Test 1 Proofs Flashcards
Nested Interval Property
In order to show that the infinite intersection is not empty, we are going to use the Axiom of Completeness (AoC) to produce a single real number x satisfying x ∈ In for every n ∈ N. Now, AoC is a statement about bounded sets, and theone we want to consider is the set of left-hand endpoints of the intervals:
A = {an : n ∈ N}
Because the intervals are nested, we see that every bn serves as an upper bound for A. Thus, we are justified in setting
x = sup A
Now, consider a particular In = [an, bn]. Because x is an upper bound for A, we have an ≤ x. The fact that each bn is an upper bound for A and that x is the least upper bound implies x ≤ bn
Altogether then, we have an ≤ x ≤ bn, which means x ∈ In for every choice of n ∈ N. Hence, x ∈ the infinite intersection and the intersection is not empty.
Monotone Convergence Theorem
Let (an) be monotone and bounded. To prove (an) converges using the definition of convergence, we are going to need a candidate for the limit. Let’s assume the sequence is increasing (the decreasing case is handled similarly), and consider the set of points {an : n ∈ N}. By assumption, this set is bounded, so
we can let
s = sup{an : n ∈ N}.
It seems reasonable to claim that lim an = s
To prove this, let ε > 0. Because s is the least upper bound for {an: n ∈ N}, s − ε is not an upper bound, so there exists a point in the sequence aN such that s − ε < aN .Now, the fact that (an) is increasing implies that if n ≥ N, then aN ≤ an. Hence,
s − ε < aN ≤ an ≤ s < s + ε
which implies |an − s| < ε, as desired
Bolzano-Weierstrass Theorem
Let (an) be a bounded sequence so that there exists M > 0 satisfying|an| ≤ M for all n ∈ N. Bisect the closed interval [−M,M] into the two closed intervals [−M, 0] and [0, M]. (The midpoint is included in both halves.) Now, it must be that at least one of these closed intervals contains an infinite number of the terms in the sequence (an). Select a half for which this is the case and label that interval as I1. Then, let an1 be some term in the sequence (an) satisfying an1 ∈ I1.
Next, we bisect I1 into closed intervals of equal length, and let I2 be a half that again contains an infinite number of terms of the original sequence. Because there are an infinite number of terms from (an) to choose from, we can select an an2 from the original sequence with n2 > n1 and an2 ∈ I2. In general, we construct the closed interval Ik by taking a half of Ik−1 containing an infinite
number of terms of (an) and then select nk > nk−1 > ··· > n2 > n1 so that ank ∈ Ik.
We want to argue that (ank ) is a convergent subsequence, but we need a candidate for the limit. The sets I1 ⊇ I2 ⊇ I3 ⊇···
form a nested sequence of closed intervals, and by the Nested Interval Property there exists at least one point x ∈ R contained in every Ik. This provides us with the candidate we were looking for. It just remains to show that (ank) → x
Let ε > 0. By construction, the length of Ik is M(1/2)k−1 which converges to zero. (This follows from Example 2.5.3 and the Algebraic Limit Theorem.) Choose N so that k ≥ N implies that the length of Ik is less than ε. Because x and ank are both in Ik, it follows that |ank − x| < ε
Let A subset of R be a set and suppose inf A = a exists. Prove there is a sequence (an) where an ∈ A for all n, and lim an = a. Hint: Specifying values of ε equal to 1/n for integers n may help.
Let s = inf A
There is an an ∈ A with an < s + 1/n
Thus s - 1/n < s <= an < s + 1/n
Thus |an - s | < 1/n
Let ε > 0 be given
choose N so that 1/N < ε.
Then if n > N, |an - s| < 1/n < 1/N < ε
Let A subset of R be a bounded nonempty set and let c > 0. Define the set cA = {ca : a ∈ A}.
Prove that inf cA = c inf A
Let s = inf A
thus s <= a implying cs <= ca, so cs is a lower bound (which is one half of the lema)
Then there is an a ∈ A with a < s + ε/c
Thus ca < cs + ε
Thus cs <= ca < cs + ε
Thus cs is the greatest lower bound
There is an infinite collection of sets Ai (subset of R) where each Ai is
uncountably infinite but for which the infinite intersection of Ai is empty.
True
if B is an uncountable set with B ∈ (0, 1), then for some n ∈ N, B ∈ [1/(n + 1), 1/n] must be uncountable.
True
Let (an) be a sequence and a ∈ R. If every epsilon neighborhood of a contains infinitely many terms of the sequence, then (an) converges to a
False
Let Ai = (ai , bi) be an open interval in R with the property that Ai+1 is a proper subset of Ai
Then the infinite intersection of Ai is the empty set
False
If (an) is unbounded, then (an) does not have a converging subsequence
False
Suppose B is a subset of R, and A is a subset of B. If A is bounded above, there is a least upper bound of A in B.
False
Cantor’s Diagonalization Method is used to show that the rationals are countable.
False
If (an) is unbounded, then (an) does not have a converging subsequence.
False
If lim(anbn) exists and lim bn = b =/= 0, then lim an exists
True
if (an) is monotonic and bounded, then it is Cauchy
True
