Exam 1 Solutions

Question 1.  (a)  Suppose that S is a set of real numbers that is bounded from below, and let x = \inf S.  (By the way, we know that this infimum exists since the set

\displaystyle -S := \{ -s : s \in S \}

can be shown to be bounded above and so has a supremum.  It is not hard to argue that \sup \left(-S\right) = -\inf S and so \inf S exists.)

Proof.  If x \in S, then we are done.  Therefore, suppose x \notin S.  To show that x = \inf S is an accumulation point, let \varepsilon > 0 be given.  We will argue that

\displaystyle \left(x-\varepsilon, x+\varepsilon\right) \bigcap S

contains at least one element that is different from s.  By the Lemma on page 35. (and the one discussed in class), this is equivalent to the original definition of accumulation point.

Since x = \inf S, it follows that x- \varepsilon < x < s for every s \in S.  By definition of infimum, the real number x + \varepsilon must fail to be a lower bound for S.  This means that there must exist an element y \in S that satisfies y < x + \varepsilon.  We therefore have

\displaystyle y \in \left(x-\varepsilon, x+\varepsilon\right) \bigcap S.

Since y \in S and, by assumption, x \notin S, it follows that y \neq x.  Since such an element y \in S exists for every \varepsilon > 0, we have that x = \inf S is an accumulation point.  \square

(b) There are lots of way to answer this question, but first let’s consider an incorrect answer.  The set T = (a, b) where a, b \in \mathbb{R} with a < b is an open interval.  The infimum of this set, \inf T = a \notin T and a is an accumulation point.

One may think to adjust this example by instead defining the interval T' = [a, b) so that a = \inf T '\in T', however this example still does not work since a is still an accumulation point for T'.

We need to construct a set that (1) has an infimum, (2) contains its infimum, and for which (3) the infimum is not an accumulation point — that is, we want S to not accumulate around \inf S, so we would like this point to be isolated.  Here is one way to take our attempt from before and “fix” it:  define the set S to be

\displaystyle S = \{-3\} \bigcup (-1, 1).

Then, clearly, \inf S = -3 \in S but there are \varepsilon-neighborhoods of the point -3 that do not intersect S (except in the point -3 itself).

Question 2.  There are many ways to prove that this sequence converges.  Here is a proof using the definition of convergence.

Claim: The sequence a_n = (5n)/(3n+2) converges to A = 5/3.

Proof.  Let \varepsilon > 0 be given.

Choose N \displaystyle > \frac{10}{9\varepsilon}-\frac{6}{9} and suppose n \geq N.  Then

\displaystyle d\left(a_n, A\right) = \left|\frac{5n}{3n+2} - \frac{5}{3}\right| = \left| \frac{-10}{9n+6} \right| = \frac{10}{9n+6} < \varepsilon

where the final inequality is a direct result of our choice for n \geq N.  \square

Question 3.  (a) Claim.  The set X = \mathbb{R} is bounded under the given (discrete) metric.

Proof.  To verify our claim we need to find a point x_0 \in X and a non-negative real number R \geq 0 so that the set containment

\displaystyle \mathbb{R} \subseteq B_R\left(x_0\right)

is true.  As it turns out, there are infinitely many choices for x_0 and infinitely many choices for R \geq 0.  In particular, for every x_0 \in \mathbb{R} and for every R \in (1, \infty), the ball B_R(x_0) contains every real number.  To prove this stronger statement, let x_0 \in \mathbb{R} and R \in (1, \infty) be arbitrary.

Let x \in \mathbb{R} be arbitrary.  To show that x \in B_R(x_0) we need to show that d(x, x_0) < R.  However, because R > 1 and d(x,x_0) = 0 \text{ or } d(x, x_0) = 1, it follows that d(x, x_0) < R.  Hence \mathbb{R} \subseteq B_R(x_0).

As a particular example, we have that \mathbb{R} = B_2(\pi).  \square.

(b)  Claim.  If \{a_n\} is a convergent sequence of points in the metric space (\mathbb{R}, d), then there exists N \in \mathbb{N} so that whenever n \geq N it follows that a_n = a.  (In other words: a convergent sequence is eventually constant.)

