# Proof the theorem of two normed space are equivalent

Let $$X$$ be vector space over field $$F$$. Two norm $$\Vert\cdot\Vert_1$$ and $$\Vert\cdot\Vert_2$$ on $$X$$ is equivalent if there exist $$k,K\in\mathbb{R}$$ such that $$k\Vert x\Vert_1\leq \Vert x\Vert_2\leq K\Vert x\Vert_1$$ or $$K^{-1}\Vert x\Vert_2\leq \Vert x\Vert_1\leq k^{-1}\Vert x\Vert_2$$ for all $$x\in X$$.

Now, I want to prove this theorem.

Let $$X$$ be vector space over field $$F$$. If $$\Vert\cdot\Vert_1$$ and $$\Vert\cdot\Vert_2$$ is equivalent on $$X$$ with metrics $$d_1$$ and $$d_2$$ respectively, then a sequence $$\{x_n\}$$ convergent to $$x$$ on metric space $$(X,d_1)$$ if and only if $$\{x_n\}$$ convergent to $$x$$ on metric space $$(X,d_2)$$.

I know, that a sequence $$\{x_n\}$$ is convergent on metric space $$(X,d_1)$$ if $$(\forall \varepsilon>0) (\exists N\in\mathbb{N}) \text{ such that }(\forall n\geq N), d_1(x_n,x)<\varepsilon.$$

Now I confused to associate the definition of $$\{x_n\}$$ convergent with equivalence of two norms.

Any idea to proof this theorem?

$$d_1(x,y)=\|x-y\|_1$$ so $$d_1(x_n,x) \to 0$$ iff $$\|x_n-x\|_1 \to 0$$. Similarly $$d_2(x_n,x) \to 0$$ iff $$\|x_n-x\|_2 \to 0$$. Can yo see from the stated inequalities that $$\|x_n-x\|_1 \to 0$$ iff $$\|x_n-x\|_2 \to 0$$.
We see that by the definition of equivalence, the the convergence of one sequence implies convergence of the other. Indeed, if we have that $$\{x_n\}_{n\geq 1}$$ converging to $$x$$, we have that $$||x_n-x||_1 \rightarrow 0$$ If we let $$\varepsilon>0$$, we take the $$K >0$$ such that $$||x||_1 \leq K||x||_2 \quad \forall x\in X$$ So, we take $$N$$ large such that $$\forall n\geq N$$, $$||x_n-x||_1 \leq \frac{\varepsilon}{K}$$ Then we see that for all $$n>N$$, $$||x_n-x||_2 \leq K||x_n-x|| < \varepsilon$$ Since $$\varepsilon$$ was arbitrary, we get convergence in the second norm.