What is the implication that $\| \cdot \|_2$ and $\| \cdot \|_\infty$ are equivalent norms on $\mathbb{R^2}$ Given $\mathbb{X}$ = $\mathbb{R^2}$, consider $\| \cdot \|_2$ and $\| \cdot \|_\infty$ 
We can show that 
$\| x \|_\infty \leq \| x \|_2 \leq \sqrt2 \| x \|_\infty$  
Hence $\| \cdot \|_2$ and $\| \cdot \|_\infty$ are equivalent norms
Is there some deeper implication regarding this particular relationship? Why do we care if two norms are equivalent in this sense?
 A: As pointed out by Clement C in the comments: Equivalent norms induce the same topology. Also the other direction of implication is true: When two norms induce the same topology then they are equivalent.
Take two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a vector space and you ask yourself: When are the topologies from those norms the same? This is the case when open sets for $\|\cdot\|_1$ are also open in $\|\cdot\|_2$ and vice versa. This is the case when in each open ball in $\|\cdot\|_1$ contains an open ball of $\|\cdot\|_2$ and the other way around. From this you can prove that there are constants $c$ and $C$ such that $c\|\cdot\|_1 \le \|\cdot\|_2 \le C \|\cdot\|_1$.
So the answer to your question is: Two norms are equivalent iff their induced topologies are the same.
What is the benefit if two topologies are the same? If a "topological property" is valid for one norm, then it is valid for the other norm. For example:


*

*Open, closed and compact sets are the same.

*If a sequence converges in one norm, it converges also in the other norm.

*If a sequence is a Cauchy sequence in one norm it is Cauchy in the other.

*...

A: Some application of this result and more generally that all norms are equivalent on finite dimensional spaces:


*

*The compact subspaces are the closed bounded spaces.

*All linear maps are continuous. More generally all multilinear maps are continuous.

*All linear maps are bounded on the unit ball.

