Download Midterm solutions

Midterm solutions
You may assume the following limits without proof :
(1) If p > 0, then limn→∞ n1p = 0.
(2) If p > 0, then limn→∞ p1/n = 1.
nα
= 0.
(3) If p > 0 and α ∈ R then lim
n→∞ (1 + p)n
(4) limn→∞ n1/n = 1.
(5) If |x| < 1 then lim |x|n = 0.
n→∞
Question 1. [10 points]- Each is 1 point State whether the following are true or false. (Just write
TRUE or FALSE in your answer booklet (with the correct number of the question).)
(a) The sequence consisting of all positive integers an = n has no subsequence that converges to a finite real number.
TRUE. (The sequence converges to infinity.)
(b) lim sup(n1/n (−1)n + (2n)1/n ) = 2.
TRUE. The given sequence is less than of equal to n1/n (1 + 21/n ) whose limit by the
limit laws and the standard limits given above is 2. The even subsequence of the given
sequence converges to 2. Hence the maximum subsequential limit is 2.
(c) In any metric space (X, d), the closure of the open ball Br (p) is the closed ball
{x|d(x, p) ≤ r}.
FALSE. Take the metric space consisting of two points (which are at a distance 1 apart)
and consider the open ball of radius 1/2. Its closure is itself.
(d) If Un are any collection of connected open subsets of R then ∪∞
n=1 Un is connected.
FALSE. Let Un be a collection of disjoint of open intervals like (2n, 2n + 1).
(e) The Cantor set is connected.
FALSE. To construct the Cantor set, you remove the middle-thirds of intervals successively thus disconnecting it.
(f ) The sequence sin(an ) where an is a sequence of all rationals has no subsequence that
converges to a finite real number.
FALSE. The given sequence is bounded and hence has a convergent subsequence by
Bolzano-Weierstrass.
(g) The set of all algebraic numbers in [0, 1] is perfect.
FALSE. Algebraic numbers are countable whereas perfect sets are uncountable.
(h) If an and bn are any two real sequences such that lim sup bn 6= −∞, then
lim sup(an + bn ) = lim sup an + lim sup bn .
FALSE. Let an = (−1)n and bn = (−1)n+1 . Then an + bn = 0 whereas the lim sups
are 1 each.
(i) Every set in a metric space is either open or closed.
FALSE. (0, 1] is neither.
(j) There exists a compact set of real numbers whose limit points form a countable set.
1
∀ n, m ≥ 1 union 0.
TRUE. A problem in Rudin. Indeed take n1 + m
1
2
Question 2. [10 points] Suppose Gn are dense open subsets of Rk . Prove that ∩∞
n=1 Gn is dense.
(This is called the Baire category theorem.)
k
Ans. Let G = ∩∞
n=1 Gn . We want to show that if p ∈ R is any point, and N is any
neighbourhood of p, that N ∩ G 6= φ. (1 point)
Since G1 is dense, G1 ∩ N 6= φ and x1 ∈ G1 ∩ N . Since G1 is open, there exists a
neighbourhood Br̃1 (x1 ) ⊂ G1 ∩ N . By taking r1 < r̃1 we see that B̄r1 (x1 ) ⊂ G1 ∩ N . (2
points)
Inductively, assume that x1 , x2 , . . . , xn−1 and B̄n−1 ⊂ Bn−2 ⊂ B̄n−2 ⊂ . . . B1 have been
constructed. (Note that xk ∈ G1 ∩ G2 ∩ . . . Gk ∩ N .) Since Gn is dense, there exists xn ∈
Gn ∩Brn−1 (xn−1 ). Therefore there exists rn > 0 such that xn ∈ B̄rn (xn ) ⊂ Gn ∩Brn−1 (xn−1 ).
