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THIS QUESTION IS NOT A DUPLICATE OF THIS ONE: INDEED, I PROPOSED TO ANALYSE AN ALTERNATIVE PROOF AND IN PARTICUAR I ASK FOR A TOPOLOGICAL PROOF WHICH SUBSTANTIALLY USE NOT MANY ALGEBRAIC MANIPULATIONS.

Definition 0

A norm on a complex/real vector space $V$ is a function $\nu$ from $V$ to $\Bbb R^+_0$ such that

  1. for any $\pmb v\in V$ the value $\nu(\pmb v)$ is zero if and only if $\pmb v$ is the zero vector;
  2. for any $\alpha\in\Bbb K$ and for any $\pmb v\in V$ the identity $$\begin{equation}\tag{2}\label{2}\nu(\alpha\cdot\pmb v)=|\alpha|\cdot\nu(\pmb v)\end{equation}$$ holds;
  3. for any $\pmb v_1,\pmb v_2\in V$ the inequality $$\begin{equation}\tag{3}\label{3}\nu(\pmb v_1+\pmb v_2)\le\nu(\pmb v_1)+\nu\pmb(v_2)\end{equation}$$ holds.

Now using $\eqref{3}$ we observe that for any $\pmb v_1,\pmb v_2\in V$ the inequalities $$ \begin{equation}\tag{4}\label{4}\nu(\pmb v_1)=\nu\big(\pmb v_1+(\pmb v_2-\pmb v_2)\big)=\nu\big((\pmb v_1-\pmb v_2)+\pmb v_2\big)\le\nu(\pmb v_1-\pmb v_2)+\nu(\pmb v_2)\end{equation} $$ $$ \begin{equation}\tag{5}\label{5}\nu(\pmb v_2)=\nu\big(\pmb v_2+(\pmb v_1-\pmb v_1)\big)=\nu\big((\pmb v_2-\pmb v_1)+\pmb v_1\big)\le\nu(\pmb v_2-\pmb v_1)+\nu(\pmb v_1)\end{equation} $$ holds and thus we conclude that actually for any $\pmb v_1,\pmb v_2\in V$ the inequality $$ \begin{equation}\tag{6}\label{6}\big|\nu(\pmb v_1)-\nu(\pmb v_2)\big|\le\nu(\pmb v_1-\pmb v_2)\end{equation} $$ holds.

Now we prove the following important lemma

Lemma 7

A Cauchy sequence $(x_n)_{n\in\omega}$ in a (metric) space $X$ converges if and only if it has a cluster point $x\in X$.

Proof. If $(x_n)_{n\in\omega}$ is a Cauchy sequence then for any $\epsilon\in\Bbb R^+$ there exists $n_\epsilon\in\omega$ such that $$ d(x_{n_1},x_{n_2})<\frac\epsilon 2 $$ for any $n_1,n_2\ge n_\epsilon$; after all, if $x$ is a cluster point for $(x_n)_{n\in\omega}$ then there exists $\nu(\epsilon)\ge n_\epsilon$ such that $$ d(x,x_{\nu(n_\epsilon)})<\frac\epsilon 2 $$ so, in this case, we conclude that for any $n\ge n_\epsilon$ the inequality $$ d(x,x_n)\le d(x,x_{\nu(n_\epsilon)})+d(x_{\nu(n_\epsilon)},x_n)<\frac\epsilon 2+\frac\epsilon 2=\epsilon $$ holds so that $(x_n)_{n\in\omega}$ converges. Now we remember that the limit of a net is a cluster point for it so that in this particular case if $(x_n)_{n\in\omega}$ converges to $x$ then this point is a cluster point for $(x_n)_{n\in\omega}$.

Clearly the previous lemma implies the following proposition

Proposition 8

Any Cauchy sequence $(x_n)_{n\in\omega}$ in a sequentially compact (metric) space $X$ converges.

Proof. If $X$ is sequentially compact then $(x_n)_{n\in\omega}$ has a convergent subsequence $(x_{n_l})_{l\in\omega}$. However a net has a cluster point if and only if it has a subnet converging there. So we conclude that $(x_n)_{n\in\omega}$ has a cluster point and thus by the previous lemma we conclude that $(x_n)_{n\in\omega}$ converges.

Finally we remember the following notable theorem

Theorem 9

For a metrizable space compactness, limit point compactness and sequential compactness are equivalent.

