Analysis – Solutions 2

1.  A sequence of real numbers is a function $f:\mathbb{N} \to \mathbb{R}$ where we notate $a_n = f(n)$.
2. skipped.
3. There are lots of differences between accumulation points of sequences and limit points.  First, the concept of an accumulation point can be defined for any subset $S \subset \mathbb{R}$, not just for sequences of real numbers.  Second, a sequence can have multiple accumulation points but if it converges, then it only converges to one limit point.  Third, it is possible for a sequence to have a limit point but no accumulation points.  For example, the (rather boring) sequence $\{a_n\}$ given by $a_n = 1$ converges to the limit point $1$, but this sequence has no accumulation points.
4. Yes, it is possible for a sequence to have two distinct accumulation points.  Consider (as discussed in class) the sequence $\{1, 2, 1/2, 3/2, 1/3, 4/3, 1/4, 5/4, \cdots \}$ — you can think of this sequence as the result of “splicing” the sequences $\{1, 1/2, 1/3, 1/4, \cdots \}$ and $\{2, 3/2, 4/3, 5/4, \cdots \}$.  This sequence has $0$ and $1$ as accumulation points.
5. A Cauchy sequence is a sequence whose terms get arbitrarily close to one another.  More precisely: a sequence $\{a_n\}$ is Cauchy if for every $\varepsilon > 0$ there exists $N \in \mathbb{N}$ so that whenever $m, n \geq N$ it follows that $d(a_m, a_n) = |a_m - a_n| < \varepsilon.$
6. This notation is just another way to say “the sequence $\{a_n\}$ converges to $A$.”
7. The Bolzano-Weierstrass Theorem claims that if $S \subseteq \mathbb{R}$ is infinite and bounded, then $S$ has at least one accumulation point.  The proof of this theorem uses the Axiom of Completeness when concluding that the intersection of “the shrinking intervals” contains at least one real number; in particular, a sequence of left-end points for closed intervals is used in the proof and it is noted that this sequence is bounded and therefore has a supremum.
8. The statement that is asserted is that this sequence is Cauchy; we have a theorem that states that Every Cauchy Sequence Converges.
9. (a) $B_r(x) = \{z \in X : d(x,z) < r\}$ — this actually defines an “open ball.”  A “closed ball”is defined by using the $\leq$ inequality.  (b) skipped — but the picture should be of the interval $(\pi-2, \pi+2)$.  (c ) The picture is the entire real line!  (d) skipped.  (e)  A point $x \in X$ is an accumulation point for a subset $S \subseteq X$ if for every $\varepsilon > 0$ the ball $B_{\varepsilon}(x)$ intersects $S$ in infinitely many points.

Proof Questions

Problem 1.  (skipped)

Problem 2.  Claim: The sequence $a_n = 3n/(2n+1)$ converges to $3/2$.

Proof.  Let $\varepsilon > 0$ be given.  Choose $N > .$  Now suppose that $n \geq N$.  We then have

$\displaystyle d(a_n, 3/2) = \left|\frac{3n}{2n+1}- \frac{3}{2}\right| = \left| \frac{-3}{4n+2}\right| = \frac{3}{4n+2} \leq \frac{3}{4N+2}.$

Because $N > 3/(4\varepsilon) - \frac{1}{2}$ it follows that

$\displaystyle \frac{3}{4N+2} < \varepsilon$

and so we have that $d(a_n, 3/2) < \varepsilon$, as desired.  $\square$

Problem 3.  Theorem.  If $\{a_n\}$ converges to $A$ and to $B$, then $A = B$.

Proof.  We will first show that for every $\varepsilon > 0$, the distance between $A$ and $B$ satisfies $d(A, B) < \varepsilon$.

To this end, let $\varepsilon > 0$ be given.  S

ince, by assumption, $\{a_n\}$ converges to $A$, there exists an index $N_1 \in \mathbb{N}$ so that whenever $n \geq N_1$ it follows that $d(a_n, A) < \varepsilon/2$.  (This follows by applying our definition of convergence with the positive number $\varepsilon/2 > 0$.)

Similarly, since $\{a_n\}$ also converges to $B$, there exists an index $N_2 \in \mathbb{N}$ so that whenever $n \geq N_2$ it follows that $d(a_n, B) < \varepsilon/2$.

Choose $N \geq \max\{N_1, N_2\}$.  Now suppose that $n \geq N$ so that $n \geq N_1$ and $n \geq N_2$.  It then follows (by use of the triangle inequality) that

$\displaystyle d(A, B) \leq d(A, a_n) + d(a_n, B) < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$

Since $d(A, B) < \varepsilon$ for every positive number $\varepsilon$, it follows that $d(A, B) = 0$.  By definition of metric space, this implies $A = B. \square$

Note: Everywhere in the proof above, we can replace $d(A, B) = |A-B|$ in the case that $\{a_n\}$ is a sequence of real numbers and $A, B \in \mathbb{R}$.  In the proof above, though, the fact that these quantities are real numbers makes no difference to the proof!  In other words, the above argument works in any metric space.

