I am reading some lectures on linear algebra which are on Russian language and few moments have confused me and I was trying to understand them correctly but I failed to do it.
Firstly, let me give you some preliminary definitions and theorems from my lectures:
Definition: A vector space $V$ is called finite-dimensional, if it has a basis consisting of finitely many vectors. Otherwise, we call the space to be infinite-dimensional.
Theorem: In finite-dimensional vector space each basis has the same number of vectors.
The proof of this theorem is based on the following lemma:
Lemma: Let $\{e_1,\dots,e_m\}$ and $\{f_1,\dots,f_n\}$ be two linearly independent system of vectors such that the second system is contained in the linear span of the first. Then $n\leq m$.
Definition: The dimension of finite-dimensional vector space $V$ is the number of elements in each basis of $V$. If $V$ is infinite-dimensional, then we write $\dim V =\infty$.
Then he is proving the following statement which confuses me.
Statement: Subspace $W$ of finite-dimensional space $V$ is finite-dimensional and $\dim W\leq \dim V$.
Proof: Since $V$ is finite-dimensional then $\dim V=m$ and let $\{e_1,\dots,e_m\}$ basis of $V$. Suppose that $\dim W>m$ then $W$ contains linearly independent vectors $f_1,\dots,f_n$ with $n>m$. Then $\{f_1,\dots,f_n\}\subset \langle e_1,\dots,e_m\rangle=V.$ But this contradicts to the above lemma. Hence $\dim W\leq \dim V$.
I understood the idea of the proof but cannot understand some technical moments.
Question 1: If $\dim W>m$ then why $W$ contains linearly independent vectors $f_1,\dots,f_n$ with $n>m$. Intuitively I know this but can anyone show it rigorously?
Question 2 (sorry for stupid question): Suppose we have shown that $\dim W\leq \dim V$ then how it follows that $W$ is also finite-dimensional?