Proof. Let $\( X \)$ be a finite dimensional normed linear space. Let $\( \mathopen{}\left\{\, {e}_{1}, {e}_{2}, \dotsc, {e}_{n}\,\right\}\mathclose{} \)$ be a basis for $\( X \)$. Then any $\( x\in X \)$ can be written $\( \sum_{i=1}^{n}{}{α}_{i}{e}_{i} \)$ with each $\( {α}_{i}\in \mathbb{C} \)$. Define a Euclidean norm $\( ρ \)$ on $\( X \)$ by $$\[ ρ\mathopen{}\left( x\right)\mathclose{}= ρ\mathopen{}\left( \sum_{i=1}^{n}{}{α}_{i}{e}_{i}\right)\mathclose{}= {\mathopen{}\left(\sum_{i=1}^{n}{}{\mathopen{}\left\lvert{}{α}_{i}\right\rvert\mathclose{}}^{2}\right)\mathclose{}}^{\frac{1}{2}} \text{.} \]$$ We want to show that the norm on $\( X \)$ and $\( ρ \)$ determine the same topology, i.e. that any $\( \mathopen{}\left\lVert{}\cdot\right\rVert\mathclose{} \)$-open set contains a $\( ρ \)$-open set and vice versa. $$\[ \mathopen{}\left\lVert{}x\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\sum_{i=1}^{n}{}{α}_{i}{e}_{i}\right\rVert\mathclose{}\leq \sum_{i=1}^{n}{}\mathopen{}\left\lVert{}{α}_{i}{e}_{i}\right\rVert\mathclose{}= \sum_{i=1}^{n}{}\mathopen{}\left\lvert{}{α}_{i}\right\rvert\mathclose{}\mathopen{}\left\lVert{}{e}_{i}\right\rVert\mathclose{}\leq {\mathopen{}\left(\sum_{i=1}^{n}{}{\mathopen{}\left\lvert{}{α}_{i}\right\rvert\mathclose{}}^{2}\right)\mathclose{}}^{\frac{1}{2}}{\mathopen{}\left(\sum_{i=1}^{n}{}{\mathopen{}\left\lVert{}{e}_{i}\right\rVert\mathclose{}}^{2}\right)\mathclose{}}^{\frac{1}{2}} \text{.} \]$$ Let $\( M= {\mathopen{}\left(\sum_{i=1}^{n}{}{\mathopen{}\left\lVert{}{e}_{i}\right\rVert\mathclose{}}^{2}\right)\mathclose{}}^{\frac{1}{2}} \)$ Thus, $\( \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\leq Mρ\mathopen{}\left( x\right)\mathclose{} \)$ and so any $\( \mathopen{}\left\lVert{}\cdot\right\rVert\mathclose{} \)$-open $\( ε \)$-ball contains a $\( ρ \)$-open $\( \frac{ε}{M} \)$-ball.

Now we want to show that any $\( ρ \)$-open ball contains a $\( \mathopen{}\left\lVert{}\cdot\right\rVert\mathclose{} \)$-open ball. Define $\( f : {\mathbb{C}}^{n} \to \mathbb{R} \)$ by $\( f\mathopen{}\left( \mathopen{}\left({α}_{1}, {α}_{2}, \dotsc, {α}_{n}\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left\lVert{}\sum_{i=1}^{n}{}{α}_{i}{e}_{i}\right\rVert\mathclose{} \)$. This function is continuous in the Euclidean topology (use the inequality above). The unit sphere in $\( {\mathbb{C}}^{n} \)$, $\( \mathopen{}\left\{\, \mathopen{}\left({α}_{1}, {α}_{2}, \dotsc, {α}_{n}\right)\mathclose{}\,\middle\vert\, , \sum_{i=1}^{n}{}{\mathopen{}\left\lvert{}{α}_{i}\right\rvert\mathclose{}}^{2}= 1, \,\right\}\mathclose{} \)$, is compact, hence $\( f \)$ attains a minimum $\( m\gt 0 \)$.

So for any $\( x\in X \)$ with $\( ρ\mathopen{}\left( x\right)\mathclose{}= 1 \)$ we have $\( \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\geq mρ\mathopen{}\left( x\right)\mathclose{} \)$. This extends to all of $\( X \)$ by properties of the norm. If $\( ρ\mathopen{}\left( x\right)\mathclose{}\gt 0 \)$, then $\( ρ\mathopen{}\left( \frac{x}{ρ\mathopen{}\left( x\right)\mathclose{}}\right)\mathclose{}= 1 \)$, and $\( \mathopen{}\left\lVert{}\frac{x}{ρ\mathopen{}\left( x\right)\mathclose{}}\right\rVert\mathclose{}\geq mρ\mathopen{}\left( \frac{x}{ρ\mathopen{}\left( x\right)\mathclose{}}\right)\mathclose{} \)$. Therefore, by multiplying both sides of the inequality by $\( ρ\mathopen{}\left( x\right)\mathclose{} \)$, we get the desired inequality $\( \mathopen{}\left\lVert{}x\right\rVert\mathclose{}\geq mρ\mathopen{}\left( x\right)\mathclose{} \)$ for all $\( x \)$ in $\( X \)$. Thus, any any $\( ρ \)$-open ball contains a $\( \mathopen{}\left\lVert{}\cdot\right\rVert\mathclose{} \)$-open ball.

