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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.

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    $\begingroup$ Ideas for (b) - Bolzano-Weierstrass theorem in $\mathbb{R}^n$ and completeness of $\mathbb{R}^n$ $\endgroup$
    – user288742
    Aug 5 '16 at 6:18
  • $\begingroup$ In particular, a vector space of dimension $n$ is isomorphic to $\mathbb R^n$. $\endgroup$
    – Math1000
    Aug 5 '16 at 6:54
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    $\begingroup$ Another idea for (b): Use induction on the dimension, and use part (a) $\endgroup$
    – Aweygan
    Aug 5 '16 at 7:00
  • $\begingroup$ I second the idea of using induction on the dimension and (a). The base case $M=\{0\}$ is trivial. $\endgroup$
    – Claudius
    Aug 5 '16 at 7:45
  • $\begingroup$ I see, by the way should I note in my proof that $u_n$ is going to zero? or is that obvious to the reader? $\endgroup$
    – Wolfy
    Aug 5 '16 at 19:36
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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 $X$.

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    $\begingroup$ I see for a.) did we already know that $f$ is continuous? $\endgroup$
    – Wolfy
    Aug 5 '16 at 20:02
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    $\begingroup$ @Wolfi, Yes, for a.) we know $f$ is a continuous linear functional. The existence of $f$ in a.) comes from the Hahn-Banach Theorem. $\endgroup$
    – Ramiro
    Aug 6 '16 at 3:27
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    $\begingroup$ @RenanWillianPrado , It is a well-known corollary of Hahn-Banach Theorem that if $M$ is a proper closed subspace of $X$ and $x\in X\setminus M$, then there exists $f\in X^*$ such that $f(x)\neq 0$ and $f(M) = \{0\}$. Of course, $f\in X^*$ means that $f$ continuous (and linear). $\endgroup$
    – Ramiro
    Mar 18 at 0:04
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    $\begingroup$ @RenanWillianPrado I see what you mean. In the proof of the corollary I mentioned, we use an "intermediate" linear function $g$ defined in span($M$, x). Such function has kernel $M$ and $g(x) \neq 0$. Such function $g$ is continuous and the simplest way to see it is continuous is because its kernel is closed. I guess it was to this intermediate function that you were talking about. I am used to apply the corollary directly. $\endgroup$
    – Ramiro
    Mar 18 at 0:22
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    $\begingroup$ @RenanWillianPrado , I understand. But it is rather subjective. In several situations, it is more direct and practical to apply the corollaries, instead of applying the theorem in all its generality. In fact, that is why it is important to know the corollaries (as well as the theorem itself). $\endgroup$
    – Ramiro
    Mar 18 at 3:14

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