Proof.  Suppose \{a_n\} converges to a \in \mathbb{R} in the given metric space.  By definition of convergence this means that for every \varepsilon > 0 there exists N \in \mathbb{N} so that

\displaystyle d\left(a_n, a\right) < \varepsilon \text{ whenever } n \geq N.

Since this happens for every positive \varepsilon, it must hold when \varepsilon = 1/2.  As in our statement above, let N denote the index beyond which we have

\displaystyle d\left(a_n, a\right) < \frac{1}{2}.

Because this metric can only return values of 0 or 1, it follows that whenever n \geq N

\displaystyle d\left(a_n, a\right) = 0.

Because this is a metric space, the above line implies that whenever n \geq N, a_n = a, as desired.  \square

(c ) The Bolzano-Weierstrass Theorem is false in this metric space.  As a concrete counter-example, consider the set

\displaystyle S = \left\{1, \frac{1}{2}, \frac{1}{3}, \cdots \right\}.

Because S \subset \mathbb{R} and \mathbb{R} is bounded, the set S is bounded, too.  The set S is infinite since there are infinitely many distinct real numbers of the form 1/n.

However, S does not have any accumulation points.  Under our usual metric for \mathbb{R}, the number 0 would be an accumulation point, but under this discrete metric all of the points of S remain of distance 1 away from 0.  Indeed, this is true for all real numbers x \notin S.

Indeed, one can prove a much more striking fact: in this metric space, no set S has any accumulation points.  The idea underlying this fact can be stated in a rather simple, English sentence: infinitely many points of any set S will fail to accumulate anywhere since they are all distance one away from every other point!

Question 4.  Oh, boy.  I am skipping this one as I would babble endlessly about failed ideas and how much they (eventually) taught me.  Trust me, I’m doing you a favor.

Question 5. Before we address either part of this problem, we establish the following helpful lemma.  It will be used in both parts (a) and (b).

Lemma 1.  If S \subseteq T \subseteq \mathbb{R} are subsets of real numbers that each have a supremum, then \sup S \leq \sup T.  (An analogous statement is true for infimuma, but with a reversed inequality.)

proof.  Suppose S \subseteq T are subsets of real numbers, each with suprema.  To prove that \sup S \leq \sup T, we need only argue that \sup T is also an upper bound for the set S.  Since, by definition, \sup S is the least upper bound for S, the inequality will have to follow.

Of course, that \sup T is an upper bound for S is instant.  After all, the \sup T is an upper bound for T and so \sup T \geq t for all t \in T.  Now let s \in S be arbitrary.  Then since s \in S \subseteq t, it follows that s \in T and so s \leq \sup T.  Therefore \sup T is an upper bound for S.  \square

(a) To prove that for a bounded sequence of real numbers, \{a_n\}, the e-sup always exist we use Lemma 1 to prove that the sequence of suprema

\displaystyle \left\{s_1, s_2, s_3, \cdots \right\}

is a decreasing sequence.

By definition we have that

\displaystyle s_1 = \sup \left\{a_1, a_2, a_3, \cdots \right\} and that \displaystyle s_2 = \sup \left\{a_2, a_3, \cdots \right\}.

Using S = \left\{a_1, a_2, a_3, \cdots \right\} and T = \left\{a_2, a_3, \cdots \right\} we have that S \subseteq T and so, by Lemma 1, s_1 \leq s_2.

Similarly, since \left\{a_n, a_{n+1}, a_{n+2}, \cdots \right\} \subseteq \left\{a_{n+1}, a_{n+2}, \cdots \right\} holds for all n \in \mathbb{N}, Lemma 1 implies that s_n \leq s_{n+1}.  Therefore the sequence \{s_n\} is decreasing.

Note that we now have an automatic upper bound for \{s_n\}, namely s_1 \geq s_n for all n \in \mathbb{N}.  To establish a lower bound for \{s_n\} we use (again) the fact that the original sequence \{a_n\} is bounded.  This means that there exist \alpha, \beta \in \mathbb{R} so that

\displaystyle \alpha \leq a_n \leq \beta \text{ for every } n \in \mathbb{N}.

