# Infinite Dimensional Vector Space

A vector space that is not finite-dimensional is called an infinite-dimensional vector space. Can you think of any off-hand? In this article, I give an example, and I prove that every vector space must have a basis.

Definition. A vector space that is not not finite-dimensional is called an infinite dimensional vector space.

Example. Prove that $k^\infty$ is infinite-dimensional. Suppose $\mathbb{F}^\infty$ is finite-dimensional with dimension $n.$ Let $v_i$ be the vector in $\mathbb{F}^\infty$ consisting of all 0’s except with 1 in the $i$-th position for $i=1,\ldots,n.$ The vectors $(v_1,\ldots,v_n)$ are linearly independent in $\mathbb{F}^\infty$ and so they must form a basis; however they do not span since the vector $v_{n+1}$ consisting of all 0’s except with $1$ in the $(n+1)$-th position. Thus $\mathbb{F}^\infty$ can not be finite-dimensional.

Example. Prove that the real vector space consisting of all continuous real-valued functions on the interval $[0,1]$ is infinite-dimensional. Notice that $f(x)=x^n$ is continuous on $[0,1]$ for every positive integer $n.$ If this vector space is finite-dimensional of say dimension $n$ then the list of vectors $(1,x,x^2,\ldots,x^{n-1})$ must form a basis since they are linearly independent and there are $n$ of them. However, $x^n$ is not in the span of this list and so this list can not be a basis. This contradiction shows that this vector space must be infinite-dimensional.

Example. Prove that $\mathbb{V}$ is infinite-dimensional if and only if there is a sequence $v_1, v_2,\ldots$ of vectors in $\mathbb{V}$ such that $(v_1, \ldots, v_n)$ is linearly independent for every positive integer $n.$ Suppose $\mathbb{V}$ is infinite-dimensional. Then $\mathbb{V}$ has no finite spanning list. Pick $v_1\neq 0.$ For each positive integer $n$ choose $v_{n+1}\not \in \text{ span}(v_1,\ldots,v_n)$, by the linear independence lemma, and for each positive integer $n$, the list $(v_1,\ldots,v_n)$ is linearly independent. Conversely, suppose there is a sequence $v_1,v_2,\ldots$ of vectors in $\mathbb{V}$ such that $(v_1,\ldots,v_n)$ is linearly independent for every positive integer $n.$ If $\mathbb{V}$ is finite-dimensional, then it has a spanning list with $M$ elements. By the previous theorem, every linearly independent list has no more than $M$ elements. Therefore, $\mathbb{V}$ is infinite-dimensional.

Theorem. Every vector space has a basis. Moreover, any two bases have the same cardinality.

Proof. Let $\mathbb{V}$ be a nonzero vector space and consider the collection $A$ of all linearly independent subsets of $\mathbb{V}.$ This collection is nonempty, since any single nonzero vector forms a linearly independent set. Now, if $I_1\subset I_2 \subset \cdots$ is a chain of linearly independent subsets of $\mathbb{V}$, then the union $U$ is also a linearly independent set. Hence, every chain in $A$ has an upper bound in $A$, and according to Zorn’s lemma, $A$ must contain a maximal element, that is, $\mathbb{V}$ has a maximal linearly independent set, which is a basis for $\mathbb{V}.$

We may assume that all bases for $\mathbb{V}$ are infinite sets, for if any basis is finite, then $\mathbb{V}$ has a finite spanning set and so is a finite-dimensional vector space. Let $B$ be a basis for $\mathbb{V}.$ We may write $B={b_i \mid i\in I}$ where $I$ is some indexed set, used to index the vectors in $B.$ Note that $|I|=|B|.$ Now let $C$ be another basis for $\mathbb{V}.$ Then any vector $c\in C$ can be written as a finite linear combination of the vectors in $B$, where all the coefficients are nonzero, say $c=\sum_{i\in U_c} r_i b_i.$ Here $U_c$ is a finite subset of the index set $I.$ Now, because $C$ is a basis for $\mathbb{V}$, the union of all of the $U_c$’s as $C$ varies over $C$ must be in $I$, in symbols, $\bigcup_{c\in C} U_c=I.$ For if all vectors in the basis $C$ can be expressed as a finite linear combination of the vectors $B-{ b_k}$ spans $\mathbb{V}$, which is not the case. Therefore, $|B|=|I|\leq |C| \alpha_0=|C|.$ But we may also reverse the roles of $B$ and $C$ we obtain the reverse inequality. Therefore, $|B|=|C|$ as desired.

Here are more examples of infinite dimensional vector spaces and with much more detail here.

David Smith (Dave) has a B.S. and M.S. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. David is the founder and CEO of Dave4Math.