- If $A$ is a nonempty bounded subset of $\R$ is it then always true that
$\inf A \le \sup A$? (give a proof or a counterexample). Is it always
true that $\inf A\lt\sup A$?
- If $A$ and $B$ are bounded and non empty subsets of $\R$ with $A\subset
B$.
- Show that $\sup A\leq \sup B$ .
- Is it always true that $\inf A\leq \sup B$ ? (proof or counterexample)
- Is it always true that $\inf B\leq \sup A$ ? (proof or counterexample)
- Let $A\subset\R$ be a bounded set of integers. Show that $\sup A$ is an
integer.
Let $x=\sup A$. We show by contradiction that $x\in A$. Suppose
$x\not\in A$. Since $x$ is the least upper bound of $A$, there is an
$n\in A$ with $x-1\lt n\lt x$. The interval $(x-1, x)$ cannot contain
more than one integer, so $n$ is the only number from $A$ that belongs
to $(x-1,x)$. Therefore $n$ is an upper bound for $A$, even though
$n\lt x$, and $x$ was the least upper bound. This is a contradiction,
and we conclude that $x=\sup A$ must belong to $A$.
- Let $A\subset\N$. Show that $A$ contains a smallest number,
i.e. there is an $n\in A$ such that every $m\in A$ satisfies $m\geq
n$.
- Let $A=\{ \frac{n}{2n+1} : n\in\N\}$. Show that $\sup A =
\frac12$.
Solution. For every $n\in\N$ we have $2n+1\gt 2n$, so that
\[
\frac{n}{2n+1} \lt \frac{n}{2n} = \frac{1}{2}.
\]
Therefore $\frac12$ is an upper bound for $A$.
Let $x\lt \frac12$, and assume that $x$ is an upper bound for $A$.
This implies that
\[
\forall n\in\N: \quad \frac{n}{2n+1} \le x
\]
Hence
\[
\forall n\in\N: \quad n \le 2nx + x,
\]
and thus
\[
\forall n\in\N: \quad (1-2x)n \le x.
\]
Since $x\lt \frac12$ we have $1-2x\gt0$ so we may divide both sides of the
inequality by $1-2x$ and conclude that
\[
\forall n\in\N:\quad n \le \frac{x}{1-2x}.
\]
This implies that $x/(1-2x)$ is an upper bound for the set of natural
numbers, in direct contradiction with the archimedean property.
The contradiction shows that no number less than $\frac 12$ can be an
upper bound for $A$.
Hence $\frac{1}{2}$ is the least upper bound for $A$.
- Show that $\sup A = 1$ for $A = \{\frac{2^n}{2^n+1} : n\in\N\}$.
- (Archimedean property and variations)
- Prove the Archimedean property of the real numbers directly from the
least upper bound axiom.
- Show that for any pair of real numbers $x\lt y$ there is a rational
number $r\in \Q$ with $x\lt r\lt y$.
- Show that the set $A=\{3^n: n\in \N\}$ is unbounded, i.e. show
that for every real number $x$ there is an integer $n$ with $3^n\gt
x$.
- Show that the set $\left\{\frac{n^2}{n+1} \mid n\in\N\right\}$ is
unbounded.
- Show that the set $\{n! \mid n\in\N\}$ is unbounded.
- Show that the set $\{\sqrt{n} \mid n\in\N\}$ is unbounded.
- Let $A=\{a_1, a_2, a_3, \ldots\}$ be a set of real numbers where
$a_{n+1}\ge a_n+1$ holds for all $n\in\N$. Show that $A$ is
unbounded.
- Let $E=\{\frac1n : n\in\N\}$, and let $F=E\cup \{0\}$. Find all the
limit points of $E$ and of $F$. Are either of $E$ or $F$ closed?
- Let $A = \{\frac{m}{m+2}| m\in\N\}$ and $B=\{\frac{m}{m-2}|m\in\N, m\geq
3\}$. Find all limit points of the set $A$ and of the set $B$.
- Let $(X, d)$ be some metric space, $a\in X$ some point in $X$, and
$r\gt0$. Show that the set $E = \{x\in X \mid d(x,a)\gt r\}$ is open.
- Let $a_n$ and $b_n$ be two sequences of real numbers.
- Show directly from the definition of “convergent sequence”
that if the sequences $a_n$ and $b_n$ both converge, and if $a_n\leq b_n$
for all $n\in\N$, then $\lim_{n\to\infty}a_n \leq \lim_{n\to\infty}b_n$
- Prove that if the sequences $a_n$ and $b_n$ are bounded,
and if $a_n\leq b_n$ for all $n\in\N$ then
\[
\liminf_{n\to\infty} a_n \leq \liminf_{n\to\infty} b_n
\quad\text{ and }\quad
\limsup_{n\to\infty} a_n\leq \limsup_{n\to\infty} b_n.
\]
-
- Show that if the series $\sum_{n=1}^\infty a_n$ converges, and if
$a_n\geq0$ for all $n\in\N$, then $\sum_{n=1}^\infty a_n^2$ also converges.
- Find a sequence $a_n$ for which $\sum_{n=1}^\infty a_n$ does not
converge, but for which $\sum_{n=1}^\infty a_n^2$ converges.
- Find a sequence $a_n$ for which $\sum_{n=1}^\infty a_n^2$ diverges even
though $\sum_{n=1}^\infty a_n$ does converge.
a. Hint: If the series $\sum a_n$ converges, then $a_n\to 0$;
if $a_n\to 0$ then the sequence $a_n$ is bounded, i.e. there is some
$M$ with $a_n\leq M$ for all $n\in\N$. If $0\leq a_n \leq M$ then $a_n^2 =a_n
\times a_n \leq Ma_n$.
b. Hint: consider $\sum_{n=1}^\infty n^{-p}$ for suitably
chosen $p$.
c. Hint: consider $\sum_{n=1}^\infty (-1)^{n+1} n^{-p}$ for
suitably chosen $p$.
- Show that if the series $\sum_{n=1}^\infty a_n^2$ and $\sum_{n=1}^\infty b_n^2$
both converge, then the series $\sum a_nb_n$ also converges.
Hint: either remember the Cauchy–Schwarz inequality, or use that
$|2ab|\leq a^2 + b^2$ for all $a,b\in\R$ (which follows from expanding
$(a-b)^2\geq 0$).
- 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 $(X,d)$ 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 $A\subset\R$, 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.
- 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 $(X,d)$ 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. Here is one:
\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
\[
\lim_{n\to\infty} \Bigl\{\sup_{k\ge n} \bigl(x_k + y_k\bigr) \Bigr\}
= X+Y.
\]