Hence, for every n \in \mathbb{N}, it follows that since s_n = \sup \{a_n, a_{n+1}, \cdots \} then s_n \geq \alpha, too.

The monotone convergence theorem implies that the decreasing, bounded sequence \{s_n\} converges.  Its limit — which is defined to be the e-sup of \{a_n\} — therefore exists.  \square

Note: A very similar proof can be used to prove that the e-inf — which is defined to be the limit of a sequence of infimuma — also exists.

(b) We will handle this iff proof with two different if-then proofs.

proof (of \Rightarrow).  Suppose that \{a_n\} converges to a.  We will show that the e-sup = a, too.  This will be accomplished by proving that for every \varepsilon > 0, it follows that

\displaystyle a \leq \text{e-sup} \leq a + \varepsilon.

The first inequality above follows from Theorems in 1.3.  In particular, note that by definition of s_n, for each n \in \mathbb{N}, it follows that s_n \geq a_n.  Since both sequences have a limit, theorems from 1.3 imply that \text{e-sup} \geq a.

For the second inequality, let \varepsilon > 0 be given.  Since, by assumption, a_n \to a, there exists N \in \mathbb{N} so that whenever n \geq N, it follows that d(a_n, a) < \varepsilon.  Worded differently,

\displaystyle \left\{a_N, a_{N+1}, a_{N+2}, \cdots \right\} \subset \left(a-\varepsilon, a+\varepsilon\right).

Lemma 1 implies that the supremum of the first set is therefore bounded above by the supremum of the second set.  The supremum of the first set is of course s_N, and the supremum for the \varepsilon-ball is a+\varepsilon.  That is, we have

\displaystyle s_N \leq a + \varepsilon.

Since \{s_n\} is a decreasing sequence we actually have that for all n \geq N, the suprema s_n \leq a + \varepsilon.  Taking a limit as n \to \infty then produces the desired inequality, namely that

\displaystyle \text{e-sup} \leq a + \varepsilon.

Since \varepsilon is arbitrary, this establishes that \text{e-sup} = a.  A similar argument can be made to conclude that \text{e-inf}=a.

proof (of \Leftarrow).  Now suppose that the e-inf and e-sup exist and equal the same number a \in \mathbb{R}.  We will prove that a_n \to a by appealing to our squeeze theorem.

Note, again, that by definition of s_n,

\displaystyle s_n = \sup \left\{a_n, a_{n+1}, a_{n+2}, \cdots \right\} \geq a_n.

Similarly, by definition of i_n,

\displaystyle i_n = \inf \left\{a_n, a_{n+1}, a_{n+2}, c\dots \right\} \leq a_n.

We therefore have for each n \in \mathbb{N},

\displaystyle i_n \leq a_n \leq s_n.

Since, by assumption, \displaystyle \lim_{n\to\infty} i_n = \text{e-inf} = a = \text{e-sup} = \lim_{n\to\infty} s_n, the squeeze theorem implies that

\displaystyle \lim_{n\to\infty} a_n = a

as desired.  \square

Question 6. (Skipped for now — but, yes, these are very similar or outright identical to homework problems, and a proof for part (a) can be used to prove one of the cases you’d naturally set up for part (b).)

Question 7.  There are lots of ways to prove this “stronger” Nested Interval Property.  My favorite way uses the Monotone Convergence Theorem.

Let I_n = [a_n, b_n] be an indexed family of closed, non-empty, bounded intervals that are nested.  Observe that we therefore have a_n < b_n for all n \in \mathbb{N}; observe also that since the intervals are nested it follows that

\displaystyle a_1 \leq a_2 \leq a_3 \leq \cdots \text{ and } b_1 \geq b_2 \geq b_3 \cdots.

(We can also note that a_1 \leq a_2 < b_2 \leq b_1 and that a_1 \leq a_2 \leq a_3 < b_3 \leq b_2 \leq b_1 and so on.)

