# If every sequence $(x_{n}) \subset X$ and $(\lambda_{n}) \subset \mathbb{R}$ we have $\lim \lambda_{n}x_{n} = 0$, then $X$ is bounded

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?

• Don't you need $\lambda_n\to0$? Jan 4, 2019 at 22:18
• 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 $n$?
– user296602
Jan 4, 2019 at 22:20
• @SmileyCraft, oh, yes! I forgot to write! Jan 4, 2019 at 22:20
• 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.
– user296602
Jan 4, 2019 at 22:21
• @T.Bongers, thanks for the hint! I'll try to correct the mistakes. Jan 4, 2019 at 22:22

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.

• 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 $A_{2}$? Jan 4, 2019 at 22:38
• 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.
– user296602
Jan 4, 2019 at 22:39

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).

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 $$0$$.

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.