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In set theory, when we talk about cardinality of a set we have notions like finite set, countably infinite and uncountably infinite sets.

Main Question

Let's talk about dimension of a vector space. In Linear Algebra I have always heard about two terms either a vector space is finite dimensional for example $\mathbb{R}^n$ or infinite dimensional for example $C[0,1]$.

Why don't we have notions like countably infinite dimensional vector space and uncountably infinite dimensional vector space.

May be, I am not able to see the bigger picture.

Extras

P.S. Long time ago, I was in a talk on Enumerative Algebraic geometry and the professor said, I always think of a positive natural number as a dimension of some vector space.

Don't we have then uncountable dimension vector space?/

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  • $\begingroup$ Some infinite-dimensional vector spaces have countable dimension, some have uncountable dimension. The dimension of a vector space is a well-defined cardinality. So, what's the question? $\endgroup$ – Lord Shark the Unknown Sep 13 '18 at 11:59
  • $\begingroup$ I asked this as I have never heard of terms like countable dimension or uncountable dimension in books. @LordSharktheUnknown $\endgroup$ – StammeringMathematician Sep 13 '18 at 12:01
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    $\begingroup$ An infinite countable dimmensionnal spaces cannot be complete but they exist. Consider the vector space of real sequences having finite support. $\endgroup$ – nicomezi Sep 13 '18 at 12:14
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    $\begingroup$ "I have never heard of terms like countable dimension or uncountable dimension in books" --- This is because beginning and intermediate level linear algebra books rarely distinguish more precisely than "finite dimension" and "infinite dimension". It's usually only in graduate level algebra courses (e.g. probably all of the "third level" books in my answer to High-level linear algebra book) where you'll find the various notions of "algebraic dimension" defined as a cardinal number. $\endgroup$ – Dave L. Renfro Sep 13 '18 at 12:31
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    $\begingroup$ Just because you don't see it in your text doesn't mean it doesn't exist. You're quite right that it makes sense! $\endgroup$ – Noah Schweber Sep 13 '18 at 12:31
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We do have the notions of countable/uncountable dimensions. Just as a set can be finite or infinite (without specifying which infinite cardinality the set as) a vector space can be finite dimensional or infinite dimensional. We can then go one step more and ask, if the dimension is infinite, which infinite cardinal is it?

The definition of dimension of a vector space is the cardinality of a basis for that vector space (it does not matter which basis you take, because they all have the same cardinality). Then for any cardinal number $\gamma$, you can have a vector space with that dimension. For example, if $\Gamma$ is a set with cardinality $\gamma$, let $c_{00}(\Gamma)$ be the space of all $\mathbb{F}$-valued functions $f$ such that $$\text{supp}(f)=\{x\in \Gamma: f(x)\neq 0\}$$ is finite. Then let $\delta_x\in c_{00}(\Gamma)$ be the function such that $\delta_x(y)=0$ if $y\neq x$ and $\delta_x(y)=1$ if $y=x$. Then $(\delta_x)_{x\in \Gamma}$ is a basis for $c_{00}(\Gamma)$ with cardinality $\gamma$. If $\Gamma=\mathbb{N}$, we have a vector space with countably infinite dimension. If $\Gamma=\mathbb{R}$, we have a vector space with dimension equal to the cardinality of the continuum.

However, for infinite dimensional topological vector spaces (and for infinite dimensional Hilbert and Banach spaces in particular) the usual notion of a basis of limited use. This is because the coordinate functionals for an infinite basis do not interact very well with the topology (one can show that if $(e_i, e^*_i)_{i\in I}$ is a basis together with its coordinate functionals for an infinite dimensional Banach space, then only finitely many of the functionals $e^*_i$ can be continuous). Since the notion of a basis is not as useful in the infinite dimensional topological space case as it is in the finite dimensional case, you can see less emphasis on what the exact dimension is in this case.

