These seems like simple questions, but I cannot seem to find straightforward answers:

(1) If $V$ and $W$ are infinite (possibly uncountable) dimensional vector spaces, does $V \bigoplus W = V \times W$? That is, is there any difference between the direct sum and direct product of infinite dimensional vector spaces?

(2) If $\{ V_i \}_{i \in I}$ is an infinite (possibly uncountable) collection of finite dimensional vector spaces, does $ \bigoplus_{i \in I} V_i = \times_{i \in I} V_i$ ? Is the direct product product of infinitely many vector spaces even defined since vectors are supposed to consist of finite linear combinations of basis vectors (I realize there are subtleties in what "basis" means in the case of an infinite dimensional vector space)?

The inspiration for this question comes from the study of Banach spaces. Namely, that the product (or direct sum?) of Banach spaces can be made into a Banach space.

  • 5
    $\begingroup$ Unless I’m missing something seriously, the sum and product of two things are the same. But an infinite product of nonzero things will never be the same as an infinite sum. For, element of an infinite sum has only finitely many nonzero entries, whereas in the product, an element may have infinitely many of these. $\endgroup$
    – Lubin
    Commented Aug 31, 2017 at 13:48
  • $\begingroup$ @Lubin So would a direct product of infinitely many vector spaces even be defined? Or would only an infinite direct sum be defined? $\endgroup$ Commented Aug 31, 2017 at 13:56
  • 1
    $\begingroup$ ^Yes. The sum is realizable as a subspace of the product for the same exact reason. $\endgroup$
    – Randall
    Commented Aug 31, 2017 at 13:56
  • $\begingroup$ @wanderingmathematician I think in category theory, direct product corresponds to product, and direct sum corresponds to coproduct, and they coincide in the category of finite-dimensional vector spaces. Intuitively, when there are (uncountably) infinite dimensions, direct product can be viewed as a set of functions from the index set to the components. $\endgroup$
    – Ziyuan
    Commented Aug 18, 2020 at 9:54

4 Answers 4


There is no difference between the direct sum and the direct product for finitely many terms, regardless of whether the terms themselves are infinite-dimensional or not. However, they are different in the case of infinitely many terms (and drastically so).

A direct product $\prod_{i = 1}^\infty V_i$ can be thought of as the set of sequences $(v_1, v_2, \ldots)$ where each $v_i \in V_i$, with usual scalar multiplication $\lambda (v_1, v_2, \ldots) = (\lambda v_1, \lambda v_2, \ldots)$ and pointwise addition $(v_1, v_2, \ldots) + (w_1, w_2, \ldots) = (v_1 + w_1, v_2 + w_2, \ldots)$. The direct sum $\bigoplus_{n=1}^\infty V_i$ on the other hand is the same set, with the extra condition that only finitely many terms are nonzero.

The direct sum behaves nicely in terms of bases. If each $V_i$ has some basis set $B_i \subseteq V_i$, then the direct sum $\bigoplus_{n=1}^\infty V_i$ has a basis identified with $\bigsqcup_{i=1}^\infty B_i$. For example, if all $V_i = \mathbb{R}$, then the basis for the direct product is just putting a 1 in the $i$th place for all $i$: $(1, 0, 0, \ldots), (0, 1, 0, 0, \ldots), \ldots$. In particular, if all the $V_i$ are countable dimension, the direct sum is also countable dimension.

With the direct product, this is not the case. It is not so hard to see (I'm sure there are many answers on this site) that the space of sequences of real numbers $\prod_{i=1}^\infty \mathbb{R}$ has uncountable dimension over $\mathbb{R}$.

Finally, there is not that much subtlety in what it means to be a basis of an infinite dimensional space. A basis is a linearly independent subset $B \subseteq V$ such that any vector $v \in V$ may be written as a finite linear combination of basis vectors. This is perhaps the best way to think about the difference between direct sum and product: start trying to write down a system of elements which can express any real sequence as a finite linear combination, and you'll soon see that in many cases, a direct sum may have been what you intended all along.

