Definition of a bounded subset in a normed vector space

Question

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...!

Answer 1

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.

Answer 2

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.