However, in this situation you get into discussions of other types of coordinate systems (such as Schauder bases, FDDs, unconditional bases, etc.), which are different from the notion of an (algebraic) basis. You also can ask about density character instead of dimension, which is the smallest cardinality of a dense subset. This encodes topological information, while the purely algebraic notion of a basis does not. For example, infinite dimensional Hilbert space $\ell_2$ has no countable basis, it does have a countable, dense subset. So the dimension is that of the continuum, but the density character is $\aleph_0$.

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The dimension of a vector space is the cardinality of a basis for that vector space. To say that a vector space has finite dimension therefore means that the cardinality of a basis for that vector space is finite. Since finite cardinalities are the same thing as natural numbers, we are safe in saying, for finite dimensional vector spaces, that the dimension is a natural number.

In general, some sets are countably infinite and some sets are uncountably infinite. So, applying this to those sets which happen to be bases of vector spaces, some vector spaces have countably infinite bases and therefore countably infinite dimension, and other vector space have uncountable infinite bases and therefore uncountably infinite dimension.

An example of a vector space over $\mathbb R$ of countably infinite dimension is $\mathbb R^{\infty}$ which is the space of infinite sequences of real numbers such that all but finitely terms in the sequence are equal to $0$. A countably infinite basis consists of $(1,0,0,0,...)$, $(0,1,0,0,...)$, $(0,0,1,0,...)$ and so on.

An example of a vector space over $\mathbb R$ of uncountably infinite dimension is the one you mention in your question, $C[0,1]$.

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    $\begingroup$ That's okay. I do often write things wrong the first time. $\endgroup$ – Lee Mosher Sep 13 '18 at 12:18
  • $\begingroup$ ... all (but finitely many) terms are zero. $\endgroup$ – Lee Mosher Sep 13 '18 at 12:19
  • $\begingroup$ Had a hard time understanding your sentence, sorry for incovenience. $\endgroup$ – nicomezi Sep 13 '18 at 12:22
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For the sake of a real world example: The electron in a hydrogen atom can take on countably many states. Each state is a basis vector for the span of possible electron states of hydrogen. There's are bound states and bound states are typically discrete. The relevant Schrodinger Equation also permits scattering states which have a continuous spectrum of possible energy states implying an uncountable vector space spanning the possible scattering states.

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The following theorem is an example where we need to distinguish between countably/uncountably infinite dimensional vector spaces. Some important theorems, such as Hilbert Nullstallensatz, can be deduced from it.

Let $A$ be an associative, not necessarily commutative, $\mathbb{C}-$algebra with unit. For $a \in A$ define $$\text{Spec } a = \{\lambda \in \mathbb{C} | a-\lambda \text{ is not invertible}\}$$ Assume that $A$ has no more than countable dimension over $\mathbb{C}$. Then

(a) If $A$ is a division algebra, then $A=\mathbb{C}$

(b) For all $a \in A$ we have $\text{Spec } A \neq \emptyset$; furthermore, $a \in A$ is nilpotent if and only if $\text{Spec } A = \{0\}$

(Adapted from Representation Theory and Complex Geometry by Chriss and Ginzburg, theorem 2.1.1.)

The proof uses the fact that for any $a \in A$, $$\{(a - \lambda)^{-1} | \lambda \in \mathbb{C}\}$$ is an uncountable family of elements of $A$. But $A$ has only countable dimension over $\mathbb{C}$, so they are not linearly indenpendent over $\mathbb{C}$.

A weak version of Hilbert Nullstellensatz:

Let $A$ be a finitely generated commutative algebra over $\mathbb{C}$. Then any maximal ideal of $A$ is the kernel of an algebra homomorphism $A \to \mathbb{C}$

This follows directly from the above theorem: $A$ finitely generated $\Longrightarrow$ $A$ has countable dimension over $\mathbb{C}$ $\Longrightarrow$ $A/\mathfrak{m}$ has countable dimension over $\mathbb{C}$ $\Longrightarrow$ $A/\mathfrak{m}=\mathbb{C}$

(We can also deduce the strong version of Nullstellensatz from it but need more argument.)

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