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For a sequence $\{a_j\}$ in $\mathbb{R}$, we say that $\{a_j\}$ is bounded if and only if there exists $M>0$ such that for all $j$, $|a_j| \le M$. What is the definition of a bounded sequence in $\mathbb{R}^n$ with the standard metric? Is it as follows?

$\{a_j\} \subseteq \mathbb{R}^{n}$ is a bounded sequence if and only if there exists some $a \in \mathbb{R}^{n}$ and there exists $\varepsilon>0$ such that for all $j$, $d(a_j, a) = \sqrt{\sum_{m=1}^{n} (a_{j,m} - a_m)^2} < \varepsilon$?

Can I also conclude that a sequence $\{a_j\} \subseteq \mathbb{R}^{n}$ is bounded if and only if its component sequence is bounded? I.e, $\{a_j\} \subseteq \mathbb{R}^{n}$ is bounded $\iff$ $\{a_{j, m}\} \subseteq \mathbb{R}$ is bounded for all $m = 1, \cdots, n$?

elbarto
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2 Answers2

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Yes, although the choice of the symbol $\epsilon$ is not normal. An alternative equivalent way to formulate it is that $d(a_j, a_k)$ is bounded, that is that there is an $D$ such that always $d(a_j, a_k)<M$.

That your observation is true can easily be proven. You have if $d(a_j, a)<L$ then you have that the component-wise distance is less than $L$ as well. Also $d(a_j, a)$ is less than $n$ times the maximal component-wise distance.

skyking
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  • What is usually the symbol for it? And how about my statement regarding the equivalence, is that also true? – elbarto Sep 13 '17 at 10:47
  • The bounding constant is normally denoted by a capital latin letter. For example $M$. – skyking Sep 13 '17 at 10:56
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$\{a_j\} \subseteq \mathbb{R}^{n}$ is bounded $ \iff$ there is $c \ge 0$ such that $||a_j|| \le c$ for all $j$

($|| \cdot||$ is a norm on $\mathbb{R}^{n}$, which norm is not relevant, since all norms on $\mathbb{R}^{n}$ are equivalent.).

Your second question. If $x=(x_1,...,x_n) \in \mathbb R^n$, then we have

$|x_j| \le \sqrt{x_1^2+...+x_n^2} \le |x_1|+...+|x_n|$.

Consequence:

$\{a_j\} \subseteq \mathbb{R}^{n}$ is bounded $ \iff \{a_{j, m}\} \subseteq \mathbb{R}$ is bounded for all $m=1,...,n$.

Fred
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