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I'm having a hard time trying to understand what is exactly a bounded set (or subset) in a normed vector space.

This is where I am so far in my "understanding":

  • Let $(V, \|\cdot\|)$ be a normed vector space. We can define $\|y-x\| =: d(x,y) =: d$ such that $(V,d)$ is a metric space.
  • In a metric space (such as in $(V,d)$), a set $A$ is bounded if it is contained in a ball of finite radius, that is: $A \subset V$ is bounded if there exists $x \in V$ and $0 < R < \infty$ such that $A \subset B(x,R)$, i.e. such that $d(x,y) \le R$ for all $y \in A$.
  • A set in a topological vector space (such as a normed vector space) is called bounded if every neighborhood of the zero vector can be inflated to include the set. (Source: Wikipedia)
  • Apparently, a subset $A$ of a normed vector space $(V, \|\cdot\|)$ is said to be bounded if there exists $C > 0$ such that $\|v\| \le C$ for all $v \in A$.

What I don't understand is that, to me, it would be "logical" to say that a subset $A$ of a normed vector space $(V, \|\cdot\|)$ is bounded if there exists $x \in V$ and $C>0$ such that $d(x,y) = \|y-x\| \le C$ for all $y \in A$. But it seems like the "real" definition of a bounded (sub)set of a normed vector space is considering only $x=0$, that is, $x$ being the zero vector. Why is that?

Furthermore, I'm not sure to understand the definition given by Wikipedia. Since a neighborhood is a set containing an open set containing a particular point, and considering that this point could be the zero vector, how can we use that information to get the actual definition of a bounded subset of a normed vector space (i.e. the last definition I mentioned in my list above)?

As you can see, I'm quite confused about all this... And I'm a little bit exhausted since I've tried to get it for hours now. Any help would be greatly appreciated...!

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It's the same. Let $(V, \|\cdot\|)$ be a normed space and let $A\subset V$. If there exist $x\in V$ and $R > 0$ such that $A \subset \overline{B}_R(x)$, then you have $$ \|y\| \leq \|y-x\| + \|x\| \leq R + \|x\| =: C \qquad \forall y \in A, $$ hence $A$ is bounded in the sense of your first definition.

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  • $\begingroup$ Jesus I'm such an idiot. I actually found that $\| y \| \le R + \| x \|$ but I didn't think of defining it as $C$. Thanks a lot! $\endgroup$ – justdoit May 26 '17 at 15:38
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I also faced the same problem. But I got it clear by understanding that from the actual definition of a bounded set A in a normed linear space it can be proved that A is contained in a closed sphere and vice versa. Let's see the proof, Our claim is in a normed linear space E , a set A is bounded iff A is contained in a closed sphere S[€,0] with centre origin . Proof:- if a is bounded then there exists a∈E and K>0 such that d(x,a)=||x-0||< or = K for all x∈A 【this is the actual definition of a bounded set in a normed linear space】 But then ||x-0|| =||x||≤||x-a||+||a||≤K+||a||for all x∈A, and hence A is a subset of S[€,0], where € = K+||a|| On the other hand if A is a subset of S[€,0], then for any x∈A, d(x,0)=||x-0||≤€. Hence A is bounded.

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There is an important distinction that has not been made here yet. While the definition for boundedness in a normed vector space and boundedness in the metric space induced by that norm do coincide, the bounded sets in a topological vector space and the bounded sets in the metric space induced by some compatible metric need not be the same.

Rudin goes into the details in Functional Analysis section 1.29.

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