I learned about Banach spaces a few weeks ago. A Banach space is a complete normed vector space. This of course made me wonder: are there unnormed vector spaces? If there are, can anyone please provide any examples?

Some thoughts:

A complete space is where all Cauchy sequences converge.

A normed vector space is a vector space (say, over $\mathbb{R})$ on some norm $N$ (which is a function that maps $N\to\mathbb{R}$), where the norm obeys the triangle inequality, the norm of a vector is non-negative, and if you have a scalar being multiplied by a vector, you can factor the scalar out, but it'll have absolute value braces.

I'm not really sure what is needed in order to have an unnormed vector space (perhaps the vector space necessarily needs to be infinite dimensional?). Perhaps something really weird like the zero space?

Thanks for any insight.

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    $\begingroup$ There are certainly incomplete spaces. $\endgroup$ – amcalde Dec 20 '14 at 4:34
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    $\begingroup$ Yes, I know there are spaces which are not Banach. I can show that such a space is not Banach by showing that some Cauchy sequence doesn't converge. But I've never shown a space is not Banach by showing (if this is even possible) that the space lacks a norm $\endgroup$ – Sujaan Kunalan Dec 20 '14 at 4:43
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    $\begingroup$ As you see, your question was quite not clear : this manifests itself in the fact that you got (so far!) four answers to four different questions :-) $\endgroup$ – Mariano Suárez-Álvarez Dec 20 '14 at 5:03
  • $\begingroup$ Yes, there are vector spaces without a norm. Yes, there are normed spaces that are not complete in the norm of the space. Some asides: there are normed spaces for which the norm is not induced by any inner product. There are normed spaces which are complete in the norm induced by the inner product. Finally, there are normed spaces which are not complete in the norm induced by the inner product. $\endgroup$ – JohnD Dec 20 '14 at 5:18
  • $\begingroup$ @Freeze_S, frankly, I find it of rather poor taste to talk about people tending to think trivialities or technicalities or be "limited" to whatever you may think they ar elimited without being explicit about what and about whom you are talking about. $\endgroup$ – Mariano Suárez-Álvarez Dec 21 '14 at 2:06

While your question could have multiple answers, perhaps the closest to what you are looking for is the notion of a non-metrizable vector space.

In the general setting of topological vector spaces, we consider (as one might guess from the name) vector spaces endowed with a topology so that we can discuss ideas like the continuity of linear operators. Normed vector spaces are examples of topological vector spaces where the topology is induced by a given norm.

A non-metrizable vector space is a topological vector space whose topology does not arise from any metric. These are rather common in functional analysis. For example, if $X$ is a Banach space, then the weak-* topology on $X^*$ is never metrizable unless $X$ is finite-dimensional. Another family of examples are locally convex spaces, a natural generalization of Banach spaces, which are not metrizable unless their topology is generated by a countable collection of seminorms that separate points.

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    $\begingroup$ Frechet spaces are another possible answer. $\endgroup$ – Kimball Dec 20 '14 at 8:07
  • $\begingroup$ Frechet spaces are examples of locally convex spaces, so I think I've covered that case. Besides, I wanted to stick to non-metrizable spaces since I can't say much about non-normable spaces. $\endgroup$ – Gyu Eun Lee Dec 20 '14 at 17:39
  • $\begingroup$ Ah, I didn't read your last sentence. $\endgroup$ – Kimball Dec 21 '14 at 13:13

Vector spaces are, by default, unnormed. A norm is extra structure we add to a vector space, to define a normed vector space.


Every (real or complex) vector space admits a norm. Indeed, every vector space has a basis you can consider the corresponding «$\ell^1$» norm.

