How to understand the concept of norm equivalence? I'm mainly dealing with $\mathbb{R}^n$.
$\nu(\cdot)$ and $\mu(\cdot)$  are equivalent iff 
there exist constants $c_1,c_2>0$ such that for every $x \in \mathbb{R}^n$, 
$c_1\nu(x)\leq \mu(x)\leq c_2\nu(x)$.
I understand the definition. What I don't understand is why can we say that every element that satisfies a property where one norm is used, then it will satisfy the same property but with the other norm. Or this is not what is meant by equivalent norms?
By property I mean for example convergence, or continuity... What allows me to say that if two different normed spaces (different norms but same vector space), whenever on sequence converges in one of the normed spaces, the same sequence converges in the other normed space?
 A: First of all, this yours definition isn't right.
Definition: We say that $\nu(\cdot)$ and $\mu(\cdot)$  are equivalent norms iff there exist constants $c_1,c_2>0$ such that $c_1\nu(x)\leq \mu(x)\leq c_2\nu(x)$ for every $x \in \Bbb{R}^n.$
What does it mean?? It means that the topology induced by $\nu(\cdot)$ and the topology induced by $\mu(\cdot)$ are the same, that is, the families $\sigma_\nu$ and $\sigma_\mu$ of open sets in $(\Bbb{R}^n, \nu)$ and $(\Bbb{R}^n, \mu)$, respectively, will be equals.
A: Let's make a concrete example of a property that is invariant for equivalent norms: if $\|\cdot\|_1\sim\|\cdot\|_2$ on the space $V$, then you have the same convergent sequences in your space. This is actually a metric condition (i.e., invariant for equivalent of metrics), more than a norm condition. In fact, recall that any normed space $(V,\|\cdot\|)$ is naturally a metric space with distance $\mathrm{d}(x,y)=\|x-y\|$. 
I will then prove the statement only for sequences $\{x_n\}_n$ tending to zero.
In particular, equivalence of norms tells you that, for any given radius and any given ball (say, centered in zero) w.r.t. the first norm, you can find another ball in the second norm containing it and a third ball in the second norm which is contained in it. 
For any given $r,s>0$, denote $B^1_r=\{x\in V\mid \|x\|_1< r\}$ and $B^2_r=\{x\in V\mid \|x\|_2< s\}$.
Given $r>0$, you can find $s,t>0$ s.t. $B_s^2\subseteq B_r^1\subseteq B_t^2$. 
Assume now that $x_n\to 0$ w.r.t. the topology induced by $\|\cdot\|_1$, i.e., by definition: for any $\epsilon >0$ there exists $N\in\mathbb{N}$ such that for all $n\geq N$ you have $x_n\in B_{\epsilon}^1$.
You want to prove a similar statement for $\|\cdot\|_2$.
Thus, given $\epsilon >0$, by the equivalence of norms, you can find $r>0$ s.t. $rB^1_{\epsilon}=\{r\cdot x\mid x\in B^1_{\epsilon}\}\subseteq B^2_{\epsilon}$ (we're just rescaling the ball). If $n$ is great enough, you have that $x_n\in rB_{\epsilon}^1$, thus also $x_n\in B_{\epsilon}^2$: now you are given $\epsilon'=r\epsilon$, hence you choose $M\in\mathbb{N}$ s.t. $x_n\in rB_{\epsilon}^1$ for $n\geq M$. Then you conclude.
Please, notice the fact that the topology induced by the norm is generated by the (open) balls. This gives the connection by the metric properties and the equivalence of norms.
