Prove: $\dim E= +\infty \Longleftrightarrow \forall \;n\geq 1,\;\exists \;\;B=\{u_1,u_2,\cdots,u_n\}$ such that $B$ is Linearly independent I wish to prove that $$\dim E= +\infty \Longleftrightarrow  \forall \;n\geq 1,\;\exists \;\{u_1,u_2,\cdots,u_n\}\; \text{such that}\; \{u_1,u_2,\cdots,u_n\} \;\;\text{ is linearly independent.}$$This looks completely new to me, so I would love if someone could give me an elaborate answer or provide a reference. 
Thanks!
 A: Note first, that your statement misses something to be technically fine. Although probably implied, the statement should constraint the $u_i$ to be pairwise distinct as otherwise, you may take them all to be a single vector all the time, i.e. being always linearly independent if it is not the null vector. Considering this adjustment:

For right to left: Suppose that $\mathrm{dim}(E)<+\infty$, i.e. $\mathrm{dim}(E)=m$ f.s. $m\in\mathbb{N}$. Then For all $n>m$, any set ${u_1,\dots,u_n}$ for pairwise distinct $u_i$ is linearly dependent. (check this yourself, with what does it contradict if there exists a linearly independent one?)

For left to right: Suppose that $\mathrm{dim}(E)=+\infty$. Suppose that there exists a $n\geq 1$ such that for all sets $U=\{u_1,\dots,u_n\}$ for pairwise distinct $u_i$ we have linear dependence. Then, any extension of any such $U$, i.e. $U\cup{W}$ for any $W$ is linearly dependent as well (can you see why? this even holds for $W$ infinite) 
Now, as $E$ is infinite dimensional, we have a basis $B=\{b_i\mid i\in I\}$ with $I$ infinite where the $b_i$ are linearly independent and span $E$. But now, as $B$ is infinite, $B$ contains an $n$-element subset of pairwise distinct vectors. (just pick $n$ different ones) But by our assumptions, this subset is linearly dependent and thus $B$ as an extension has to be linearly dependent either. Contradiction
