If $\beta$ is a basis of $V$ then does that mean that $V=span{\beta}$?

Question

I am confused with the definition of 'basis'.
A basis $\beta$ for a vector space $V$ is a linearly independent subset of $V$ that generates $V$. And span($\beta$) is the set consisting of all linear combinations of the vectors in $\beta$.
So from my understanding, just because all vectors in $V$ can be generated by $\beta$ doesn't necessarily mean that V=span($\beta$), since there might exist $b\in span(\beta)$ s.t. $b\notin V$
But I have learned that if $W\leq V$ and $\beta$ is a basis for both $V, W$, then $V=W$ since $W=V=span(\beta)$. ($V, W$ are finite dimensional vector space)
This seems to imply that a vector space is equal to the span of its basis, which contradicts my understanding of its definition.
I'd like to know which part of my understanding above is flawed.

Answer 1

Since $V$ is the whole space, the possibility “there might exist $b\in\operatorname{span}(\beta)$ s.t. $b\notin V$” isn't real. So, $V$ being generated by $\beta$ is the same thing as $V=\operatorname{span}(\beta)$.

Answer 2

Yes, it is true that $$V = \text{span}\ \beta$$ To address your concern, suppose $\beta = \{v_1,v_2,\ldots,v_n\}$. If $b\in\text{span}\ \beta$, then $b = \alpha_1v_1 + \ldots + \alpha_nv_n$ for scalars $\alpha_1,\alpha_2,\ldots,\alpha_n$. Since $V$ is a vector space, and $v_1,v_2,\ldots,v_n \in V$, $\alpha_1v_1 + \ldots + \alpha_nv_n\in V$. This is from vector space axioms!

Not only is a vector space a span of its basis, but the basis of a vector space is also oftentimes defined as a minimal spanning set! In other words, the basis of a vector space (finite-dimensional) is a set of minimal/least possible size, such that the span of vectors in this set is exactly the entire space $V$.