Introduction to Analysis: Fall 2008
Practice problems IV
MTH 4101/5101 10/13/2008
1. Let K={0,1
n, n = 1,2· · ·}. Prove that Kis compact, directly from the
definition.
Solution: Let {Gα}be an open cover for K. That is K⊂ ∪αGα.Since 0 ∈K
we have 0 ∈Gα0for some α0. Also, since Gα0is an open set, 0 is an interior
point of Gα0. So, there exists a r > 0 such that Nr(0) ⊂Gα0.By Archimedean
principle, there exists a Nsuch that r > 1
N. Therefore, 1
n∈Nr(0) for all
n≥N. So,
{1
N,1
N+ 1,· · ·} ⊂ Nr(0) ⊂Gα0.
Now let, 1
n∈Gαn,for n= 1,2, ..(N−1).Clearly,
K⊂Gα0∪Gα1∪ · ·· ∪ Gα(N−1) .
We have thus produced a finite subcover for Kfor every open cover {Gα}.
Thus, Kis compact.
2. Construct a compact set of real numbers whose limit points form a countable
set.
Solution: Let S={1
n+ 1m|n, m ∈IN}.For each n∈INN, 1
n+1
m→1
nas
m→ ∞.Thus, {1
n, n ∈IN}constitutes a countable set of limit points of S.
3. Give an example of an open cover of the segment (0,1) which has no finite
subcover.
Solution: Let Gn= ( 1
n,1−1
n), n ∈IN. Clearly, (0,1) ⊂ ∪∞
n=3Gn. But there
exists no finite subcover. For, if (0,1) ⊂ ∪N
n=3Gnfor any positive integer
N > 3, then we have (0,1) ⊂(1
N,1−1
N).But, for any n>N we have
1
n∈(0,1) and 1
n/∈(1
N,1−1
N).Hence, {Gn, n = 3,4· · · N}cannot be an cover
for (0,1) for any N.
4. Regard Q, the set of rational numbers, as a metric space with d(p, q) = |p−q|.
Let Ebe the set of all p∈Qsuch that 2 < p2<3. Show that Eis closed and
bounded in Qis not compact. Is Eopen in Q?
Solution: From the definition of E, clearly E⊂ {p∈Q| − 3< p < 3}. So, it
is bounded. We show that Eis closed by showing Ecis open. Let p∈Q∩Ec.
Then, either p2<2 or p2>3. Suppose p2<2. Choose, r=(√2−p)
3. Let
U=Nr(p)∩Q. Then, U∩E= Φ.Thus, U⊂Ec∩Q. Thus, pis an interior
point of Ecproving that Ecis open. To show that Eis not compact, consider
1
