- Consider the sequence of real numbers given by $a_n=n^{-1}$.
- Show that $a_n\to 0$, directly from the definition.
Given $\varepsilon\gt0$ choose an integer $N$ such that $N\gt 1/\varepsilon$. Then for all $n\ge N$ one has $0\lt a_n \le 1/N \lt \varepsilon$.
- Show that if a sequence $x_n$ converges, then any subsequence of $x_n$ also converges and has the same limit.
Let $y_k = x_{n_k}$ be a subsequence of $x_n$, and let $L$ be the limit of $x_n$. In particular this implies that $n_k\ge k$ for all $k$. Given
$\varepsilon\gt0$ we choose $N\in\N$ such that $|x_n-L|\lt \varepsilon$ for all $n\ge N$. Then, if $k\ge N$ we have $n_k\ge k\ge N$ and thus
$|y_k-L| = |x_{n_k}-L| \lt \varepsilon$.
- Show that $b_n=n^{-2}\to 0$ without using the definition of
convergence.
$b_n = a_{n^2}$, so $b_n$ is a subsequence of the sequence $a_n=1/n$. Therefore it converges, and it has the same limit:
$\lim_{n\to\infty} 1/n^2 = 0$.
- Let $x_n$ be a convergent sequence of points in a metric space $(X,d)$. Show that the sequence is bounded.
Let $x_n\to p$, i.e. let $p$ be the limit of the sequence $x_n$. Choose $\varepsilon=1$. There is an $N\in\N$ such that for
all $n\ge N$ on has $d(x_n, p)\lt \varepsilon=1$. Thus, if \[ R=\max\left\{1, d(x_1, p), \dots, d(x_{N-1}, p)\right\}, \] then all $x_k$ satisfy
$d(x_k, p)\le R$.
- Let $x_n$ be a sequence of points in a metric space $(X,d)$. Show that $x_n\to x$ holds if and only if $d(x_n, x)\to0$.
Consider the sequence of real numbers $a_n = d(x_n, p)$. We are asked to show that $a_n\to0 \iff x_n\to p$.
Assume $a_n\to 0$. To show $x_n\to p$, let $\varepsilon\gt 0$ be given. By assumption there is an $N\in\N$ such that $|a_n|\lt \varepsilon$ for all
$n\ge N$. This implies $d(x_n, p)\lt \varepsilon$ for all $n\ge N$. Therefore $x_n\to p$.
Assume $x_n\to p$. To show $a_n\to 0$, let $\varepsilon\gt 0$ be given. By assumption there is an $N\in\N$ such that $d(x_n, p)\lt \varepsilon$ for
all $n\ge N$. This implies $|a_n|\lt \varepsilon$ for all $n\ge N$. Therefore $a_n\to 0$.
- Let $(X,d)$ be a metric space, $C\subset X$ a closed subset, and $x_n\in C$ a sequence of points with $x_n\to x$. Show that $x\in C$.
In order to reach a contradiction, assume that $x\not\in C$. The complement $C^c$ of $C$ is open in $X$, so there is an $\varepsilon\gt0$ such that
$B_\varepsilon(x)\subset C^c$. Since $x_n\to x$, there is an $N_\varepsilon\in\N$ for such that $x_n\in B_\varepsilon(x)$ for all $n\ge N$. But then
we would have $x_n\not\in C$ for $n\ge N$, which is a contradiction. We conclude that $x\in C$ after all.
- Let $x_n\in\R^k$ and $y_n\in \R^k$ be two sequences of vectors for which $x_n\to x$ and $y_n\to y$. Show that $(x_n, y_n) \to (x,y)$. Here $(x,y)$ is
the dot product, or inner product of the two vectors. In Rudin’s notation (page 16): the assignment is to show that $x_n\cdot y_n \to x\cdot y$.
