Cardinality of a linearly independent set in an infinite-dimensional vector space

linear algebravector-spaces

I know the following statement is true for finite-dimensional vector spaces:

If $S$ is a vector subspace of $V$ and there is a linearly independent set of vectors in $S$ with the same cardinality as a basis of $V,$ then $S = V.$

I'd like to know if it's true for infinite-dimensional spaces as well and, if so, how it can be proven.

Best Answer

It is not true.

The set $S=\{(0, x_1,x_2\dots)| \forall i > 0: x_i\in\mathbb R\}$ is a vector subspace of $V = \{(x_0, x_1,x_2,\dots)|\forall i\geq 0: x_i\in\mathbb R\}$.