1. For any two bounded sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ of real numbers, show that

$$

\limsup _{n \rightarrow \infty}\left(a_{n}+b_{n}\right) \leq \limsup _{n \rightarrow \infty} a_{n}+\limsup _{n \rightarrow \infty} b_{n}

$$

Give an example of strict inequality.

Note. The inequality also remains true for unbounded sequences provided the right side is not of the form $\infty-\infty$ or $-\infty+\infty$.

Solution: Let $A=\limsup _{n \rightarrow \infty} a_{n}, B=\limsup _{n \rightarrow \infty} b_{n}$ and $L=\limsup _{n \rightarrow \infty}\left(a_{n}+b_{n}\right) .$ Suppose $L>A+B .$ Choose an $\varepsilon>0$ such that $L-\varepsilon>A+B+\varepsilon .$ For any $N>0$ there exists an $n>N$ such that

$$

a_{n}+b_{n}>L-\varepsilon \text {. }

$$

On the other hand, there exists $N_{1}$ such that for all $n>N_{1}$,

$$

a_{n}<A+\frac{\varepsilon}{2}

$$

and there exists $N_{2}$ such that for all $n>N_{2}$,

$$

b_{n}<B+\frac{\varepsilon}{2}

$$

But then if $N=\max \left(N_{1}, N_{2}\right)$, then for any $n>N$ we have

$$

a_{n}+b_{n}<A+B+\varepsilon<L-\varepsilon

$$

contradicting $(0.1)$.

2. This exercise shows that even in a complete metric, a closed and bounded set need not be compact. Let

$$

l^{\infty}(\mathbb{C}) ;=\left\{\left\{a_{k}\right\}_{k=1}^{\infty} \mid \exists M>0 \text { such that }\left|a_{k}\right|<M \text {, for all } k\right\} .

$$

That is, $l^{\infty}(\mathbb{C})$ is the set of all bounded sequences of complex numbers. Note that the $M$ will vary from sequence to sequence. For two sequences $A=\left\{a_{n}\right\}$ and $B=\left\{b_{n}\right\}$, define

$$

d(A, B)=\sup _{k}\left|a_{k}-b_{k}\right|

$$

(a) For any two sequences $A, B \in l^{\infty}(\mathbb{C})$, show that $d(A, B)$ is a finite number.

Solution: Suppose $A=\left\{a_{n}\right\}$ and $B=\left\{b_{n}\right\} .$ Let $\alpha=\sup \left|a_{n}\right|$ and $\beta=\sup \left|b_{n}\right|$, which are finite since $A, B \in l^{\infty}(\mathbb{C}) .$ Then for any $n$,

$$

\left|a_{n}-b_{n}\right| \leq\left|a_{n}\right|+\left|b_{n}\right| \leq \alpha+\beta

$$

Hence the sup is finite and $d(A, B)$ is finite. (b) Let $E_{n}$ be the sequence with 1 at the $n^{t h}$ place and zero everywhere else, and let $O$ be the sequence with zeroes everywhere. What is $d\left(E_{n}, O\right) ? d\left(E_{n}, E_{m}\right)$ for $n \neq m ?$

Solution: From the definition, $d\left(E_{n}, O\right)=1$ and $d\left(E_{n}, E_{m}\right)=1$ whenever $n \neq m$

(c) Show that $\left(l^{\infty}, \mathbb{C}\right)$ is a complete metric space.

Solution: The symmetry axiom is trivial. Suppose $A=\left\{a_{k}\right\}$ and $B=\left\{b_{k}\right\}$. THen clearly $d(A, B) \geq 0 .$ Also if $d(A, B)=0$, then clearly $a_{k}=b_{k}$ for all $k$ and so $A=B .$ This shows that $d$ is positive definite. For the triangle inequality, note that if $A=\left\{a_{k}\right\}, B=\left\{b_{k}\right\}$ and $C=\left\{c_{k}\right\}$ are three elements, for each fixed $k$,

$$

\left|a_{k}-c_{k}\right| \leq\left|a_{k}-b_{k}\right|+\left|b_{k}-c_{k}\right|

$$

Taking sup it follows that $d(A, C) \leq d(A, B)+d(B, C)$ which verifies the triangle inequality. Next we show completeness.

Let $\left\{A_{n}\right\}$ be a Cauchy sequence in $l^{\infty}(\mathbb{C}) .$ So for every $\varepsilon>0$, there exists an $N>0$ such that whenever $n, m>N$, we have

$$

d\left(A_{n}, A_{m}\right)<\varepsilon .

$$

If we denote $A_{n}=\left\{a_{n k}\right\}$, then for all $n, m>N$ and all $k$,

$$

\left|a_{n k}-a_{m k}\right|<\varepsilon .

$$

For each $k$, we see that $\left\{a_{n k}\right\}_{n=1}^{\infty}$ is a Cauchy sequence, and since $\mathbb{C}$ is complete, the sequence converges to a $p_{k} \in \mathbb{C}$. We denote the sequence $\left\{p_{1}, p_{2}, \cdots, p_{k}, \cdots\right\}$ by $P$. Since $\left\{A_{n}\right\}$ is Cauchy, it is also bounded, and so there is some $M$ such that $\left|a_{n k}\right|<M$ for all $n, k$. In particular $\left|p_{k}\right|<M$ for each $k$, and so $P \in l^{\infty}(\mathbb{C})$.