The sequences $x_n\in\R^k$ and $y_n\in\R^k$ converge, so they
are bounded, i.e. there is an $R\gt0$ such that $\|x_n\|\le
R$, and $\|y_n\|\le R$ for all $n\in\N$. We also have
\begin{align*}
\left|x_n\cdot y_n - x\cdot y\right|
&=\left| x_n\cdot y_n - x_n\cdot y + x_n\cdot y - x\cdot y\right| \\
&\le \left| x_n\cdot y_n - x_n\cdot y\right| + \left|x_n\cdot y - x\cdot y\right|
& \text{triangle }\le\\
&\le \left| x_n\cdot (y_n - y)\right| + \left|(x_n - x)\cdot y\right| \\
&\le \| x_n\|\,\| y_n - y\| + \|x_n - x\|\,\| y\|
& \text{Cauchy }\le \\
&\le R\| y_n - y\| + R \|x_n - x\|
\end{align*}
Since $x_n$ and $y_n$ converge we have
\[
\lim_{n\to\infty} R\| y_n - y\| + R \|x_n - x\| = 0
\]
and hence
\[
\lim_{n\to\infty} \left|x_n\cdot y_n - x\cdot y\right| =0.
\]
- Let $A\subset\R$, and assume that every term in the sequence $\{x_n\}_{n\in\N}$ is an upper bound for $A$. Show that if $x_n\to x$, then $x$ is also
an upper bound for $A$.
Suppose $x$ is not an upper bound for $A$. Then there is a number $a\in A$ with $a\gt X$. Let $\varepsilon= a-x$. Since $x_n\to
X$, there is an $N$ such that for all $n\ge N$ one has $|x_n-x|\lt \varepsilon$. This implies, for $n\ge N$, that $x_n \lt x+\varepsilon = a$. Hence
$x_n$ is not an upper bound for $A$ if $n\ge N$, contradicting what was given.
Instead of the definition in Rudin (3.16, page 56) we will use the following alternative definition of “lim sup”,
which was discussed in class:
\[
\limsup_{n\to\infty} x_n = \lim_{n\to\infty}\Bigl\{\sup_{k\ge n} x_k\Bigr\}.
\]
This is currently the more common definition: see, for example,
the wikipedia
page on lim inf and lim sup.
- Prove that if $\limsup_{n\to\infty} a_n =A$ then for every $\varepsilon\gt0$ and any $n\in\N$ there is an $m\gt n$ such that $a_m\gt
A-\varepsilon$.
Let $\varepsilon\gt0$ and $n\in\N$ be given.
By definition of “lim sup” we have
\[
\lim_{m\to\infty} \Bigl(\sup\bigl\{a_k : k\ge m\bigr\}\Bigr) =A.
\]
By definition of “limit” this implies that there is an
$N_\varepsilon\in\N$ such that for all $m\ge N_\varepsilon$ we have
\[
A - \varepsilon \lt
\sup\bigl\{a_k : k\ge m\bigr\}\lt
A + \varepsilon
\]
We choose $m \gt \max \{N_\varepsilon, n\}$. Then
\[
\sup\{a_k : k\ge m\} \gt A - \varepsilon.
\]
Since $\sup \{\dots\}$ is the least upper bound it follows that $A-
\varepsilon$ is not an upper bound for $\{a_k : k\ge m\}$. Thus there is an
$k\ge m$ such that $a_k \gt A-\varepsilon$. We had chosen $m\gt n$, and
thus $k\gt n$, so that we have found a $k\gt n$ for which $a_k\gt A-
\varepsilon$.
- Compute $\limsup_{n\to\infty} a_n$ and $\liminf_{n\to\infty} a_n$ for the
following sequences, showing how your result follows from the definition of
limsup/inf (given below):
- $a_n = (-1)^n$
- $a_n = (-1)^n+ \frac 2n$
- $a_n = (-1)^n \frac {n+2}n$
- $a_n = n$
- $a_n = (-1)^n n$
- $a_n = \bigl(1 + (-1)^n\bigr) n$
In each case compute $m_n = \inf \{a_k : k\ge n\}$, and $M_n = \sup \{a_k :
k\ge n\}$. Then $\lim_{n\to\infty} m_n$ and $\lim_{n\to\infty}M_n$ are the
lim inf and lim sup, respectively.
- $a_n = (-1)^n$: $m_n=-1$ and $M_n=+1$ for all $n$, so
$\liminf_{n\to\infty} a_n = -1$, while $\limsup_{n\to\infty}a_n = +1$.