  • $\begingroup$ can I ask an example? suppose $I$ is an infinite set. if we consider $\prod_I \mathbb{R}$ as a real vector space, is it isomorphic to the direct sum of $\mathbb{R}$ over the index set $2^I$? $\endgroup$
    – youknowwho
    Commented Dec 15, 2019 at 2:42
  • $\begingroup$ @tooweaktolearnmathematics $\prod_I \mathbb{R}$ is isomorphic to the vector space of functions $f: I \to \mathbb{R}$. I assume that you would need to invoke the axiom of choice in order to pick a basis for this space, and then figure out the cardinality of that basis. You could look around for the size of basis of $\ell_\infty(\mathbb{R})$ sequence space (although this space is a bit smaller than $\prod_I \mathbb{R}$). $\endgroup$
    – Joppy
    Commented Dec 16, 2019 at 4:45

The definition of finite direct sum and the definition of finite direct product is exactly the same definition. (Unless you are working in categories and then the definitions aren't the same, but as vector spaces they are isomorphic).

Infinite direct sum of $\{V_i\}_{i\in I}$ is the set of all finite sums of vectors, while the direct product is as a set, the product of all $V_i$, so, in particular, the direct sum does not contain the element $(1,1,1,1,....)$ while the direct product does.

For your questions

1)V and W are infinite dimensional, but since you only take the direct sum/product once the definitions are equal.

2) No, because the element (1,1,1,1,...) is not an element in the direct sum, but it is an element in the direct product.

  1. $V\times W$ and $V\oplus W$ are isomorphic, as are any finite sums/products of spaces. This is true for any category of modules.

  2. When $I$ is infinite $ \bigoplus_{i \in I} V_i $ and $ \times_{i \in I} V_i$ can be very different. For example, when $V_i=F_2$ the field of two elements, and $I=\mathbb N$, then the left hand one is countable while the right hand one is uncountable. So they obviously cannot be isomorphic. I think this is probably always the case by some cardinal arithmetic, but at this very moment I can't definitively say.

  • $\begingroup$ Thanks, this answer gave me a nice example to make things concrete. The direct sum here is the set of all finite bit strings (which is isomorphic to N), while the direct product is the set of all bit strings: put a dot in front, and it becomes the reals in [0, 1]. $\endgroup$
    – A_P
    Commented Oct 2, 2018 at 18:53
  • $\begingroup$ @A_P Not quite exactly all decimals: in $\prod_{i=1}^\infty \mathbb Z$ there are a countable number of pairs of infinite decimal representations which actually represent the same real number. But yes, very close to the real numbers. $\endgroup$
    – rschwieb
    Commented Oct 2, 2018 at 19:29
  • $\begingroup$ Ah, right, thanks. For $F_2$, all representations with finitely many nonzero terms map to the same real by setting the the last term to zero and appending repeating ones. I'm not sure I understand your $\mathbb{Z}$ example though. $\endgroup$
    – A_P
    Commented Oct 2, 2018 at 20:19
  • $\begingroup$ @A_P Sorry, I meant $\mathbb N$. It seemed to me you were describing decimal representations of reals from 0 to 1. It's the same as you described except $0.\bar{9}=1.0$, and all the other things that entails (like $0.0\bar{9}=0.1$ and so on.) $\endgroup$
    – rschwieb
    Commented Oct 2, 2018 at 20:33
  • $\begingroup$ I was using your example of $F_2 = \{0, 1\}$, so the binary representations, but yes, that was the idea. Isn't this written $\prod_{i=1}^{\infty} F_2$ (or $\times_{i \in \mathbb{N}} F_2$) though? I'm confused by your $\prod_{i=1}^{\infty} \mathbb{Z}$ (or $\prod_{i=1}^{\infty} \mathbb{N}$). $\endgroup$
    – A_P
    Commented Oct 2, 2018 at 21:17

A vector space is anything that satisfies the axioms of a vector space which say nothing at all about bases.

Finite coproducts (direct sums) and finite products coincide in any Abelian category which includes categories of vector spaces. So, the answer to your first question is, "yes", they are the same (as in isomorphic).

However, infinite coproducts and products do not coincide. (They do exist.) The result won't be a finite dimensional vector even if the summands/multiplicands are finite dimensional, but I don't think you were suggesting that. The (countable for simplicity) direct sum is the space of finite linear combinations of vectors from the summands, or, given a basis, you can think of them as infinite sequences of scalars that are $0$ at all but finitely many indices. The product, on the other hand, is simply the space of functions from $I$ into the appropriate $\bigcup_{i\in I}V_i$ with the vector space operations defined pointwise. For the countable case, given a basis, this can be viewed as arbitrary infinite sequences of scalars.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .