If $A$ is bounded, then $\overline{A}$ is bounded

Question

Let $A,B$ be subsets of a normed vector space $X$ and $\alpha\in K$. Show that

a) $\alpha\overline{A} = \overline{\alpha A}$

b) $\overline{A}+\overline{B}\subseteq \overline{A+B}$

c) If $A$ is bounded, then $\overline{A}$ is bounded

a)

Pick an element $\overline{a}$ in $\overline{A}$. By definition it is an element such that it has a sequence $a_n$ such that $a_n\to \overline{a}$ which means $\alpha a_n \to \alpha\overline{a}$. Isn't an element of the space $\overline{\alpha A}$ one such that there exists a sequence $\alpha a_n$ such that $\alpha a_n$ converges to it? So it should prove $\alpha\overline{A}\subseteq \overline{\alpha A}$

Now for $ \overline{\alpha A} \subseteq \alpha\overline{A}$, we must pick an element of the set $\overline{\alpha A}$ which is simply an element such that there exists a sequence $\alpha a_n$ that converges to it. Hw should I conclude here?

b)

Pick an element $\overline{a}\in \overline{A}$ and $\overline{b}\in \overline{B}$. By definitiion there exists sequences $a_n, b_n$ such that $a_n\to \overline{a}$ and $b_n\to \overline{b}$.

$$(a_n+b_n)\to \overline{a}+\overline{b}$$

by the property of limits. So we showed that for every element $A+B$ there is a sequence $a_n+b_n$ that converges to it, therefore it's proved.

c)

Suppose that $\overline{A}$ is unbounded. Then for sufficiently large $M>0$, there exists $a$ such that $a_n\to a$ with $a_n\in A$, but $$||a||\ge M$$

Since $a_n\to a$, $n>n_0\implies ||a_n-a||<\epsilon$

How to use the triangular inequality in $||a_n-a||$ to derive that $A$ is unbounded?

Answer 1

I think it would be better to go on the other direction. Suppose $A$ is bounded. Then there is $M>0$ such that $||x||\leq M$ for all $x\in A$. Now let $x\in \bar{A}$. There is a sequence $\{x_n\}_{n=1}^\infty\subseteq A$ such that $x_n\to x$. Now given any $\epsilon>0$ there is $n\in\mathbb{N}$ such that $||x_n-x||<\epsilon$. Hence:

$||x||\leq ||x_n||+||x-x_n||< M+\epsilon$

Since this is true for all $\epsilon>0$ we conclude that $||x||\leq M$.

Answer 2

Your argument for part (b) looks OK except that the last two sentences are redundant and even wrong. What you have shown is that $\bar{a}+\bar{b} \in A+B$ for all $\bar{a}\in\overline{A}$ and $\bar{b}\in\overline{A}$ which implies that $\overline{A}+\overline{B} \subseteq \overline{A+B}$.

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To conclude part (a), just use the same argument you used again.

if $\alpha=0$, as pointed out by Daniel Schepler, both sides are empty when $A=\emptyset$ and both sides are $\{0\}$ when $A$ is not-empty. So, we have $\overline{\alpha A}=\alpha\overline{A}$ when $\alpha=0$.

So, assume $\alpha \neq 0$. If $x\in\overline{\alpha A}$ then you will find a sequence of elements $\{\alpha a_n\}_{n=1}^{\infty}$ in $\alpha A$ such that $\lim_{n\to\infty}\alpha a_n = \alpha\lim_{n\to\infty} a_n = x$. This shows that $\frac{1}{\alpha}x \in \overline{A}$ which implies that $x \in \alpha \overline{A}$. Hence, $\overline{\alpha A} \subseteq \alpha \overline{A}$.

$\fbox{Q.E.D.}$

To show (c), let's say that $A$ is bounded. So, there exists a natural number $M$ such that $\forall a\in A: \|a\| \leq M$. Now for any $x \in \overline{A}$ there exists a sequence $\{a_n(x)\}_{n=1}^{\infty}$ in $A$ such that $\lim_{n\to\infty}a_n(x) = x$. Because of the continuity of norm and because $a_n(x) \in A$, we have

$$\forall x \in \overline{A}, \forall n\in\mathbb{N}: \|a_n(x)\|\leq M$$ $$\forall x \in \overline{A}: \lim_{n\to\infty} \|a_n(x)\| = \| \lim_{n\to\infty} a_n(x) \|\leq M$$ $$\forall x \in \overline{A}: \|x \|\leq M$$

which proves that $\overline{A}$ is bounded. $\fbox{Q.E.D.}$

Remark: (uniform) continuity of norms is an implication of the reverse triangle inequality: $$\bigg| \|x\| - \|y\| \bigg| \leq \|x-y\|$$

Answer 3

For c) suppose $\forall a\in A\;(\|a\|< r).$ Let $r'=1+r.$ Now if $\bar a\in \bar A,$ there exists $a\in A$ such that $\|\bar a -a\|<1,$ and $\|a\|<r,$ so $$\|\bar a\|=\|(\bar a -a)+a\|\le \|\bar a - a\|+\|a\|<1+r=r'.$$

Alternatively $$\lim_{n\to \infty}\|a_n-\bar a\|=0 \implies \lim_{n\to \infty}\|a_n\|=\|\bar a\|$$ so if $\{\|a_n\|:n\in \Bbb N\}\subset [0,s]$ then $\|\bar a\|=\lim_{n\to \infty}\|a_n\|\in [0,s].$ In particular if $s_0$ is the least $s\in [0,\infty)$ such that $\forall a\in A\;(\|a\|\le s)$ then $\forall \bar a \in \bar A\;(\|\bar a\|\le s_0).$

Remark: We have $\|\bar a\|=\|(\bar a -a)+a\|\le \|\bar a -a\|+\|a\|,$ so $$\|\bar a\|-\|a\|\le \|\bar a -a\|.$$ Interchanging $a$ with $\bar a$ we have $$\|a\|-\|\bar a\|\le \|a-\bar a\|.$$ Therefore $|\;\|\bar a\|-\|a\|\;|\le \|\bar a -a\|.$ Hence if $\|\bar a -a_n\|\to 0$ then $(\|\bar a\|-\|a_n\|)\to 0.$