is nondecreasing and
bounded above the sequence must converge, say to L. That is,
for each ϵ > 0 there is some N so that |xk − L| < ϵ for all k ≥ N .
Now consider some subsequence yk = xnk ; note that n1 < n2 <
n3 < · · · by definition; in fact, clearly nk ≥ k. If nk ≥ N then
|yk − L| < ϵ. But of course if k ≥ N then nk ≥ N so that
|xnk − L| = |yk − L| < ϵ. In other words, yk (or xnk ) converges to
L.
(b) In any metric space, an arbitrary union of compact sets is compact.
Solution: Totally false. Take En = [n, n + 1] (each compact)
in the reals, but note ∪n∈Z En = R is not compact.
(c) In any metric space, a finite union of compact sets is compact.
Solution: This is true. Let Kn , 1 ≤ n ≤ N , be compact sets
in X, and let K = ∪N
n=1 Kn . Consider an open cover ∪α Gα of K.
Then the Gα cover each Kn . A finite subset of the Gα thus covers
Kn ; let An denote those indices α of the Gα that provide this finite
cover. Then the set
∪N
n=1 ∪α∈An Gα
is a finite subcover of K. This works for any open cover, and so
K is compact.
2. (30 points) Circle all that are appropriate for each set S ⊂ R below.
For anything you circle, give a brief reason why you circled it!
(a) S = [0, 5] is closed (it contains its limit points), and compact (it’s
closed and bounded). It is not open (0, 5 are not interior points).
(b) S = {1/k; k ∈ Z, k ≥ 1} is not closed (0 is a limit point, but not
in S). It’s not open either, e.g., 1 ∈ S but 1 is not an interior
point. It’s not compact, since it’s not closed.
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(c) S = {0} ∪ {1/k; k ∈ Z, k ≥ 1} is closed, for each point is isolated
(not a limit point) except for 0, which is a limit point and is
contained in S. But S is not open—same argument as last part.
Of course S is bounded, and so compact.
(d) S = R \ {4} is not closed, since 4 is a limit point not in S, but S
is open since every point in S contains a neighborhood contained
in S (each point is an interior point). Since S is not closed (nor
bounded) it is not compact.
(e) S = Z is closed, since all points are isolated. It’s not open, since
no point has a neighborhood contained in S. It’s not bounded, so
not compact.
3. (30 points) Use the formal definition for limn→∞ xn = L to show that
xn = n/(n + 7) converges.
Solution: Here’s some scratch work. Clearly xn → 1, so consider
−7
n+7
7
=
.
n+7
|xn − 1| =
Given ϵ > 0 if we choose N large enough that 7/(N + 7) < ϵ we should
be fine. A bit of algebra turns this last inequality into N > 7/ϵ − 7.
Certainly we’re on solid ground if choose N = ⌈7/ϵ⌉.
So here’s the formal proof: Given ϵ > 0 let N = ⌈7/ϵ⌉. If n ≥ N we
have
|xn − 1| =
=
≤
≤
<
=
−7
n+7
7
n+7
7
(since N ≤ n)
N +7
7
(since N ≥ 7/ϵ)
7/ϵ + 7
7
7/ϵ
ϵ.
We conclude that limn xn = 1.
4. (30 points) Let ak denote the sequence ak = sin(2k), k ∈ N.
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(a) Prove that ak contains a convergent subsequence. Cite any theorem you use.
Solution: The sequence ak is clearly contained in the compact set
[−1, 1], and so contained a convergent subsequence (since [−1, 1]
is sequentially compact).
(b) (Bonus) Show that ak itself does not converge.
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