$X$ is a Banach space and $Y$ is a closed subspace of $X$ such that $\operatorname{dim}(X/Y)$ is finite. Then show that $Y$ is complemented. I have two ideas.
First one–
Claim– $Z= \operatorname{span} \{x_1, x_2, x_3, \ldots, x_n\}$ is the complement of $Y$, where $[x_i]$, $i= 1,\ldots, n$ are equivalence classes:then, since $\operatorname{dim}(X/Y)$ is finite, the equivalence classes $[x_n]$ are finite.
I want to prove that $Z$ is closed and $Z$ intersect with $Y$ is empty. Can we do this by a contradiction? Then I need to show $X= Z+Y$.
Second one–
Or else I thought of finding a projection $P$ from $X$ to $X$ such that $\operatorname{range}(P)=Y$. I could not come up with such a projection.
Which idea is good for a proof?
Best Answer
Your first idea is correct but you have not written the proof correctly. Let $[x_1],[x_2],..,[x_n]$ be basis for $X/Y$. Define $Z$ to be the span of $x_1,x_2,...,x_n$. [$x_1,x_2,...,x_n$ are not unique but that doesn't matter. Just pick some $x_i$'s in the equivalence classes].
If $x \in X$ then $[x]=\sum c_i[x_i]$ for some $c_i$'s. This means $x-\sum c_ix_i \in Y$. Hence, $x=(x-\sum c_ix_i)+\sum c_ix_i \in Y+Z$.
It remains to check that $Z\cap Y=\{0\}$. If $x \in Z \cap Y$ then (by definition of $Z$) $x =\sum a_ix_i$ for some $a_i$'s. But $x \in Y$ so $\sum a_ix_i \in Y$ implying that $[\sum a_ix_i]=0$, so $\sum a_i[x_i]=0$. By independence of $[x_i]$'s this implies $a_i=0$ for all $i$, so $x=0$.