- $a_n = (-1)^n+ \frac 2n$: $m_n = -1$ for all $n$. $M_{n} = 1+\frac2n$
if $n$ is even, and $M_n=M_{n+1}$ if $n$ is odd. Hence
$\liminf_{n\to\infty} a_n = -1$, while $\limsup_{n\to\infty}a_n = +1$.
- $a_n = (-1)^n \frac {n+2}n$: $m_n=-\frac{n+2}n$ if $n$ is odd, and
$m_{n}=m_{n+1}$ if $n$ is even, so $\liminf_{n\to\infty} a_n = -1$;
$M_n=\frac{n+2}n$ if $n$ is even, and
$M_{n}=M_{n+1}$ if $n$ is odd, so $\limsup_{n\to\infty} a_n = 1$.
- $a_n = n$: $m_n=n$, and $M_n=+\infty$ for all $n$. Both lim sup and
lim inf are $+\infty$.
- $a_n = (-1)^n n$: $m_n=-\infty$, $M_n=+\infty$.
$\liminf_{n\to\infty} a_n = -\infty$, while $\limsup_{n\to\infty}a_n =
+\infty$.
- $a_n = \bigl(1 + (-1)^n\bigr) n$ : $m_n=0$, $M_n= 2n$ if $n$ is even,
$M_n=M_{n+1}$ if $n$ is odd.
$\liminf_{n\to\infty} a_n = 0$, while $\limsup_{n\to\infty}a_n =\infty$.
- Let $x_n=\cos(\theta + n\frac\pi3)$ where $\theta\in(0, \frac\pi3)$ is some
constant. For this sequence, compute
- $\sup_{n\in\N} x_n$
- all possible limits of subsequences of $x_n$
- $\limsup_{n\to\infty} x_n$
See student-solution on Piazza (type “@52”
in the search box at the top left of the Piazza page).
- Let $(X,d)$ be a metric space, and let $x_n$ be a convergent sequence in
$X$. Show that $x_n$ also is a Cauchy sequence.
Assume $x_n\to p$. Let $\varepsilon\gt0$ be given.
Then there is an $N_\varepsilon\in\N$ such that for all $n\ge N_\varepsilon$
one has $d(x_n, p)\lt \varepsilon$. If $n, m\ge N_\varepsilon$, then
\[
d(x_n, x_m) \le d(x_n, p)+d(p, x_m) \le \frac \varepsilon2 +
\frac\varepsilon2 =\varepsilon.
\]
Thus $\{x_n\}$ is a Cauchy sequence.
- Give an example of a metric space $(X,d)$ with a Cauchy sequence that
does not converge.
There are many examples. Perhaps the simplest example is to
take any metric space $(X,d)$ and a convergent sequence $x_n\in X$,
$p=\lim_{n\to\infty}x_n$, and then consider the space $(Y, d)$ where
$Y=X\setminus \{p\}$, and $d$ is the same metric as on $X$. For instance,
let $Y=\R\setminus\{0\}$, with distance $d(x, y) = |x-y|$; then the sequence
$x_n = \frac1n$ is a Cauchy sequence, but it does not converge in $Y$.
- Give an example of a complete metric space that is not compact. Provide
a brief explanation for your answer.
The real line. Compact metric spaces are bounded, and
while $\R$ is complete, it is not bounded.
- Let $x_n$ and $y_n$ be two bounded sequences of real numbers. Show that
\[
\limsup_{n\to\infty} (x_n+y_n) \le
\limsup_{n\to\infty} x_n + \limsup_{n\to\infty} y_n.
\]
By definition
\[
\limsup_{n\to\infty}(x_n+y_n)
= \lim_{n\to\infty} \Bigl(\sup\{x_k+y_k : k\ge n\}\Bigr).
\]
For every $k\ge m$ we have
\[
x_k \le \sup\{x_l : l\ge m\}, \qquad
y_k \le \sup\{y_l : l\ge m\}.
\]
Hence, for all $k\ge m$,
\[
x_k+y_k \le \sup\{x_l : l\ge m\} + \sup\{y_l : l\ge m\}.
