# Understanding of the theorem that all norms are equivalent in finite dimensional vector spaces

The following is a well-known result in functional analysis:

If the vector space $X$ is finite dimensional, all norms are equivalent.

Here is the standard proof in one textbook. First, pick a norm for $X$, say

where $x=\sum_{i=1}^n\alpha_ix_i$, and $(x_i)_{i=1}^n$ is a basis for $X$. Then show that every norm for $X$ is equivalent to $\|\cdot\|_1$, i.e.,

For the first inequality, one can easily get $c$ by triangle inequality for the norm. For the second inequality, instead of constructing $C$, the Bolzano-Weierstrass theorem is applied to construct a contradiction.

The strategies for proving these two inequalities are so different. Here is my question,

Can one prove this theorem without Bolzano-Weierstrass theorem?

UPDATE:

Is the converse of the theorem true? In other words, if all norms for a vector space $X$ are equivalent, then can one conclude that $X$ is of finite dimension?

If $(X,\|\cdot\|)$ is a normed space of infinite dimension, we can produce a non-continuous linear functional: Choose an algebraic basis $\{e_{i}\}_{i \in I}$ which we may assume to be normalized, i.e., $\|e_{i}\| = 1$ for all $i$. Every vector $x \in X$ has a unique representation $x = \sum_{i \in I} x_i \, e_i$ with only finitely many nonzero entries (by definition of a basis).
Now choose a countable subset $i_1,i_2, \ldots$ of $I$. Then $\phi(x) = \sum_{k=1}^{\infty} k \cdot x_{i_k}$ defines a linear functional on $x$. Note that $\phi$ is not continuous, as $\frac{1}{\sqrt{k}} e_{i_k} \to 0$ while $\phi(\frac{1}{\sqrt{k}}e_{i_k}) = \sqrt{k} \to \infty$.
There can’t be a $C \gt 0$ such that the norm $\|x\|_{\phi} = \|x\| + |\phi(x)|$ satisfies $\|x\|_\phi \leq C \|x\|$ since otherwise $\|\frac{1}{\sqrt{k}}e_k\| \to 0$ would imply $|\phi(\frac{1}{\sqrt{k}}e_k)| \to 0$ contrary to the previous paragraph.