I had learned about quotient spaces several times, but I always have a hard time with them in practice.
For a vector space $V$ over $\Bbb{R}$ with $\dim V=n$ and a subspace $U$ of dimension $k$, find $\operatorname{codim}(U)$.
I thought I proved it, but then realized I didn't refer to the field $\Bbb{R}$.
I argued that cosets $[u_1],...,[u_k]$ ($u_i$-s linearly independent vectors from $U$) are all $[0]$ since $[x]=[y]$ iff $x-y\in U$.
I then said $u_1,...,u_k$ could be extended to a set of $n$ linearly independent vectors $u_1,...,u_k, v_1,..., v_{n-k}$ and that for $[x]=\{x+U\}$ I can take $a_i$ with $x=\sum^{k}_{i=1}a_iu_i+\sum^{n-k}_{i=1}a_iv_i$ and that way $x-\sum^{n-k}_{i=1}a_iv_i\in U$ so $[x]=[\sum^{n-k}_{i=1}a_iv_i]=\sum^{n-k}_{i=1}a_i[v_i]$. Is that enough to show the dimension is $n-k$? How can the fact the $\Bbb{F}=\Bbb{R}$ be used? Am I missing something?
You must still check that the cosets $[v_1],\ldots,[v_{n-k}]$ are linearly independent. All this is true for any field $F$, including the field $\mathbb{R}$.