Vtyjkyuk

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.

Take an arbitrary $v_1ne 0.$ Now for $nin Bbb N$ suppose that $S(n)=v_j:jleq n$ 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 $S(n+1)=v_n+1cup S(n)$ is a linearly independentset

Without the Axiom of Choice we obtain, by induction on $n$, that for each $nin Bbb N$ 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 $(v_n)_nin Bbb N$ 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 $n in mathbbN$, $leftlbrace v_1, v_2, cdots, v_n rightrbrace$ 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.