Codimension inequality Let $N$ and $R$ be subspace of vector space $X$(not necessary finite dimension)$\DeclareMathOperator{\codim}{codim}$
Assume $N\subset R \subset X$, prove the codimension inequality:
$$\codim R \le \codim N$$
This is very simple if we use the dimension formula: $\dim X = \dim N + \codim N = \dim R + \codim  R\:$ (where codimension is defined as $\codim  N = \dim(X/N)$ )
How to prove it without this formula?
 A: Consider the map  $f: X/N\to X/R $ defined via  $x+N\mapsto x+R $. It is a well defined surjective map. So by rank-nullity theorem
$$\dim(X/R)=\dim(X/N)-\dim(\ker f)\leq \dim(X/N).$$
A: If $N\subseteq R$ then $X/N$ is a vector space 'containing' $R/N$, in which $(X/N)/(R/N)\cong X/R$, by one of the isomorphism theorems ($x+N\mapsto x+R$ as in Shivering Soldier's answer).
In general, $\dim(V/W)\le\dim V$, hence it follows that $\mathrm{codim}(R)=\dim(X/R)\le\dim(X/N)=\mathrm{codim}(N)$.
Proof of this fact: If $v_i$ is a basis for $V$ then $v_i+W$ span $V/W$ (obvious) and thus the dimension of $V/W$ can only be less than that of $V$.
A: I will change the notation to something which is more in the spirit of linear algebra. Let $K$ be an arbitrary field and $V$ a left $K$-vector space with subspaces $U \leqslant_K U' \leqslant_K V$.
The identity map $\mathbf{1}_V$ maps $U$ to $U'$ so it induces a quotient morphism $f \in \mathrm{Hom}_{\operatorname{K-\mathbf{Mod}}}(V/U, V/U')$ which satisfies the relation $f \circ \sigma=\sigma'$, where $\sigma \colon V \to V/U$ and $\sigma' \colon V \to V/U'$ denote the respective canonical surjections. As it is the quotient of a surjective map, $f$ is itself surjective and we can also easily obtain the description $\mathrm{Ker}f=\sigma[U']=U'/U \leqslant_K V/U$.
Hence, by virtue of the fundamental (iso)morphism theorem one infers that:
$$\left(V/U\right)/\left(U'/U\right) \approx V/U' \quad (\operatorname{K-\mathbf{Mod}}),$$
which in particular entails $\mathrm{codim}_{V/U}(U'/U)=\mathrm{codim}_V(U')$.
From the general dimension-codimension relation (for any subspace, the sum between its dimension and codimension in the ambient space is the dimension of the ambient) we derive the following:
$$\mathrm{dim}_K (V/U)=\mathrm{dim}_{K}(V/U')+\mathrm{dim}_K(U'/U),$$
which -- by taking into account the definition of codimensions -- can be more clearly written as:
$$\mathrm{codim}_VU=\mathrm{codim}_{V}U'+\mathrm{codim}_{U'}U,$$
the well-known relation of transitivity of codimensions.

I will mention here (without proof) the fundamental theorem for existence of quotient morphisms:

Let $K$ be an arbitrary field with $V$, $V'$ two left $K$-vector spaces. Let $f \in \mathrm{Hom}_{\operatorname{K-\mathbf{Mod}}}(V, V')$ be a $K$-linear map (morphism of $K$-vector spaces) and $U \leqslant_K \mathrm{Ker}f$ be a subspace of $U$ included in the kernel of $f$. Let $\sigma \colon V \to V/U$ denote the canonical surjection. There exists a unique morphism $g \in \mathrm{Hom}_{\operatorname{K-\mathbf{Mod}}}(V/U, V')$ such that $f=\sigma \circ g$, morphism which has the properties:
$$\begin{align*}
\mathrm{Ker}g&=(\mathrm{Ker}f)/U\\
\mathrm{Im}g&=\mathrm{Im}f.
\end{align*}$$

A: Based on the linear map $f: X/N\to X/R$ ,which is :

*

*well-defined

*surjective

*linear

Now given basis set for $(X/R)$ i.e. $\{b_i|i\in I\}$,we can find it's one of preimage point correspond to it i.e. $\mathcal{X} = \{x_i|i\in I\}$ such that $f(x_i) = b_i$
Now this set $\mathcal{X} $ is linear independent,since finite sume $\sum c_ix_i = 0$ then  act $f$ on both side we have $\sum c_ib_i = 0$ which means all of $c_i = 0$.
Since for vector space,we can always expand linear independent set $L$ to basis set of this vector space,we have the inclusion map between $\mathcal{X}$ and basis of $X/N$ (denoted it $\mathcal{B}$).such that $i:\mathcal{X} \to \mathcal{B}$.so:
$$|I| = |\mathcal{X}| \le |\mathcal{B}|$$ which complete the proof
A: Perhaps easier (though really similar under the hood to the various quotient-based arguments, above), using Extension of linearly independent set to a basis in an infinite dimensional vector space .
A basis of $N$, $\mathcal{N}$, can be extended to a basis of $R$, $\mathcal{R}$, which can be extended to a basis of $X$, $\mathcal{X}$.  Then $\mathcal{X} \smallsetminus \mathcal{R} \subseteq \mathcal{X} \smallsetminus \mathcal{N}$, as was to be shown.
