In Conway's A Course in Functional Analysis, the author stated the following Proposition $1.4$ at page $377.$
Let $X$ be a vector space over $\mathbb{R}.$ Given $f,f_1,f_2,...,f_n$ are linear functionals on $X.$ If $$\bigcap_{j=1}^n ker(f_j) \subseteq ker(f),$$ then there exist scalars $a_1,...,a_n$ such that $$f(x) = \sum_{j=1}^n a_j f_j.$$
The proof starts as follows:
It may be assumed that without loss of generality that for $1\leq k\leq n,$ $$\bigcap_{j\neq k}ker(f_j) \neq \bigcap_{j=1}^n ker(f_j).$$
Question: Why can we assume above? In general, we have $$\bigcap_{j\neq k} ker(f_j) \supseteq \bigcap_{j=1}^n ker(f_j).$$ So in the proof, we can safely assume that the inclusion is strict. If the inclusion is not strict, meaning the two intersections are equal, then kernel of $f_j$ is a subset of kernel of $f_i$ for some $i.$