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Let $X$ be an $n$-dimensional vector space with a basis $\{e_1,\dots,e_n\}$. Consider the norm $\|\sum_{i=1}^n \alpha_ie_i\|=\max_{i\leqslant n} |\alpha_i|$ for $x=\sum_{i=1}^n\alpha_ie_i\in X$.

We say that two normed spaces $(X,\|\cdot\|_X$ and $(Y,\|\cdot\|_Y)$ (over the same field) are isometrically isomorphic if there is a bijective linear map $\psi\colon X\to Y$ such that $\|\psi(x)\|_Y=\|x\|_X$ for all $x\in X$.

We want to prove that $(X,\|\cdot\|)$ is isometrically isomorphic to $(\mathbb R^n, \|\cdot\|_\infty)$, where $\|\cdot\|_\infty$ is defined by $\|\mathbf x\|=\max_{i\leqslant n}|x_i|$ for $\mathbf x=(x_1,\dots,x_n)\in\mathbb R^n.$

Firstly, I think the lecturer forgot to say that $X$ is a vector space over $\mathbb R$, otherwise the result does not hold. Is this correct?

Then, the proof we were given considers the map $\psi(\sum_{i=1}^n\alpha_ie_i)=(\alpha_1,\dots,a_n)$. This is clearly linear, injective and an isometry.

However, I do not see how it is surjective. How can we prove that it is surjective when we are not given its co-domain? Which co-domain would make it surjective?

As you see this is not homework, so a complete answer would be appreciated. Thank you in advance.

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The co-domain here is $\mathbb{R}^n$, because we are trying to construct an isomorphism $\psi: X\to \mathbb{R}^n$. To see that the map is surjective, let $(x_1, \dots, x_n)$ be a vector in $\mathbb{R}^n$. Now, $\psi(x_1 e_1 + \dots + x_n e_n) = (x_1, \dots, x_n)$, so every vector in $\mathbb{R}^n$ is in the image of $\psi$.

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  • $\begingroup$ You are right. That was silly of me. Thanks! $\endgroup$ – Ryuky Dec 26 '15 at 20:58

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