Claim. $\lim _{n \rightarrow \infty} A_{n}=P .$

Proof. Fix $\varepsilon>0$ and consider $N$ as above. For each $k$, since $a_{n k} \stackrel{n \rightarrow \infty}{\longrightarrow} p_{k}$, there exists a $n_{k}>N$ such that $\left|a_{n_{k} k}-p_{k}\right|<\varepsilon .$ But then for any $n>N$, from the above inequality, we know that $\left|a_{n_{k} k}-a_{n k}\right|<\varepsilon$, and so by triangle inequality,

$$

\left|a_{n k}-p_{k}\right|<\left|a_{n_{k} k}-p_{k}\right|+\left|a_{n_{k} k}-a_{n k}\right|<2 \varepsilon .

$$

This is true for all $k$ and all $n>N$. That is for all $n>N$, we have that

$$

d\left(A_{n}, P\right)=\sup _{k}\left|a_{n k}-p_{k}\right|<2 \varepsilon

$$

and this proves the claim.

We have thus shown that any Cauchy sequence in $l^{\infty}(\mathbb{C})$ converges.

(d) Show that the set $B_{1}(O)$ is closed and bounded, but not compact. Hint. Show that the sequence $E_{n}$ from above has no limit point.

Solution: Consider the sequence $\left\{E_{n}\right\}$ from above. Then $d\left(E_{n}, 0\right)=1$, the sequence is contained in $B_{1}(O) .$ Next, since $d\left(E_{n}, E_{m}\right)=1$ for all $n \neq m$, clearly the sequence $E_{n}$ is not Cauchy and cannot have a limit point. But every infinite sequence in a compact metric space has a limit point, and so $\overline{B_{1}(O)}$ cannot be compact.

3. The aim of this exercise is to generalize Bolzano-Weierstrass theorem. A metric space $X$ is called totally bounded if for any $r>0$, there is a finite covering of $X$ by balls of radius $r .$ That is, given any $r>0$, we can find finitely many points $p_{1}, p_{2}, \cdots, p_{N}$ (where $N$ will in general depend on $r$ ) such that

$$

X=\cup_{k=1}^{N} B_{r}\left(p_{k}\right)

$$

(a) Show that a set in $\mathbb{R}^{n}$ (considered as a metric subspace) is totally bounded if and only if it is bounded.

Solution: Trivially any totally bounded set is bounded (in any metric space). For the converse, suppose $E \subset \mathbb{R}^{n}$ be a bounded set and let $r>0$ be given. Then $\bar{E}$ being closed and bounded is also compact. Then we can find points (see exercise 3 in assignment 2) $q_{1}, \cdots q_{N} \in \bar{E}$ such that

$$

\bar{E} \subset \cup_{k=1}^{N} B_{\frac{r}{2}}\left(q_{k}\right)

$$

The only problem is that some of the points $q_{k}$ might not be in $E .$ If $q_{k} \in E$, we let $p_{k}=q_{k}$. Trivially $B_{r / 2}\left(q_{k}\right) \subset B_{r}\left(p_{k}\right) .$ If $q_{k} \notin E$, then $q_{k}$ has to be a limit point of $E .$ So there exists a $p_{k} \in B_{r / 2}\left(q_{k}\right) \cap E .$ But then by the triangle inequality,

$$

B_{r / 2}\left(q_{k}\right) \subset B_{r}\left(p_{k}\right) .

$$

Then clearly

$$

E \subset \cup_{k=1}^{N} B_{r}\left(p_{k}\right)

$$

and $p_{k} \in E$ for all $k$.

(b) A metric space is compact if and only if it is complete and totally bounded.

Solution: We have already seen in class that a compact metric space is complete. It is also totally bounded. A finite covering by balls of radius $r$ can be obtained by extracting a finite sub-cover of any covering by balls of radius $r$. For the converse, suppose there is a covering $\left\{G_{\alpha}\right\}$ such that no finite sub-collection covers $X$. Take a finite covering by balls of radius $r=1$ (possible since $X$ is totally bounded). The there is at least one ball such that no finite sub-collection from $\left\{G_{\alpha}\right\}$ covers it. Call this ball $B_{0}$. Since subsets of totally bounded sets are also totally bounded, there is a finite covering of $B_{0}$ by balls of radius $1 / 2$. Again there is at least one ball such that no finite sub-collection from $\left\{G_{\alpha}\right\}$ covers it. Call this ball $B_{1}$. Repeating this process, we end up with a sequence of balls $\left\{B_{n}\right\}$ such that

- $B_{0} \supset B_{1} \supset B_{2} \cdots \supset B_{n} \cdots .$

- diameter of $B_{n}$ is $2^{-(n-1)}$.

- No ball $B_{n}$ is covered by any finite sub-collection of $\left\{G_{\alpha}\right\}$.

Now pick a point $x_{n} \in B_{n}$. We claim that the sequence $\left\{x_{n}\right\}$ is Cauchy. To see this, observe that $m>n$, then $x_{n}, x_{m} \in B_{n}$, and so $d\left(x_{n}, x_{m}\right)<2^{-(n-1)}$. So given any $\varepsilon$ if $N$ such that $2^{-(N-1)}<\varepsilon$, then for any $n, m>N, d\left(x_{n}, x_{m}\right)<\varepsilon .$ So the sequence is Cauchy and by completeness there is a limit, which we denote by $p$. Now $p \in X$ and hence $p \in G_{\alpha}$ for some index $\alpha$. Since $G_{\alpha}$ is open, $B_{r}(p) \subset G_{\alpha}$ for some $r>0$. Since $x_{n} \rightarrow p$, there is some $n$ such that $2^{-(n-1)}<r / 2$ and $x_{n} \in B_{r / 2}(p) .$ By triangle inequality and the fact that $\operatorname{diam}\left(\mathrm{B}_{\mathrm{n}}\right)=2^{-(\mathrm{n}-1)}$, it is easy to see that $B_{n} \subset B_{r}(p)$, and so is covered by $G_{\alpha}$. But this contradicts the third property above.