Is $V$ infinite-dimensional?

Question

$V$ is a vector space and for any positive integer $n$, there exists a linearly independent subset $S_n \subseteq V$ of size $n$.

Is $V$ infinite-dimensional and how do I go about proving this?

Answer 1

This is an exercise in verifying definitions and using some basic theorems, but I am guessing that this is a bit confusing when taking the first few steps into mathematics. So let's go through this together.

(Let me point out that I don't know what exactly are the definitions and theorems that you've seen. But I imagine these will be along the lines of the ones I write here. In case this is not so, you should consider this an invaluable exercise to see how to apply what we've done here to your given definitions and theorems.)

First, the definitions:

Definition. A set $X$ is finite if and only if there is a natural number $n$ and a bijection $f\colon X\to\{0,\ldots,n-1\}$.

And,

Definition. A vector space $V$ is $n$-dimensional if every basis of $V$ has size $n$. We say that $V$ is finite dimensional if for some $n$, $V$ is $n$-dimensional, and otherwise we say that $V$ is infinite dimensional.

Now we have a theorem:

Theorem. The dimension of a vector space is at most $n$ if and only if every set of more than $n$ vectors is linearly dependent.

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Now we can finally prove what we want.

Theorem. Suppose that for all $n$, there is $S_n\subseteq V$ which is linearly independent and of size $n$, then $V$ is infinite dimensional.

Proof. If $V$ is finite dimensional, then there is some $n$ such that $V$ is $n$-dimensional, but then $S_{n+1}\subseteq V$ is a set of size $n+1>n$ of linearly independent vectors. Therefore for all $n$, $V$ is not $n$-dimensional, and thus not finite dimensional, i.e. $V$ is infinite dimensional as wanted. $\quad\square$

Answer 2

I moved my detailed answer to the question which this question duplicates. Proving a vector space is infinite-dimensional