In other words, we have that the sequence \{a_n\} is increasing, and the sequence \{b_n\} is decreasing.  Since both of these sequences are bounded — a_1 \leq a_n < b_1 \text{ for all } n and b_1 \geq b_n > a_1 \text{ for all } n — the Monotone Convergence Theorem implies that each sequence converges.

Let us write \displaystyle a = \lim_{n\to\infty} a_n and \displaystyle b = \lim_{n\to\infty} b_n, and recall from the proof of the Monotone Convergence Theorem, that these two limits satisfy a = \sup \{a_n\} and b = \inf \{b_n\}.  We now treat the following sub-claim:

Sub-Claim:  If x \in \bigcap I_n then a \leq x \leq b.

sub-proof.  Suppose x \in \bigcap I_n but that, say, x < a.  Since a is the supremum of \{a_n\}, this means that there exists a term in the sequence, a_N so that a_N > x.  This inequality implies that

\displaystyle x \notin \left[a_N, b_N\right]

and so x \notin \bigcap I_n as assumed.  The case when x > b is handled similarly.  \Rightarrow\Leftarrow.

We will now use the assumption that \ell_n \to 0 to prove that a = b.  From useful Theorems in sectoin 1.3, we know that since each sequence \{a_n\} and \{b_n\} converges to a and b (respectively), the sequence \{b_n - a_n \} also converges to b-a.  However, we also know that

\displaystyle \ell_n = \text{Length}\left([a_n, b_n]\right) = b_n - a_n

and so

\displaystyle 0 = \lim_{n\to\infty} \ell_n = \lim_{n\to\infty} (b_n - a_n) = \left(\lim_{n\to\infty} b_n\right) - \left(\lim_{n\to\infty} a_n \right) = b - a.

Therefore b = a.

The Nested Interval Property tells us that there exists at least one point x \in \bigcap I_n, and by our sub-claim this point must lie inbetween a and b.  However, since \ell_n \to 0 implies that b = a, there is only one value that x can equal.  In particular, this all shows

\displaystyle \bigcap_{n=1}^{\infty} I_n = [a, a] = \{a\} = \{b\} = [b, b]

which completes the proof.  \square

Question 8.  Let a_n = \sqrt{n} and define b_n = \sqrt{n+1} - \sqrt{n}.  There are lots of ways to argue that b_n \to 0.  My personal favorite uses an algebraic “trick” that let’s us rewrite the terms as

\displaystyle b_n = \sqrt{n+1} - \sqrt{n} = \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\,\left(\sqrt{n+1}+\sqrt{n}\right) = \frac{1}{\sqrt{n+1}+\sqrt{n}}

We can now use various theorems and techniques from Section 1.3 to argue that this sequence converges to zero.  For example one can prove (either by induction or more direct methods) that

\displaystyle 2\sqrt{n} \leq \sqrt{n+1} + \sqrt{n} \leq 2\sqrt{n+1} \, \text{ for all } n \in \mathbb{N}

from which we can conclude that

\displaystyle \frac{1}{2\sqrt{n}} \geq b_n \geq \frac{1}{2\sqrt{n+1}}.

Some of our results from Section 1.3 then imply that the sequences 1/(2\sqrt{n}) and 1/(2\sqrt{n+1}) both converge to zero.  By our Squeeze Theorem, this implies that \{b_n\} converges to 0, too.  \square

We have shown that the sequence of successive distances d\left(a_n, a_{n+1}\right) converges to zero, and this seems like it should imply that the original sequence is Cauchy.  However, this is not the case!  In particular, that these distances converge to zero does not necessarily tell us anything about the distances between other terms further out in the sequence; that is, we do not necessarily know that

\displaystyle d\left(a_n, a_{n+2}\right) \to 0

or that, more generally, for any pair m \geq n,

\displaystyle d\left(a_n, a_m\right) \to 0.

Indeed, we can argue that while the sequence \{b_n\} converges to zero, the original sequence is not Cauchy.  We can accomplish this via a proof by contradiction.

