Recover the topology of a Banach space from its 1-dimensional subspacess Any infinite dimensional vector space can be written as direct sum of 1-dimensional vector subspaces like that $$V=\oplus_{i\in I}\text{span}(e_i)$$ where $\{e_i\}$ is a Hamel basis.
$\space$   I am taking Hamel basis because infinite direct sum contains tuples having at most finitely many non-zero entry. Now my question is if $V$ is a normed space like Banach space, then what can we say about the topology of the direct sum. Can we retrieve our topology from topology on 1-dimensional spaces. Or any relation between them?
 A: No, there is no relation between the topology of a normed space and the topology of its $1$-dimensional subspaces in the sense that its $1$-dimensional subspaces is homeomorphic to $\Bbb R$ or $\Bbb C$ always if the normed space is over $\Bbb R$ or $\Bbb C$ respectively.
In particular, we cannot retrieve the former from the latter unless the normed space is finite-dimensional, in which case we know its topology already: if it is $n$-dimensional over $\Bbb R$, any $\Bbb R$-linear isomorphism from it to $\Bbb R^n$ is also a homeomorphism; if over $\Bbb C$, any $\Bbb C$-linear isomorphism from it to $\Bbb C^n$ is also a homeomorphism.

In other words, the topology of a normed space is not necessarily determined by the subspace-topologies on its finite-dimensional linear-subspaces, since the subspace-topology of every finite-dimensional linear-subspaces is completely determine by its linear structure alone.
It can happen that there are open sets that cannot be "pieced together" by any number of subsets in finite-dimensional linear-subspaces that are open in the entire normed space. In fact, in an infinite-dimensional normed space no subsets in any finite-dimensional linear-subspace is open. For any point $a$ in a linear-subspace $S$,  the open (w.r.t. to the entire normed space) ball $B(a, \epsilon)$ contains the point $a+\frac\epsilon2\frac e{\|e\|}$ that is not in $S$, where $e$ is any point in the infinite-dimensional space but not in $S$.

$\Bbb{R, C, R^n, C^n}$ are endowed with the usual topologies when they are viewed as topological spaces in this answer.
