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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?

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  • $\begingroup$ Don't you need $\lambda_n\to0$? $\endgroup$ Jan 4, 2019 at 22:18
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    $\begingroup$ There are a couple issues with your part (c); one is your assumption that you can write $x_n$ in coordinates in this way (and why would you need to?), and second is that there is a quite large gap - some magic has happened when you say "Thus." What would happen if $\lambda_n = 1$ for all $n$? $\endgroup$
    – user296602
    Jan 4, 2019 at 22:20
  • $\begingroup$ @SmileyCraft, oh, yes! I forgot to write! $\endgroup$
    – Lucas
    Jan 4, 2019 at 22:20
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    $\begingroup$ For another issue, note that in part (b) you're implicitly assuming that $V = \mathbb{R}$. This is supposed to be a general normed space, although your idea is on the right track. $\endgroup$
    – user296602
    Jan 4, 2019 at 22:21
  • $\begingroup$ @T.Bongers, thanks for the hint! I'll try to correct the mistakes. $\endgroup$
    – Lucas
    Jan 4, 2019 at 22:22

4 Answers 4

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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 $\|\lambda_n x_n\| \not\to 0$. Constant norm works.

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  • $\begingroup$ Awesome, I got it! But for (b), I have some troubles. I thinking in $A_{1}, A_{2}$ such that $\partial A_{1} \cap \partial A_{2} \neq \emptyset$ and $V\setminus(A_{1}\cup A_{2}) = \partial A_{1} \cap \partial A_{2}$. How can I ensure that $A_{1}$ and $A_{2}$ exist? Or how can I make explicit $A_{1}$ and $A_{2}$? $\endgroup$
    – Lucas
    Jan 4, 2019 at 22:38
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    $\begingroup$ For your attempt, you used "positive" and "negative" as the separation condition. This doesn't work for the norm, because you cannot have a negative norm value... but there's no need to cut at zero. Try cutting somewhere else. $\endgroup$
    – user296602
    Jan 4, 2019 at 22:39
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For the converse: suppose $X$ is not bounded. Then it contains some sequence $(x_n)$ such that $(\|x_n\|)\to\infty$ (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 $(\lambda_nx_n)\not\to 0$, since $(\|\lambda_nx_n\|) = (1) \not\to 0$. 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 $\lim\lambda_n = 0$ bit).

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

Assume that $X$ is unbounded. Hence for every $n\in\mathbb{N}$ exists $x_n \in X$ such that $\|x_n\| \ge n$. Consider the sequence $(\lambda_n)_n = \left(\frac1n\right)_n$. We have $\lim_{n\to\infty} \lambda_n = 0$ but $ \|\lambda_nx_n\| \ge 1, \forall n\in\mathbb{N}$ so it cannot converge to $0$.

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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 $z\in F_1$ and $z\in F_2$ since $F_1\supset \overline{A_1}$ and $F_2\supset\overline{A_2}$, 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.

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