Suppose $M$ is a normed linear space. $L$ and $N$ are two closed subspaces of $M$ such that $L \subseteq N$. Then $L$ is a closed subspace of $N$. Let $\text{dim}(M/N)=r$ and $\text{dim}(N/L)=s$.
My question is:
Can we get a conclusion that $\text{dim}(M/L)=r+s$ ?
Supplement：
We can give a positive answer when $\text{dim}(M) < \infty$. Since $$\text{dim}(M) = \text{dim}(M/N) + \text{dim}(N)= r+n$$ and $$\text{dim}(L) = \text{dim}(N) - \text{dim}(N/L) = n-s,$$ we have $$\text{dim}(M/L) = \text{dim}(M) - \text{dim}(L) = r+s.$$ However, I can't find a way to verify it when $\text{dim}(M)=\infty$. Thanks in advance.
Yes this is true. Suppose $v_1,...,v_s$ is a set of vectors in $N$ so that $v_1 + L,...,$ is a basis for the space $N/L$. Let $w_1,...,w_r$ be a set of vectors in $M$ so that $w_1 + N,...$ is a basis for the space $M/N$. I claim that $v_1+L,...,v_s+L,w_1+L,...,w_r+L$ is a basis for $M/L$.
First we show this set spans the space. Let $v \in M$. Then $v $ lies in some $ \lambda_1 w_1 + ... \lambda_r w_r + N$ for some constants $\lambda_1,...$, since the $w_i$ form a basis for $M/N$. Thus $$ v = \lambda_1 w_1 + ... \lambda_r w_r + n, $$ for some $n \in N$. But then $n$ must lie in some $ \rho_1 v_1 +... \rho_s v_s + L$ for constants $\rho_1,...$, since the $v_i$ form a basis for $N/L$. Therefore $$ n = \rho_1 v_1 +... \rho_s v_s + \ell $$ for some $\ell \in L$. Combining this with the above, we find that \begin{align*} v&=\lambda_1 w_1 + ... \lambda_r w_r + n \\ & = \lambda_1 w_1 + ... \lambda_r w_r + \rho_1 v_1 +... \rho_s v_s + l. \end{align*} Thus $v + L$ must lie in the span of the $v_i+L$ and $w_i+L$.
Now, we show that this set is linearly independent. Suppose $$ 0 \in \lambda_1 v_1 + ... + \lambda_s v_s + \rho_1 w_1 + ... + \rho_r w_r + L. $$ Then $\lambda_1 v_1 + ... + \lambda_s v_s + L$ lies in $N$, since $v_1 + L,$ is a basis for $N/L$. Thus $$ 0 \in \rho_1 w_1 + ... + \rho_r w_r + N. $$ However, since the $w_1+N,...,w_r+N$ form a basis of $M/N$, they must be linearly independent in this space. This implies that $\rho_1 = ... = \rho_r = 0 $
Now, to show that $\lambda_1 = ... = \lambda_s = 0$, we apply the same trick. Since $$ 0 \in \lambda_1 v_1 + ... + \lambda_s v_s + \rho_1 w_1 + ... + \rho_r w_r + L = \lambda_1 v_1 + ... + \lambda_s v_s+L. $$ by the linear independence of the $v_i + L$, we must have that each $\lambda_i = 0$. Therefore, this set spans the space, and is linearly independent. This basis has precisely $r+s$ elements, so $\dim M/L = r+s$, as desired.