Now let's we suppose that $(x_n)_{n\in\omega}$ is a Cauchy sequence for a (real) normed space $(V,\nu)$ so that there exists $n_1\in\omega$ such that the inequality $$ \begin{equation}\tag{10}\label{10}\Big|(\pmb v_{h_1})-\nu(\pmb v_{h_2})\Big|\le\nu(\pmb v_{h_1}-\pmb v_{h_2})<1\end{equation} $$ holds for any $h_1,h_2\ge n_1$: so putting $$ M:=\max_{n\le n_1}\nu(\pmb v_n) $$ we first observe that $$ \nu(\pmb v_n)\le M<M+1 $$ for any $n< n_1$ and then by $\eqref{10}$ we observe that $$ \nu(\pmb v_n)<1+\nu(\pmb v_{n_1})\le M+1 $$ so that we conclude that for any $n\in\omega$ the inequality $$ \begin{equation}\tag{11}\label{11}\nu(\pmb v_n)< M+1\end{equation} $$ holds and thus we finally conclude that any Cauchy sequence is bounded and in particular it is contained in $B(\pmb 0,M+1)$ where $\pmb 0$ is the zero vector. Now if the dimension of $V$ is finite then it is homeomorphic to $\Bbb R^n$ via any linear isomorphism so that in this case the clousure of $B(\pmb 0,M+1)$ is compact (it is closed and bounded) and then for $9$ sequentially compact: so $(x_n)_{n\in\omega}$ has a convergent subsequence and thus by $7$ even $(x_n)_{n\in\omega}$ converges, since a limit point for a subsequence is a cluster point for the sequence. So we conclude that $V$ is complete.

So we actually proved that any (real) normed space with finite dimension is complete! and this is surely (at least in my opinion) a very strong result but unfortunately I do not know if the proof I gave is correct so that I thought to put a specific question where I ask to analyze it: substantially I ask if the closure of $B(\pmb 0,M+1)$ is compact because I do not see it directly via any homeomorphism with $\Bbb R^n$ with $n\in\omega$ dimension of $V$. So could someone help me, please?

Antonio Maria Di Mauro
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  • The norms give equivalent topologies so the map $(V,|;|2)\rightarrow (V,|;|^*)$ given by $x\mapsto x$ and its inverse are bounded linear operators. Then there are contacts $a,b>0$ such that $a|x|\leq |x|_2\leq b|x|^$. – Mittens Oct 18 '22 at 18:16
  • @OliverDíaz Sorry, could you edit the latex? – Antonio Maria Di Mauro Oct 18 '22 at 18:16
  • I just did it! Hope it's fine now. – Mittens Oct 18 '22 at 18:17
  • @OliverDíaz Yes, I saw. Anyway I do not understand completely what you want mean. So you observer that the identity is a linear operator, right? But why this implies the existence of $a$ and $b$? – Antonio Maria Di Mauro Oct 18 '22 at 18:18
  • Check the reference (this is a standard result) The point is that if $n$ and $m$ are norms in $V$ then the linear operator $x\mapsto x$ is bounded (continuous) since the topologies are the same. What does it mean for an operator to be bounded? Now $|;|_2$ is known to be complete. – Mittens Oct 18 '22 at 18:21
  • @OliverDíaz The adjective bounded is equivalent to the adjective continuous? – Antonio Maria Di Mauro Oct 18 '22 at 18:23
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    for linear operators between normed spaces,yes. – Mittens Oct 18 '22 at 18:24
  • @OliverDíaz Well, I know that a linear operator is continuous if and only if it is continuous at 00 but this it seems to me does not help now. Perhaps I have to work with neighborhoods using continuity? Sorry, but I have no many ideas now. – Antonio Maria Di Mauro Oct 18 '22 at 18:33
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    It is not difficult to show that a linear operator $A$ between normed space $(V_1,n_1)$ and $(V_2,n_2)$ is continuous if it send bounded sets into bounded sets. This means that there is a constant $K>0$ such that $n_2(Ax)\leq K n_1(x)$. – Mittens Oct 18 '22 at 18:34
  • @OliverDíaz Okay, so if $V_1$ is complete with respect $n_1$ then using your last inequality with $V_1=V_2$ I can prove that $V_1$ is even complete with respect $n_2$, right? However I have to show that there exists at least one complete norm, right? – Antonio Maria Di Mauro Oct 18 '22 at 18:38
  • Let us continue this discussion in chat. – Mittens Oct 18 '22 at 18:39

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