\]
The quantity on the right does not depend on $k$ and thus it is an upper
bound for $\{x_k+y_k : k\ge m\}$. It is therefore not smaller than the
least upper bound:
\[
\sup\{x_k+y_k : k\ge m\} \le \sup\{x_l : l\ge m\} + \sup\{y_l : l\ge m\}.
\]
If we now take the limit for $m\to\infty $ on both sides we get
\[
\limsup_{k\to\infty} x_k+y_k \le \limsup_{k\to\infty} x_k +
\limsup_{k\to\infty} y_k.
\]
- Find two sequences $x_n$, $y_n$, of real numbers for which
\[
\limsup_{n\to\infty} (x_n+y_n) \lt
\limsup_{n\to\infty} x_n + \limsup_{n\to\infty} y_n.
\]
There are many possibilities. The following was
suggested by a student in my office hour:
\begin{align*}
\{x_n\} &= \{+1, 0, +1, 0, +1, 0, +1, 0, \dots \}\\
\{y_n\} &= \{-1, 0, -1, 0, -1, 0, -1, 0, \dots \}
\end{align*}
For these sequences $x_n+y_n = 0$ for all $n$, so
$\limsup_{n\to\infty}x_n+y_n =0$. But
\[
\limsup x_n = 1, \qquad \limsup y_n =0,
\]
so $\limsup(x_n+y_n) \lt \limsup x_n+ \limsup y_n$.
- If $x_n$ and $y_n$ are bounded sequences of real numbers for which
$x_n$ converges, then show that
\[
\limsup_{n\to\infty} (x_n+y_n) =
\lim_{n\to\infty} x_n + \limsup_{n\to\infty} y_n.
\]
Abbreviate
\[
\lim_{n\to\infty} x_n = X, \qquad
\limsup_{n\to\infty}y_n = Y.
\]
By definition
\[
\limsup_{n\to\infty} \bigl(x_n +y_n\bigr)
= \lim_{n\to\infty} \Bigl\{\sup_{k\ge n} \bigl(x_n+y_n\bigr) \Bigr\}.
\]
Let $\varepsilon\gt 0$ be given. Then there is an $N_\varepsilon$ such that
for all $n\ge N_\varepsilon$ we have
\[
X-\varepsilon \lt x_n \lt X+\varepsilon,
\]
and
\[
Y - \varepsilon \lt \sup_{k\ge n} y_k \lt Y + \varepsilon.
\]
For any $k\ge N_\varepsilon$ we then have
\[
X-\varepsilon + y_k \lt x_k + y_k \lt X+\varepsilon +y_k,
\]
and thus for any $n\ge N_\varepsilon$,
\[
X-\varepsilon + \sup_{k\ge n}y_k
\lt \sup_{k\ge n} \bigl(x_k + y_k\bigr)
\lt X+\varepsilon +\sup_{k\ge n}y_k,
\]
and therefore, for all $n\ge N_\varepsilon$,
\[
X+ Y - 2\varepsilon \lt \sup_{k\ge n} \bigl(x_k + y_k\bigr) \lt X + Y + 2\varepsilon.
\]
This implies, by definition of “$\lim_{n\to \infty} (\dots)$”
that
\[
\lim_{n\to\infty} \Bigl\{\sup_{k\ge n} \bigl(x_k + y_k\bigr) \Bigr\}
= X+Y.
\]
- If $f:J\to\R$ is a function on some open interval $J\subset \R$, and if
$f$ is differentiable at $a\in J$, then show that $f$ is continuous at $a$.
- Suppose $f:[a,b]\to\R$ is a continuous function which is differentiable on
$(a,b)$, and assume that $f'(x)\lt 0$ for all $x\in (a,b)$. Show that $f(a)
\gt f(b)$.
- Suppose $f:\R\to\R$ is a continuous function which is twice
differentiable.
Assume there are $a\lt b\lt c$ such that $f(a) = f(b)= f(c)= 0$.
Show that there is an $x\in(a, c)$ such that $f''(x)=0$.
Solution: Apply Rolle’s theorem to $f$ on
the interval $[a,b]$ to conclude that there is a $p\in(a,b)$ with $f'(p)=0$.