Suppose that \{a_n\} = \{\sqrt{n}\} is Cauchy.  This is means that \{a_n\}  converges.  As a consequence, we know that this sequence must therefore be bounded.  That is, there must exist a real number B \in \mathbb{R} so that

\displaystyle \sqrt{n} \leq B \, \text{ for all } n \in \mathbb{N}.

This inequality is equivalent to claiming that there exists B \in \mathbb{R} so that

\displaystyle n \leq B^2 \, \text{ for all } n \in \mathbb{N}.

Given our familiarity with natural numbers, we can cite a contradiction right here.  Therefore, the sequence a_n = \sqrt{n} does not converge and so is not Cauchy.  \square

Note: Some of you very cleverly proved directly that a_n = \sqrt{n} is not Cauchy.  If you would like to see one of these proofs, ask around!

Question 9.  We will approach this problem by proving two claims.  Assume the hypotheses of the question, then

Claim 1.  If a_n converges, then it must converge to zero.

Claim 2.  The sequence a_n converges.

The second claim is actually the trickier of the two (imo).  The first claim can be proven by appealing to our arithmetic laws that govern convergent sequences.

Proof (of claim 1).  For a contradiction, suppose that a_n converges to a non-zero number, A \in \mathbb{R} with A \neq 0.  Note further that, by assumption, none of the terms of a_n are equal to zero, and so we may form the sequence

\displaystyle b_n = \frac{a_{n+1}}{a_n}.

Moreover, by using our results from section 1.3, since a_n converges to A \neq 0, and since a_{n+1} also converges to A we have that

\displaystyle b_n = \frac{a_{n+1}}{a_n} \text{ converges to } \frac{A}{A} = 1.

The claim that A/A = 1 only holds if, as we’ve assumed, A \neq 0.  This conclusion, however, contradicts the assumption that b_n converges to a number b < 1.

Therefore, if a_n converges, it must converge to A = 0.  \square

There are also many ways to prove claim 2.  The proof below will show that the sequence \{a_n\} is monotone and bounded, and therefore must converge by the Monotone Convergence Theorem.  Actually, we will prove that the sequence is eventually monotone and bounded.

Proof (of claim 2).  We are told that the sequence contains only positive numbers, and so we immediately have a lower bound: 0 < a_n for all n.

Now we will show that this sequence is eventually decreasing.  That is, that there exists N \in \mathbb{N} so that whenever n \geq N, a_{n+1} \leq a_n.  The idea for this claim comes from the fact that the sequence b_n converges to b \in (0, 1), and so the fractions a_{n+1}/a_n eventually look very close to the number b.  This implies that, far enough out in the sequence, a_{n+1} \approx b\,a_n and so a_{n+1} \leq a_n.

To make this idea more precise, we know that for any \varepsilon > 0 there exists N \in \mathbb{N} so that

\displaystyle \left|\frac{a_{n+1}}{a_n} - b \right| < \varepsilon

holds whenever n \geq N.  Since b \in (0, 1) there exists a positive number \varepsilon > 0 so that b + \varepsilon \in (0, 1) — choose such an \varepsilon, and denote the associated index by N_1.  We have that whenever n \geq N_1

\displaystyle -\varepsilon < \frac{a_{n+1}}{a_n} - b < \varepsilon

and this inequality is equivalent to

\displaystyle b-\varepsilon < \frac{a_{n+1}}{a_n} < b + \varepsilon.

The upper bound inequality above is what we focus on now.  By choice of \varepsilon, we have

\displaystyle n \geq N_1 \Rightarrow \frac{a_{n+1}}{a_n} < b + \varepsilon < 1

and so

\displaystyle n \geq N_1 \Rightarrow \frac{a_{n+1}}{a_n} < 1.

Since this last inequality is equivalent to a_{n+1} < a_n, we have that beyond the index N_1, our remaining sequence is decreasing.  Note that this provides for us an upper bound for the sequence \{a_{N_1}, a_{N_1 + 1}, \cdots \}, namely the number a_{N_1}.

The Monotone Convergence Theorem implies that this sequence \{a_{N_1}, \cdots \} converges.  This, of course, implies that the original sequence \{a_1, a_2, \cdots \} converges, too.  \square

Question 10. (skipped for now)

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