# Show that a infinite dimensional vector space must have linearly independent vectors for every positive integer [closed]

Show that V is infinite dimensional vector space if and only if there is a sequence $v_1,v_2,...$ of vectors in V such that $\{v_1,v_2,...,v_n\}$ is linearly independent for every positive integer n.

## closed as off-topic by Hurkyl, Arnaud D., Namaste, Brahadeesh, NosratiAug 17 '18 at 17:34

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• What have you done on this? Where are you having difficulty? What are your thoughts? – saulspatz Aug 17 '18 at 5:08
• What exactly is your definition of "infinite dimensional"? – Arthur Aug 17 '18 at 5:10

Take an arbitrary $v_1\ne 0.$ Now for $n\in \Bbb N$ suppose that $S(n)=\{v_j:j\leq n\}$ is a set of $n$ linearly independent vectors. Since $V$ is not finite dimensional we can take some $v_{n+1}$ that does not belong to the linear span of $S(n),$ so $S(n+1)=\{v_{n+1}\}\cup S(n)$ is a linearly independentset

Without the Axiom of Choice we obtain, by induction on $n$, that for each $n\in \Bbb N$ there exists a linearly independent set of $n$ vectors. We MUST have the Axiom of Choice (or at least a corollary of it called Dependent Choice) to justify th existence of an infinite sequence of "Takes" to get an infinite sequence $(v_n)_{n\in \Bbb N}$ with the desired property.

Dimension of a vector space is defined as the number of elements in the basis, provided the basis has a finite number of elements. In such cases, the vector space is said to be finite dimensional.

Therefore, a vector space is said to be infinite dimensional if the basis of the vector space has infinitely many elements. Now, a basis should be linearly independent.

An infinite subset $S$ of a vector space $V$ is said to be linearly independent, iff every finite subset $A \subset S$ is linearly independent in $V$. This gives directly that for every $n \in \mathbb{N}$, $\left\lbrace v_1, v_2, \cdots, v_n \right\rbrace$ is linearly independent in $V$. Here, however, each $v_i$ must belong to the basis. Otherwise, the construction of the linearly independent set would be different.

• I doubt that the OP has seen a proof that every vector space has a basis. – saulspatz Aug 17 '18 at 5:12
• But then, how can one talk about "dimension" without talking about basis? Is there some other way? – Aniruddha Deshmukh Aug 17 '18 at 5:14
• Well, you can define "infinite-dimensional" in an elementary course to mean that the is no finite basis. Then you can construct a countable linearly independent set by induction. – saulspatz Aug 17 '18 at 5:16
• @saulspatz: ... using the axiom of choice! I remark this because it would be a clue to the OP on how to proceed. I wonder if it's consistent with ZF that there is a infinite-dimensional vector space that doesn't have a countable linearly independent set? – Hurkyl Aug 17 '18 at 5:23
• @Hurkyl Right. I meant to say that for every $n$ there is a set of $n$ independent vectors. That doesn't use the axiom of choice, right? – saulspatz Aug 17 '18 at 5:26