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?


  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


    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!

  • 4
    $\begingroup$ What does "inflated" mean in a general metric space? $\endgroup$ – Chris Eagle Feb 24 '12 at 15:59
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    $\begingroup$ Note that, despite the difference in the definitions, whenever $V$ is normed, they agree (if you turn $V$ into a metric space by defining the metric $d(x, y) = \|x - y\|$). $\endgroup$ – student Feb 24 '12 at 16:04
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    $\begingroup$ @Tim: Consider $\mathbb Z$ with the usual distance. Then $B(0,\frac12)$ and $B(0,\frac16)$ are the same set, namely $\{0\}$. This means that it is not well-defined what it would mean to inflate the set $\{0\}$ by a factor of 5. $\endgroup$ – Henning Makholm Feb 24 '12 at 16:56
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    $\begingroup$ Henning's point is a good one, but somewhat different from mine: Consider $\{0,3,4,7\}$ with the usual distance. Then $B(3,2)=B(4,2)$. $\endgroup$ – Harald Hanche-Olsen Feb 24 '12 at 17:31
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    $\begingroup$ In an ultrametric space (a metric space where the triangle inequality is replaced by the much more restrictive ultrametric inequality), any element of your ball is "at the center" of the ball. An example of such a space is the p-adic numbers (which, by the way, form a complete metric space). Hence, you now have a class of "counter-examples", as pointed out by @HaraldHanche-Olsen $\endgroup$ – M Turgeon Feb 24 '12 at 18:43

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 $0$.

  • $\begingroup$ Thanks a lot! +1. I am reading it slowly like a snail now ... $\endgroup$ – Tim Feb 27 '12 at 21:51
  • $\begingroup$ @Tim: Have you finally managed to go through my wannabe English? $\endgroup$ – savick01 Feb 28 '12 at 18:46
  • $\begingroup$ Nothing bad about your English! It is my math that is not good. Did you come up the analogy between boundness and convergence by yourself, or find it somewhere else? $\endgroup$ – Tim Feb 28 '12 at 22:25
  • $\begingroup$ @Tim: Sorry, I can't give you any references, it's my own idea. I added a little clarification to the proposition - previously multiplying by $0$ was allowed as a way of deflating, which wasn't particularly reasonable... $\endgroup$ – savick01 Feb 29 '12 at 21:23

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.


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.

  • $\begingroup$ +1 Thanks! What does "a lettering "B(a,r)"" mean? In particular, what is "lettering"? $\endgroup$ – Tim Feb 24 '12 at 19:27
  • $\begingroup$ @Tim: Thanks and... you're welcome. "Lettering" is from a dictionary, I meant something like "word" or "writing". Not the set itself, but one of its possible descriptions. $\endgroup$ – savick01 Feb 24 '12 at 19:33

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