Let $X$ be a Banach space and $W\subset V \subset X$ are two subspaces of $X$. Suppose now we know $\dim(V/ W) = 1$, then what can we say about the dimension $\dim(\overline{V}/\overline{W})$? Here $\overline{V}$ and $\overline{W}$ are the closures of $V$ and $W$ in $X$, respectively. Is it true that $\dim(\overline{V}/\overline{W})\leq1$?
Any hint will be appreciated. Thanks!
Indeed it follows that $\dim (\overline{V}/\overline{W}) \leqslant 1$.
Let $x \in V \setminus W$. Then, since $\dim (V/W) = 1$, we have $V = W + \mathbb{K}\cdot x$ (where $\mathbb{K}$ is the scalar field, be that $\mathbb{R}$ or $\mathbb{C}$). Note that this is just the vector space sum, it need not be a topological sum.
Now, the sum of a closed subspace and a finite-dimensional subspace is closed, hence $U = \overline{W} + \mathbb{K}\cdot x$ is a closed subspace of $X$. And $V \subseteq U$, thus $\overline{V}\subseteq U$. On the other hand we have $\overline{W} \subseteq \overline{V}$ and $\mathbb{K}\cdot x \subseteq V \subseteq \overline{V}$, hence $U \subseteq \overline{V}$. Together it follows that $\overline{V} = \overline{W} + \mathbb{K}\cdot x$, whence $\overline{V}/\overline{W}$ is generated by $x + \overline{W}$, hence its dimension at most $1$.
More precisely, we see that $\dim (\overline{V}/\overline{W}) = 1$ if and only if $W$ is relatively closed in $V$, i.e. $W = \overline{W} \cap V$, and $\overline{W} = \overline{V}$ otherwise.