This proves (by induction) that the sets B̄k = B̄xk (rk ) ⊂ G1 ∩ G2 . . . Gk ∩ N are such that
any finite intersection of them is not empty. (4 points)
Since B̄k are compact (by Bolzano-Weierstrass), and any finite intersection of them is not
empty, there exists x ∈ ∩∞
k=1 B̄k . (2 points)
Since x ∈ B̄n ⊂ Gn ∩ N ∀ n, ∩Gn ∩ N 6= φ. (1 point)
Question 3. [10 points] Do the following sequences converge ? If so, find their limits.
(1) an = (n1/n − 1)n . (4 points)
By the standard limits, for any 1 > > 0, there exists an N such that n > N implies
that |n1/n − 1| < . Fix an < 1. Therefore |an | = |(n1/n − 1)n | < n ∀ n > N . (2
points)
Taking n → ∞, using the standard limits, noticing that the limit of a sequence
does not depend on the first N terms for any fixed N , and the Sandwich/squeeze
lemma/observation we see that limn→∞ |an | = 0. So yes it converges and does so to 0.
(2 points)
r
1
(2) x1 = 1 and xn+1 = x2n + n . (6 points)
2
Firstly, x2n+1 = x2n + 21n (1 point) and hence x2n+1 = 12 + 12 + 212 + . . . + 21n =
1
2n .
1
2n+1
1− 12
1−
=
2−
(2 points)
Therefore by the limit laws and standard limits, limn→∞ x2n+1 = limn→∞ x2n = 2. (2
points)
√
This is not enough to conclude that limn→∞ xn = 2. You need to prove this either
directly from definition, or notice that xn+1 ≥ xn , i.e., xn is monotonic. Since x2n
converges, it is bounded. Thus so is xn . Hence xn converges (to a positive number).
√
By limit laws x2n converges to (lim xn )2 . Thus xn converges and does so to 2. (2
points)
Question 4. [10 points] Let E be the set of all real numbers in [0, 1] having only 0, 2, and 3 in
their decimal expansion. Is E
(1) Countable ? (2 points)
Ans. No. The set of all binary sequences is uncountable by Cantor’s diagonalisation.
(Proved in the class.) (1 point)
Since E consists of all binary sequences (it has 2 and 3), it is definitely not countable.
(1 point)
(2) Dense in [0, 1] ?(2 points)
No. E is bounded above by 31 = 0.3333 . . .. Therefore 1 is not a limit point of E for
1
1
instance. Indeed, the neighbourhood (1 − 100
, 1 + 100
) does not contain any points
from E. (2 points).
3
(3) Compact ?(5 points)
E is definitely bounded. If we show that E is closed then by Heine-Borel it is compact.
(1 point)
Indeed, to show E is closed, we need to show that E c is open, i.e. if x ∈ E c then
(x − , x + ) ⊂ E c for some small > 0. (1 point)
Firstly, assume that x is represented in such a way that it does not end with 9s (i.e.
0....99999999999 . . . is not allowed). Let n be the first digit of x that is not equal to
0, 2, or 3. Choose any digit after n (let’s say the (n + m)th digit) which is not a 9 (it
1
1
exists by assumption). Now (x − 10n+m
, x + 10n+m
) ∈ E c because either
(a) The number x does not have 0 in the (n + m)th digit and thus any element
1
1
of (x − 10n+m
, x + 10n+m
) differs from x only in the (n + m)th digit or higher.
Therefore such numbers all contain the nth digit of x which is assumed to not be
0, 2, or 3.
(b) The number x does have 0 in the (n + m)th digit. This means that if x +
1
> y > x then it differs from x in only the (n + m)th digit or higher. If
10n+m
1
x − 10n+m
< y < x then either if differs from x in only the the (n + m)th digit or
higher or the (n + m)th digit is 9 which of course differs from 0, 2, or 3.
(3 points)
(4) Connected ? (1 point)
No. 0, 0.3 ∈ E and 0.1 ∈
/ E. If E was connected (0, 0.3) ⊂ E which is not the case. (1
point)