so to recap: the more modern definition
1) $X$ is a Baire space iff every countable union of closed nowhere dense sets has empty interior.
Equivalent by taking complements (note that a set $A$ is nowhere dense iff its complement $X \setminus A$ contains an open dense subset) to my favourite formulation, which seems to be more commonly used among topologists:
1') $X$ is a Baire space iff every countable intersection of open and dense subsets is dense.
(note that in any space a finite intersection of open and dense subsets is open and dense, so the countable intersection is the first "interesting" question, in a way.)
And what they call the historical definition:
2) Every non-empty open subset of $X$ is of second category.
The article call it historical because it uses a notion "category" of a subset (a subset is either first category or second category, and not both, by definition), which has fallen in some disuse. Nowhere dense sets and meagre sets (the countable unions of nowhere dense subsets) are still normal usage. Note that a first category subset is now called meagre, and the notion of "second category" is not used as much (but still occurs), so it's good to know it. But definition 1) and 2) are easily proved to be equivalent, so they give rise to the same spaces being called Baire. So we have a trivial reformulation of the "classical" definition 2 as:
2') Every non-empty open subset of $X$ is non-meagre.
Or, stated more "positively"
2'') Every meagre set has empty interior.
(otherwise the non-empty interior is a subset of a meagre set, and thus meagre etc.)
which brings us back to definition 1) again.
It's just that the Wikipedians do not like the category terminology (because it might confuse people with category theory as a branch of maths) and so choose to reformulate everything using meagre and non-meagre instead.
If $X$ is an infinite-dimensional vector space over some field $F,$ then any basis $B$ must be an infinite set.
It's true that any $v\in X$ can be written as a finite linear combination $p_1 b_1 +\dots + p_n b_n,$ where the $p_k$ are in the underlying field $F$ and the $b_k$ are in the basis $B.$
This doesn't say that $B$ is finite though. Different $v$ in $X$ will require different basis elements to write them in the above format — only finitely many basis elements for any particular $v,$ but (assuming that $X$ is infinite-dimensional so that $B$ is infinite) as you let $v$ vary, you're going to need the infinitely many basis elements to write the various linear combinations (even though each linear combination is itself, individually, a finite sum).
Here's an example:
Let $X$ be the set of all infinite sequences of real numbers that are eventually $0;$ in other words, a member of $X$ is a function $f\colon\mathbb{N}\to\mathbb{R}$ such that for some $n\in\mathbb{N}$, for all $k\gt n,$ $f(k)=0.$ Of course, $X$ is a vector space over $\mathbb{R}$ under pointwise addition and pointwise scalar multiplication.
For each $n\in\mathbb{N},$ let $b_n\in X$ be defined by setting $$b_n(k)=\begin{cases}1,\text{ if }k=n,\\0,\text{ if }k\ne n.\end{cases}$$
Then you can see that $\{b_n\mid n\in\mathbb{N}\}$ is a basis for $X$ over $\mathbb{R}.$ Any member of $X$ can be written as a finite linear combination of the $b_n\text{'s}$ (since each member of $X$ is eventually $0).$ But you need all the $b_n\text{'s}$ (infinitely many) to write all the members of $X$ in that way.
[By the way, the theorem that every vector space has a basis uses the axiom of choice, as does the theorem that any two bases for the same vector space must have the same cardinality. Without the axiom of choice, there can be vector spaces without a basis, and also vector spaces which have a basis but which don't have a well-defined dimension, because different bases can have different cardinalities. I wouldn't worry about any of this when you're just starting to study vector spaces though.]
Best Answer
Let $\{e_n :n\in\mathbb{N}\}$ is countable basis of $F$ and let denote by $$F_n =\{\lambda_1 e_1 +\lambda_2 e_2 +...+\lambda_n e_n :\lambda_1 ,\lambda_2 ,...,\lambda_n \in\mathbb{R}\} .$$ Then $$F=\bigcup_{j=1}^{\infty} F_j .$$ Since $F_n$ are closed and $F$ is complete then by Baire theorem we have that $F\subset F_k $ for some $k\in\mathbb{N} .$