[Math] Proving a vector space is infinite-dimensional

Question

Let $S$ be a vector space. Suppose that for any positive integer $n$, there exists a linearly independent subset $S_n ⊆ V$ of size $n$.

How do you show that $S$ is not finite dimensional but infinite-dimensional.

I understand that to be finite dimensional then it must have a finite then there exists a finite set that such at $\operatorname{Span}(T)=S$. So would it be right to prove by contradiction that there is not a finite spanning set that satisfies $\operatorname{Span}(T)=S$, and hence I could argue that the vector space $S$ is infinite dimensional.

How would I show that $S$ does not have a spanning set $T$, such that $\operatorname{Span}(T)=S$?

Answer 1

Try to prove the following statement:

If $V$ is a vectorspace and $A = \{x_1, \ldots, x_n\}$ is a subset of $V$ containing $n$ (different) vectors. Suppose $A$ is a generating set for $V$, then every subset of $V$ with more than $n$ elements is a linearly dependent subset.

Given: a vector space $V$ such that for every $n \in \{1, 2, 3, \ldots\}$ there is a subset $S_n$ of $n$ linearly independent vectors.

To prove: $V$ is infinite dimensional.

Proof: Let us prove this statement by contradiction: suppose that $V$ has finite dimension $k$. Fix a basis $\{v_1, \ldots, v_k\}$, then it is given that we can find a set $S_{k+1}$ of $k +1 < \infty$ vectors which are linearly independent. However, the above statement shows that the elements have to be linearly dependent, a contradiction. Hence it must be that $V$ is infinite dimensional.