We know that if $V$ is a normed vector space and $W$ is a finite dimensional subspace of $V$, then $W$ is closed. One way to prove this is to show that $W$ is actually complete. Since complete space has to be closed, the result follows.
However, I am wondering if there is a "more direct" proof. For example, is it possible to prove the proposition using definition of "closedness"? Or perhaps that would end up being a even more complicated argument?
Best Answer
An alternate approach is induction on $n=\dim(W)$. The base case $n=0$ is clear, so the hard part is the induction step.
For this, it's enough to prove the following result: if $M$ is a closed subspace of $V$ and $x\in V,x\not\in M$, then $M+\mathbb{C}x$ is also a closed subspace.
Indeed, by the Hahn-Banach theorem there is a continuous linear functional $f$ on $V$ such that $f=0$ on $M$ and $f(x)\neq 0$. Let $\{y_j\}$ be a sequence in $M+\mathbb{C}x$, and suppose that $y_j\to y$. Define $$z_j=y_j-\frac{f(y_j)}{f(x)}x$$ Then $z_j\in M+\mathbb{C}x$ and $f(z_j)=0$, so $z_j\in M$. Since $y_j\to y$ and $f$ is continuous, $z_j\to z=y-\frac{f(y)}{f(x)}x$, and $z\in M$ because $M$ is closed. Therefore $$ y=z+\frac{f(y)}{f(x)}x\in M+\mathbb{C}x$$ so $M+\mathbb{C}x$ is closed.