Problem 4.  The Theorem states the following: If $\{a_n\}, \{b_n\},$ and $\{c_n\}$ are sequences of real numbers where $a_n$ converges to $A$, $b_n$ converges to $A$, and

$a_n \leq c_n \leq b_n$

for all $n$, then $c_n$ converges to $A$, too.

The triangle inequality is used three times.

The index is chosen to be the maximum of $N_1$ and $N_2$ so that both the $a_n$ terms and the $b_n$ terms are within $\varepsilon/2$ of $A$ at the same time.

Problem 5.  (sketch)  The main observation to make is that the function $[x]$ is bounded above and below by the following quantities:

$\displaystyle x-1 \leq [x] \leq x$.

After using the given summation formula, this allows us to construct two new sequences, one that bounds $\{a_n\}$ from above, and the other that bounds it from below.  In particular, we have that

$z_n \leq a_n \leq w_n$

where

$\displaystyle z_n = \frac{x}{2}\,\frac{n+1}{n}\, - \frac{1}{n} \text{ and } w_n = \frac{x}{2}\,\frac{n+1}{n}$.

Standard $\varepsilon$ arguments show that $w_n \to x/2$ and $z_n \to x/2$, too.  Hence, by Problem 4., this sequence converges to $x/2$.

Problem 6.  Theorem (1.4): Every Cauchy sequence of real numbers is bounded.

Proof.  Suppose $\{a_n\}$ is a Cauchy sequence of real numbers.  This means that for every $\varepsilon > 0$ there exists an index $N \in \mathbb{N}$ so that whenever $m, n \geq N$ it follows that $d(a_m, a_n) < \varepsilon$.

Let us apply this definition to a specific choice of $\varepsilon$, say when $\varepsilon = 1$.  We then have that there exists an index $N \in \mathbb{N}$ so that whenever $m, n \geq N$, it follows that $d(a_m, a_n) < 1$.

Since the index $m$ is required to be bigger than or equal to $N$, we can go ahead and choose $m = N$ if we like.  Applying this choice to the above definition, we have that whenever $n \geq N$, all of the remaining terms in our sequence are no more than distance $1$ from the fixed term $a_m = a_N$.  That is, we have that if $n \geq N$, then

$a_N - 1 < a_n < a_N + 1$.

This implies that all but finitely many terms of the sequence $\{a_n\}$ are bounded below by $a_N - 1$ and bounded above by $a_N + 1$.  In particular, we have that the set

$\{a_N, a_{N+1}, a_{N+2}, \cdots \}$

is bounded below by $B = a_N-1$ and bounded above by $A = a_N+1$.  To bound the entire Cauchy sequence, then, we also need to bound the first $N-1$ terms of the sequence; that is, we need to bound the set

$\{a_1, a_2, \cdots, a_{N-1} \}$

This finite list of numbers is bounded below by $m_1 = \min \{a_1, a_2, \cdots, a_{N-1}\}$ and bounded above by $M_1 = \max \{a_1, a_2, \cdots, a_{N-1}\}$.

Therefore, the entire sequence is bounded above by

$\max \{M_1, A\}$

and is bounded below by

$\min \{m_1, B\}$

completing the proof.  $\square$

Problem 7.  There are lots of ways to do this, and probably the best way to indicate an answer is with a picture.  Still, for what its worth, here is a set of points (expressed as a union of sequences) whose limit points is the set of naturals:

$\displaystyle S = \left\{1-\frac{1}{n}\right\} \bigcup \left\{2- \frac{1}{n}\right\} \bigcup \left\{3 - \frac{1}{n}\right\} \bigcup \cdots = \bigcup_{k=1}^{\infty}\, \left\{k-\frac{1}{n}\right\}.$

Problem 8.  (sketch)  Let $S \subseteq \mathbb{R}$ be a non-empty set that is bounded from above.  By the Completeness Axiom, the supremum of $S$ exists — let’s call it $x = \sup S$.

If $x \in S$, then we are done.  Therefore suppose that $x \notin S$ and let $\varepsilon > 0$ be given.  By definition of supremum, $x - \varepsilon$ is not an upper bound for $S$.  This means there exists an element $y \in S$ so that $y > x - \varepsilon$.  In particular, we have that

$x - \varepsilon < y < x < x + \varepsilon \iff y \in (x-\varepsilon, x+\varepsilon)$.

Since $y \in S$ and, by assumption, $x \notin S$, it follows that $y \neq x$.  Therefore, we have shown that every $\varepsilon$-nbhd of $x$ contains at least one point $y \in S$ where $y \neq x$.  By the Lemma on page 35, this shows that $x$ is an accumulation point of $S$, as desired.