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    $\begingroup$ It might be worth clarifying that the admitted norm may not be compatible with the topology (if the vector space has a topology). $\endgroup$ – copper.hat Dec 20 '14 at 5:53
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    $\begingroup$ It should be noted that for the "every vector space has a basis", you need the Axiom of Choice (if I remember correctly). $\endgroup$ – Taladris Dec 20 '14 at 17:31
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    $\begingroup$ @copper.hat Vector spaces do not have topologies. Topological vector spaces do. $\endgroup$ – Mariano Suárez-Álvarez Dec 20 '14 at 22:07
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    $\begingroup$ @leftaroundabout, that the norm is not canonical is a rather absurd complaint, as I started with nothing but a vector space. If the question had been «does every vector space have a canonical norm?» or «does it have a natural norm?» or something along those lines, I would understand, but as the question was most certainly not that, I don't. Similarly, the fact that the norm be useful or not is absolutely irrelevant to its existance; compaibility with a topology that might also exist on the vector space is, likewise, quite irrelevant. $\endgroup$ – Mariano Suárez-Álvarez Dec 21 '14 at 1:58
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    $\begingroup$ Honestly, I cannot conceive what can possibly be wrong with the "point of view" that vector spaces have bases. Likewise, commutative rings have maximal ideals, and the unit ball in Banach spaces is weakly* compact. IMO the point of view that there may be something "bad" about this point of view is pretty silly. $\endgroup$ – Mariano Suárez-Álvarez Dec 21 '14 at 2:03

Outlining the construction by Mariano Suárez-Alvare...


Does every plain vector space admit a norm?


Given a plain vector space $V$.

Choose a Hamel basis $\mathcal{B}$.

Denote functions with finite support by $\mathbb{R}^\mathcal{B}_0$.

Decide for a norm there $\|(\lambda_b)_{b\in\mathcal{B}}\|$.

Now, regard the isomorphism: $$\Phi:V\to\mathbb{R}^\mathcal{B}_0:\sum_{b\in\mathcal{B}}\lambda_bb\mapsto\left(\lambda_b\right)_{b\in\mathcal{B}}$$

Then, it pulls back the norm: $\|\sum_{b\in\mathcal{B}}\lambda_bb\|:=\|(\lambda_b)_{b\in\mathcal{B}}\|$

Special Example

Take the norm of $\ell^2$.

The norm becomes: $\|\sum_{b\in\mathcal{B}}\lambda_bb\|_2^2=\sum_{b\in\mathcal{B}}|\lambda_b|^2$

Especially, it becomes an orthonormal basis: $\langle b,b'\rangle=\delta_{b,b'}$


Every plain vector space admits a norm - no matter of its dimension.


Does every algebra with unit admit a compatible norm: Existence of Norm for C*-Algebras


If $V$ is finite dimensional, it is normable, in the sense that you can use an isomorphism into $\mathbb R^n$ to pull back the $\mathbb R^n$-norm . And then this norm is equivalent to any other norm topology-wise, i.e., in a finite- dimensional space, all norms are equivalent in this sense.

  • $\begingroup$ Not every vector space is isomorphic to $\mathbb{R}^n$; a finite dimensional vector space is isomorphic to $\mathbb{K}^n$, where $\mathbb{K}$ is some (possibly finite) field $\endgroup$ – Zach Effman Dec 20 '14 at 18:21
  • $\begingroup$ You're right, for some reason I was assuming we were working over the Reals for some reason. $\endgroup$ – RikOsuave Dec 21 '14 at 1:09
  • $\begingroup$ @ZachEffman, in a context where norms are being discussed, it is a standrd assumption that the field is either $\mathbb R$ or $\mathbb C$. One can consider normed vector spaces over other fields, but as this is rather non-standard when one does so one is explicit about it. $\endgroup$ – Mariano Suárez-Álvarez Dec 21 '14 at 18:41

Of course there are. :-) First of all, there are linear spaces which are not endowed by a topology, and there are topological vector spaces which are not endowed by a norm. So a wise question is: when a topology of a topological vector space $X$ is generated by a norm? It seems it is iff $X$ is a locally convex Hausdorff space containing such a neighborhood $U$ of the zero that for each neighborhood $V$ of the zero there exists a scalar $\lambda>0$ such that $\lambda U\subset V$.


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