Similarly, there is a $q\in(b, c)$ with $f'(q)=0$. Next, apply
Rolle’s theorem to $f'$ on the interval $[p,q]$, and you find that
there is an $r\in(p,q)$ such that $f''(r) = 0$.
- Suppose $f:\R\to\R$ is a continuous function which is twice
differentiable, and suppose the graph of $f$ intersects a given straight line
in three different points. Show that $f''(x)=0$ for at least one $x\in \R$.
Solution:
Assume the straight line has equation $y=mx+n$, and assume that the graph of
$f$ intersects the line at $x=a$, $x=b$, and $x=c$, where we may assume that
$a\lt b\lt c$.
Apply the Mean Value Theorem theorem to $f$ on
the interval $[a,b]$ to conclude that there is a $p\in(a,b)$ with $f'(p)=m$.
Similarly, there is a $q\in(b, c)$ with $f'(q)=m$. Next, apply
the MVTheorem to $f'$ on the interval $[p,q]$, and you find that
there is an $r\in(p,q)$ such that $f''(r) = 0$.
- Let $f:[a,b]\to\R$ be a continuous function which is three times
continuously differentiable, and assume its graph intersects the graph of
the parabola $y=x^2$ in four different points. Show that there is an $x$ with
$f^{(3)}(x) = 0$.
Solution:
Let the $x$ coordinates of the intersections points be $a\lt b\lt c\lt d$.
Let the parabola be given by $y=P(x) = kx^2+\ell x+m$.
Consider the function $g(x) = f(x) - P(x)$. This function vanishes at
$a,b,c,d$, so by Rolle’s theorem there are $p,q,r$ with
\[
a\lt p\lt b\lt q \lt c \lt r\lt d
\]
such that $g'(p)=g'(q)=g'(r)=0$. Apply Rolle’s theorem again, but now
to $g'$, and you find that there are $s\in(p,q)$ and $t\in(q,r)$ with
$g''(s) = g''(t) = 0$. Applying Rolle one more time then gives you a
$u\in(s,t)$ with $g'''(u)=0$.
This implies $f'''(u) = P'''(u)$. But $P(x)$ is a quadratic function, so
its third derivative vanishes. Therefore $f'''(u) = 0$.
- Let $f:\R\to\R$ be a differentiable
function. Assume that there are four
numbers $a\lt b\lt c\lt d$ such that $f(b)\lt f(a)$ and $f(c)\lt f(d)$. Show
that there is an $x\in (a, d)$ such that $f'(x)=0$.
Solution: There are two theorems that we will
use:
a continuous function on a compact metric space attains its
minimum and
the derivative of a differentiable function on an
interval vanishes at any interior minimum or maximum point.
The interval $[a,d]$ is compact, and the function $f$ is continuous, so
it attains its minimum at some $x\in[a,d]$. Since $f(b)\lt f(a)$ the
minimum is not attained at $a$, so $x\in(a, d]$. Similarly, $f(c) \lt
f(d)$, so the minimum is also not attained at $x=d$. Therefore $x\in(a,d)$
so the point $x$ at which the minimum is attained is an interior point, and
we conclude that $f'(x)=0$.
- Let $f:\R\to\R$ be a twice differentiable function. Assume that there are four
numbers $a\lt b\lt c\lt d$ such that $f(b)\lt f(a)$, $f(c)\gt f(b)$, and
$f(d)\lt f(c)$. Show that there is an $x\in (a, d)$ such that $f''(x)=0$.
Solution:
The Mean Value Theorem applied to $f$ on each of the three intervals
$(a,b)$, $(b,c)$, $(c, d)$, implies that there are $p\in(a,b)$, $q\in(b,c)$,
and $r\in (c,d)$ with
\[
f'(p)\lt 0, \quad f'(q)\gt0, \quad f'(r)\lt0.
\]
Since we are given that $f$ has two derivatives, we know that $f'$ is a
differentiable function, and thus that $f'$ is continuous. The Intermediate
Value Theorem then implies that there exist $s\in(p,q)$ and $t\in(q, r)$
with
\[
f'(s) = 0, \quad f'(t)=0.
\]
Finally, we apply Rolle’s theorem to $f'$ on the interval $[s,t]$, and
conclude that there is a $u\in(s,t)$ with $f''(u)=0$.