Why is boundedness defined so differently in a topological vector space and in a metric space?

Question

From Wikipedia

A subset S of a metric space (M, d) is bounded if it is contained in a ball of finite radius

Also from Wikipedia

a set in a topological vector space is called bounded or von Neumann bounded, if every neighborhood of the zero vector can be inflated to include the set.

I was wondering why boundedness in a topological vector space is defined so differently from in a metric space?

Specifically,

1. Can boundedess of a subset in a metric space be defined in the same way as in a TVS as

   a subset of a metric space is bounded, if every open ball with any finite radius can be "inflated " (in your proper definition, please see comments below) to include the subset?

2. Can a bounded subset of a TVS be defined by

   there exists a subset of the TVS with some "special property" that can be inflated to include the set

   or

   there exists a subset of the TVS with some "special property" to include the set

   Added: Note: The purpose for looking for alternative definitions in TVS case, is that the original definition requires to check for every neighbourhood of $0$ and their all possible inflations, which is more complicated than finding an open ball in metric case. Do you also think this way?

Thanks and regards!

Answer 1

I have some new ideas regarding the second part of your question after those three days:)

Intuition, definition of convergence

Note that by bounded we mean 'small' in some sense or 'close' to zero (if we have zero). In topology, open (nonempty ;-) ) sets are usually considered big (contrary to sets with empty interior or nowhere-dense [Baire category theorem]). Let's consider convergence - convergent sequence is also 'close' to the limit. Look at the definition of convergence: it is very similar to the definition of a bounded set in a TVS. We need all the neighbourhoods there (or at least base or subbase neighbourhoods), because one neighbourhood is just too big - even if we allow deflating it (say, a sequence is convergent to zero if there is some special neighbourhood such that - no matter how much deflated - contains the sequence without a finite number of elements). I don't know about any special subset that could be used here (of course an open ball in a normed space is great, but that's not the case).

Example justifying this intuition and how to define a subset with a 'special property'

Let $F=\{f: [0,1] \to \mathbb{R} \}$. It is a vector space over $\mathbb{R}$. Let's make it a topological vector space by establishing the topology of pointwise convergence (equivalently: product topology $\prod_{i\in [0,1]}\mathbb{R}$). A subbase of such topology is $\{A_{i,U}\}_{i \in[0,1], U=U^\circ\subseteq\mathbb{R}},$ where $$A_{i,U} = \{f \in F | f(i) \in U\}.$$ We can see, that in this space open base sets (even base ones!) are indeed very big: we put restrictions only on a finite number of values of $f$ - on the rest of the domain it may be whatever (in particular, arbitrarily big). And what do we get by applying the definition of a bounded set? A set $K\subseteq F$ is bounded if for each $i$ and for each subbase neighbourhood $A_{i,U}$ of the zero function, we can inflate $A_{i,U}$ to contain $K$, which means, that for each $i$ $\{f(i) | f \in K\}$ has to be bounded. In other words, $K$ is bounded, if there is $g:[0,1]\to\mathbb{R}_+$ and a set with a 'special property' $B_g$ defined as $$B_g=\left \{ f \in F \ \Big| \ \forall_{i \in [0,1]} \ |f(i)| \leq g(i) \right \}$$ such that $K \subseteq B_g$. We can see, that the definition is very reasonable - from the topology of pointwise convergence, we get pointwise boundness.

Nonetheless, I don't know how to define 'a set with "special property"' in a general case. Actually, it seems hard even to define $B_g$ in terms of the topology on $F$ if we do not use our specific subbase $\{A_{i,U}\}_{i \in[0,1], U=U^\circ\subseteq\mathbb{R}}$ appriopriately.

One more remark about convergence

Assume that the topology on a TVS over $\mathbb{F}$ is generated by translations of balanced sets (somewhat less than local convexity). Assume moreover for clarity, that the absolute value on $\mathbb{F}$ is nontrivial (if it is trivial, then the sequence in the proposition doesn't have to be convergent to $0$).

A set $K$ is bounded iff for (any)/(each)/(one canonical) sequence $(t_n)\in \mathbb{F}^*$ st. $t_n \to 0$ and for any neighbourhood $O$ of the $0$-vector the sequence of sets $t_n K$ falls into $O$ after finitely many steps.

Equivalently: iff for (any)/(each)/(one canonical) sequence $(t_n)\in \mathbb{F}^*$ st. $t_n \to 0$ any sequence $t_n k_n$ converges to zero (where $k_n \in K$).

This shows that boundness and convergence are very close to each other and we probably can't avoid examining all the neighbourhoods of $0$.

Answer 2

I don't know if my ideas aren't just trivial, but maybe someone will found a bit of sense in them.

As it's been already said, there is no natural way of defining inflation in a non-linear space. In comments you wrote about inflation of a ball - it also is not well defined if you mean the set, not just a lettering "$B(a,r)$". But that's not the key point - what is important is the thing said by Leandro: if the space is a metric space (or - more specifically - normed), then both definitions are trivially equivalent (every neighbourhood contains a ball, and we inflate that ball just by taking a ball with some big radius [it's not well defined, but I hope that what I want to express is clear]).

So, here we have the answer: the definition is different, in order to give some definition of boundness, when there is no metric. And if we have a metric, then the chosen definition is equivalent to the traditional one - it suggests that the new definition is reasonable.

The above ideas shouldn't be new to you, since you've recently asked some questions about uniform spaces - it is closeness (convergence) without metric. On the other hand, there is boundness without metric, and that's what coarse spaces are for.

Unfortunately, I don't know how to address the second part of your question, the one about 'special properties'. One thing is quite clear: we don't need to worry too much about inflating or neighbourhoods of $0$, because multiplying by a non-zero scalar or translation by a vector is a homeomorphism between $V$ and $V$. In particular, both your ideas from the second point are equivalent.

Answer 3

Boundedness is always relative to the non-topological structure we have. For example if $(X,d)$ is a metric space, then $d_b(x,y) = \min\{d(x,y),1\}$ defines another metric on $X$ which is uniformly equivalent to $d$ and under which all subsets of $X$ are bounded.

Thus, if we care about topological invariants (and we do!) boundedness is not a particularly useful notion. So when we talk about boundedness, it's usually because this extra structure that gives us a notion of boundedness comes with some "nice" consequences thereof, e.g. a vector space $X$ might be a normed linear space or a Montel space.