Proof. Including the above proof, we know the following about $\( Y \)$: Any norm of $\( Y \)$ is equivalent to the Euclidean norm on $\( {\mathbb{C}}^{n} \)$; Every Cauchy sequence in $\( Y \)$ is Cauchy in $\( {\mathbb{C}}^{n} \)$; $\( {\mathbb{C}}^{n} \)$ is a complete metric space, thus every Cauchy sequence converges; Convergence in $\( {\mathbb{C}}^{n} \)$ implies convergence in $\( Y \)$; $\( Y \)$ is a complete subspace of the metric space $\( X \)$, thus $\( Y \)$ is closed.

Proof. Let $\( {x}_{0}\in X\setminus Y \)$. The distance from $\( {x}_{0} \)$ to $\( Y \)$, $\( \operatorname{d}\mathopen{}\left( {x}_{0}, Y\right)\mathclose{} \)$, is positive (else you can find a sequence of elements in $\( Y \)$ converging to $\( {x}_{0} \)$, which would contradict that $\( Y \)$ is a closed subspace). Let $\( ε \)$ be a positive real number. There exists $\( {y}_{0}\in Y \)$ such that $\( 0\lt \mathopen{}\left\lVert{}{x}_{0}-{y}_{0}\right\rVert\mathclose{}\leq \operatorname{d}\mathopen{}\left( {x}_{0}, Y\right)\mathclose{}+ε \)$. Let $\( x= \frac{{x}_{0}-{y}_{0}}{\mathopen{}\left\lVert{}{x}_{0}-{y}_{0}\right\rVert\mathclose{}} \)$. Then $\( \mathopen{}\left\lVert{}x\right\rVert\mathclose{}= 1 \)$. We want to show that with the right choice of $\( ε \)$ we get the inequality we want. Because, for any $\( y\in Y \)$, $$\[ \mathopen{}\left\lVert{}x, y\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\frac{{x}_{0}-{y}_{0}}{\mathopen{}\left\lVert{}{x}_{0}-{y}_{0}\right\rVert\mathclose{}}-y\right\rVert\mathclose{}= \frac{1}{\mathopen{}\left\lVert{}{x}_{0}-{y}_{0}\right\rVert\mathclose{}}\mathopen{}\left\lVert{}\mathopen{}\left({x}_{0}-{y}_{0}\right)\mathclose{}-\mathopen{}\left(\mathopen{}\left\lVert{}{x}_{0}-{y}_{0}\right\rVert\mathclose{}y\right)\mathclose{}\right\rVert\mathclose{}\gt \frac{\operatorname{d}\mathopen{}\left( x, Y\right)\mathclose{}}{\operatorname{d}\mathopen{}\left( x, Y\right)\mathclose{}+ε}= 1-\frac{ε}{\operatorname{d}\mathopen{}\left( x, Y\right)\mathclose{}+ε} \text{,} \]$$ we just need to choose $\( ε \)$ so that $\( \frac{ε}{\operatorname{d}\mathopen{}\left( x, Y\right)\mathclose{}+ε}\lt r \)$.

Proof. Let $\( X \)$ be a finite dimensional normed linear space. Let $\( S\subseteq X \)$ be a closed and bounded subset. By Proposition I.62, closed and bounded in the norm implies closed and bounded in the Euclidean norm. So, the Heine-Borel Theorem applies and $\( S \)$ is compact.

Suppose every closed, bounded subset of $\( X \)$ is compact. By way of contradiction, suppose that $\( X \)$ is infinite dimensional. We'll construct a bounded sequence with no convergent subsequence to get the contradiction.

Let $\( {x}_{1}\in X \)$ such that $\( \mathopen{}\left\lVert{}{x}_{1}\right\rVert\mathclose{}= 1 \)$. Let $\( {Y}_{1}= \operatorname{span}\mathopen{}\left( \mathopen{}\left\{\, {x}_{1}\,\right\}\mathclose{}\right)\mathclose{} \)$. By Proposition I.63, $\( {Y}_{1} \)$ is closed and by Lemma I.65, there exists an $\( {x}_{2}\in X\setminus {Y}_{1} \)$ such that $\( \mathopen{}\left\lVert{}{x}_{2}\right\rVert\mathclose{}= 1 \)$ and $\( \mathopen{}\left\lVert{}{x}_{1}-{x}_{2}\right\rVert\mathclose{}\gt \frac{1}{2} \)$. Let $\( {Y}_{2}= \operatorname{span}\mathopen{}\left( \mathopen{}\left\{\, {x}_{1}, {x}_{2}\,\right\}\mathclose{}\right)\mathclose{} \)$. As before, $\( {Y}_{2} \)$ is closed and there exists an $\( {x}_{3}\in X\setminus {Y}_{2} \)$ such that $\( \mathopen{}\left\lVert{}{x}_{3}\right\rVert\mathclose{}= 1 \)$, $\( \mathopen{}\left\lVert{}{x}_{1}-{x}_{3}\right\rVert\mathclose{}\gt \frac{1}{2} \)$, and $\( \mathopen{}\left\lVert{}{x}_{2}-{x}_{3}\right\rVert\mathclose{}\gt \frac{1}{2} \)$. Continue forever, which you can do because $\( X \)$ is infinite dimensional. In this way, we get a bounded infinite sequence $\( \mathopen{}\left({x}_{n}\right)\mathclose{}_{n=1}^{\infty} \)$ where $\( \mathopen{}\left\lVert{}{x}_{n}\right\rVert\mathclose{}= 1 \)$ for all $\( n \)$, and $\( \mathopen{}\left\lVert{}{x}_{i}-{x}_{j}\right\rVert\mathclose{}\gt \frac{1}{2} \)$ when $\( i\neq j \)$. Therefore, $\( \mathopen{}\left({x}_{n}\right)\mathclose{}_{n=1}^{\infty} \)$ has no convergent subsequence, which contradicts the assumption that every closed, bounded subset is compact.