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Let $(V,Vert cdot Vert)$ a normed vector space.

(a) Prove that if $A,B subset V$ with $A$ open, then $A + B$ is open.

(b) Is there disjoint open sets $A_{1},A_{2}$ for which there is no disjoint closed sets $F_{1},F_{2}$ such that $A_{1} subset F_{1}$ and $A_{2} subset F_{2}$?

(c) Prove that if $X subset V$ is bounded, then for every sequence $(x_{n}) subset X$ and $(lambda_{n}) subset mathbb{R}$ with $lim lambda_{n} = 0$ we have $lim lambda_{n}x_{n} = 0$. What about the converse?

My attempt.

(a) Let $A$ be an open set. So, $A + b = {a+b mid a in A}$ is translation, therefore, is open. But $displaystyle A + B = bigcup_{b in B}(A+b)$, then $A+B$ is open.

(b) Take $A_{1} = mathbb{R}_{>0}$ and $A_{2} = mathbb{R}_{<0}$, because $overline{A_{1}} = A_{1}cup{0}$ and $overline{A_{2}} = A_{2}cup{0}$

(c) If $(x_{n})$ is a sequence in $X$, then $(x_{n})$ is bounded. But
$$(x_{n}) = ((x_{1,n}),(x_{2,n}),...,(x_{k,n}))$$
where each $(x_{i,n})$ is a bounded sequence in $mathbb{R}$. Thus, $lim lambda_{n}x_{i,n} = 0$ for each $i$, then $lim lambda_{n}x_{n} = 0$.

The converse seems true, but I cannot prove. Can someone help me?

Your part (a) is fine, assuming that you know translation is continuous.

Your part (b) is on the right track, but it assumes that $V$ is the continuum, whereas it should be a normed vector space. You have the right idea of separating two sets by a very small set, but you shouldn't rely on coordinates (since you don't have any coordinates...). Use the norm instead.

Your part (c) has a similar issue: unless you're in a finite dimensional space, there is neither reason nor justification for writing the sequence in coordinates. Again, you need to be using the norm. Notice that

$$|lambda_n x_n| = |lambda_n| cdot |x_n| le |lambda_n| sup_{x in X} |x| to 0.$$

The converse is in fact true, and can be handled similarly; choose an unbounded sequence ${x_n}$ and craft $lambda_n$ depending on $x_n$ so that $|lambda_n x_n| notto 0$. Constant norm works.

The converse is true, we'll prove it by contraposition:

Assume that $X$ is unbounded. Hence for every $ninmathbb{N}$ exists $x_n in X$ such that $|x_n| ge n$. Consider the sequence $(lambda_n)_n = left(frac1nright)_n$. We have $lim_{ntoinfty} lambda_n = 0$ but
$ |lambda_nx_n| ge 1, forall ninmathbb{N}$ so it cannot converge to $0$.

For the converse: suppose $X$ is not bounded. Then it contains some sequence $(x_n)$ such that $(|x_n|)toinfty$ (take $x_n$ to be an example given by the negation of the definition of "bounded" with the constant set to $n$). Take $lambda_n = frac{1}{|x_n|}$. Then $(lambda_n)to 0$, but $(lambda_nx_n)notto 0$, since $(|lambda_nx_n|) = (1) notto 0$. Thus, taking the contrapositive, we have the converse (NB: you've got a bit missing out of the contrapositive you've given in the title: you've dropped the $limlambda_n = 0$ bit).

Your approach to part (a) is correct; to be more explicit, you could show why $A+b$ is open.

You have the right idea for part (b), but it needs to be stated a bit more generally. Let $A_1$ and $A_2$ be disjoint open balls that both have some point $z$ in their closure. Then if $F_1$, $F_2$ are closed sets containing $A_1$, $A_2$, respectively, we have $zin F_1$ and $zin F_2$ since $F_1supset overline{A_1}$ and $F_2supsetoverline{A_2}$, so $F_1$ and $F_2$ cannot be disjoint. To construct such $A_1$ and $A_2$, let $x,y$ be distinct points in $V$ with $|x|=|y|$ and consider the open balls centered at $x$ and $y$ with radius $frac 12 |x-y|$. Then the point $frac12(x+y)$ is in the closure of both of these balls.

For part (c) I will defer to the accepted answer.