Every finite-dimension subspace of $\mathcal{X}$ is closed.

Background Information:

(Folland)Theorem 5.8 - Let $\mathcal{X}$ be a normed vector space.

a.) If $M$ is a closed subspace of $\mathcal{X}$ and $x\in \mathcal{X}\setminus M$, there exists $f\in\mathcal{X}^*$ such that $f(x)\neq 0$ and $f(M) = \{0\}$. In fact, if $\delta = \inf_{y\in M}\|x - y\|$, $f$ can be taken to satisfy $\|f\| = 1$ and $f(x) = \delta$.

Question:

Let $\mathcal{X}$ be a normed vector space.

a.) If $M$ is a closed subspace and $x\in\mathcal{X}\setminus M$ then $M + \mathbb{C}x$ is closed. (Use Theorem 5.8a.)).

b.) Every finite-dimensional subspace of $\mathcal{X}$ is closed.

Proof a.) - Let $M$ be a proper closed subspace of $X$ and let $x\in X\setminus M$. There existts $f\in X^*$ such that $f(x)\neq 0$ and $f(M) = \{0\}$. Let $\{u_n + a_nx \}_{1}^{\infty}$ be a sequence in $M + Kx$ that converges to $y\in X$. Then $$f(y) = \lim\limits_{n\rightarrow \infty}f(u_n + a_n x) = \lim\limits_{n\rightarrow \infty} a_nf(x)$$ since $f$ is continuous, so $\{a_n\}_{1}^{\infty}$ converges to $a:= f(y)/f(x)$, which implies that $\{a_nx \}_{1}^{\infty}$ converges to $ax$. Therefore $$\{u_n\}_{1}^{\infty} = \{(u_n + a_nx) - a_nx\}_{1}^{\infty} \to (y - ax)$$ which lies in $M$ because $M$ is closed. It follows that $y\in M + Kx$, which shows that $M + Kx$ is closed.

Proof b.) - I have spent a considerable amount of time thinking about this but I am not sure how to proceed or how to show this result. Any hints or suggestions are greatly appreciated.

Solution 1:

We already know (Background Information):

(Folland)Theorem 5.8 - Let $\mathcal{X}$ be a normed vector space.

a.) If $M$ is a closed subspace of $\mathcal{X}$ and $x\in \mathcal{X}\setminus M$, there exists $f\in\mathcal{X}^*$ such that $f(x)\neq 0$ and $f(M) = \{0\}$. In fact, if $\delta = \inf_{y\in M}\|x - y\|$, $f$ can be taken to satisfy $\|f\| = 1$ and $f(x) = \delta$.

Question:

Let $X$ be a normed vector space.

a.) If $M$ is a closed subspace and $x\in X \setminus M$ then $M + \mathbb{C}x$ is closed. (Use Theorem 5.8a.)).

b.) Every finite-dimensional subspace of $X$ is closed.

Proof a.) - Let $M$ be a proper closed subspace of $X$ and let $x\in X\setminus M$. There exists $f\in X^*$ such that $f(x)\neq 0$ and $f(M) = \{0\}$. Let $\{u_n + a_nx \}_{1}^{\infty}$ be a sequence in $M + Kx$ that converges to $y\in X$. Then $$f(y) = f(\lim_{n\rightarrow \infty} (u_n + a_n x))=\lim\limits_{n\rightarrow \infty}f(u_n + a_n x) = \lim\limits_{n\rightarrow \infty} a_nf(x)$$ since $f$ is continuous, so $\{a_n\}_{1}^{\infty}$ converges to $a:= f(y)/f(x)$, which implies that $\{a_nx \}_{1}^{\infty}$ converges to $ax$. Therefore $$\{u_n\}_{1}^{\infty} = \{(u_n + a_nx) - a_nx\}_{1}^{\infty} \to (y - ax)$$ which lies in $M$ because $M$ is closed. It follows that $y\in M + Kx$, which shows that $M + Kx$ is closed.

Proof b.) - Let $Y$ be a finite dimensional subspace of $X$. Let $\{e_1, \ldots, e_n\}$ be a basis of $Y$. Now, note that $M=\{0\}$ is a closed subspace of $X$. So, by item a.), $M + \mathbb{C}e_1$ is a closed subspace of $X$. Again, by item a.) we have that $M + \mathbb{C}e_1+\mathbb{C}e_2$ is a closed subspace of $X$. Repeating the argument enough times, we have that $M + \mathbb{C}e_1+\ldots+\mathbb{C}e_n$ is a closed subspace of $X$. But $M + \mathbb{C}e_1+\ldots+\mathbb{C}e_n=Y$. So we have proved that $Y$ is a closed subspace of $X$.