TITLE: Definition of Equivalent Norms
QUESTION [10 upvotes]: Two norms $F,G$ are equivalent when there are constants $a,b$ such that $aF \le G \le bF$.  I'm reading about this idea, and so far I've seen that equivalence of norms implies that the underlying space $X$ has the same topology with respect to either norm.  Maybe it preserves even more properties than this too.  
But I'm finding it very difficult to use this property when doing proofs or problems because although it's very simple to state, I don't immediately see what it is saying.  In comparison, when you define 'equivalence' in other settings, like in the definition of an isomorphism of abelian groups or a continuous map, it's very clear that a certain operation or object is being preserved as you pass across a map.  
To be concrete, here are my questions: 
(1) Is there another way to characterize when norms are equivalent that might provide more intuition for what it says about is being preserved 
and (2) is there a way to show that this definition is the one you want by starting with something more fundamental (like saying that the norms induce the same topology) and then proving that it's equivalent to the stated definition?  
Any intuition for the definition would be helpful for either of these questions.

REPLY [6 votes]: PROBLEM
Answer

 Equivalent norms define the same uniform vector space.

Explanation

 Completeness is a concept by uniform vector spaces.

Demonstration
Given the real line $\mathbb{R}$.
Consider the metrics:
$$d(x,y):=|y-x|$$
$$d'(x,y):=\arctan|y-x|$$
Then one obtains:
$$\mathcal{N}=\mathcal{N}'\quad\mathcal{U}\neq\mathcal{U}$$
Concluding problem.

PROOF
Identification
Given normed spaces $\Omega$ and $\Omega'$.
Regard the category:
$$\mathrm{UVS}:\quad\mathrm{Hom}(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_U(\Omega,\Omega')$$
Identification:
$$\Phi:\Omega\leftrightarrow\Omega':x\mapsto x$$
Linearity follows:
$$\Phi(x+y)=x+y=x+'y=\Phi(x)+'\Phi(y)$$
$$\Phi(\lambda\cdot x)=\lambda\cdot x=\lambda\cdot'x=\lambda\cdot'\Phi(x)$$
But by the below:
$$\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_U(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_L(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{B}(\Omega,\Omega')$$
Explicitely that is:
$$\|x\|'=\|\Phi x\|'\leq\|\Phi\|\cdot\|x\|$$
$$\|x'\|=\|\Phi^{-1}x'\|\leq\|\Phi^{-1}\|\cdot\|x'\|'$$
Concluding proof.

CATEGORIES
Topological Vector Spaces
Note for linear maps:
$$\Phi\in\mathcal{L}(\Omega,\Omega')\implies\Phi^{-1}\in\mathcal{L}(\Omega',\Omega')$$
Continuous at zero:
$$\mathcal{C}_0(\Omega,\Omega'):=\{\Phi:\Omega\to\Omega:\Phi^{-1}(\mathcal{N}_{\Phi0})\subseteq\mathcal{N}_0\}$$
It holds equality:
$$\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_0(\Omega,\Omega')$$
Homomorphisms:
$$\mathrm{Hom}(\Omega,\Omega'):=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}(\Omega,\Omega')$$
Isomorphic spaces:
$$\Omega\cong\Omega':\iff\Phi:\Omega\leftrightarrow\Omega':\quad\Phi(\mathcal{N})=\mathcal{N}'$$
$$\Phi(x+y)=\Phi(x)+'\Phi(y)\quad((x,y)\in\Omega\times\Omega)$$
$$\Phi(\lambda\cdot x)=\lambda\cdot'\Phi(x)\quad((x,\lambda)\in\Omega\times\mathbb{C})$$
Basic entourages:
$$B_N:=\{(x,y):(y-x)\in N\}\subseteq\Omega\times\Omega$$
Uniform structure:
$$\mathcal{U}:=\uparrow\{B_N: N\in\mathcal{N}_0\}$$
Going first step up:
Uniform Vector Spaces
Neighborhoods:
$$\mathcal{N}_z:=\{U[z]:U\in\mathcal{U}\}$$
Uniform maps:
$$\mathcal{C}_U(\Omega,\Omega'):=\{\Phi:\Omega\to\Omega':\Phi^{-1}(\mathcal{U}')\subseteq\mathcal{U}\}$$
It holds equality:
$$\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_U(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}(\Omega,\Omega')$$
Homomorphisms:
$$\mathrm{Hom}(\Omega,\Omega'):=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_U(\Omega,\Omega')$$
Isomorphic spaces:
$$\Omega\cong\Omega':\iff\Phi:\Omega\leftrightarrow\Omega':\quad\Phi(\mathcal{U})=\mathcal{U}'$$
$$\Phi(x+y)=\Phi(x)+'\Phi(y)\quad((x,y)\in\Omega\times\Omega)$$
$$\Phi(\lambda\cdot x)=\lambda\cdot'\Phi(x)\quad((x,\lambda)\in\Omega\times\mathbb{C})$$
Suppose one finds:
$$\text{Locally Convex Base}$$
Induced seminorms:
$$\mu_U(x):=\inf\{r\geq0:x\in rU\}$$
Going next step up:
Locally Convex Spaces
Basic entourages:
$$B_{\mu\varepsilon}:=\{(x,y):\mu(y-x)<\varepsilon\}\subseteq\Omega\times\Omega$$
Uniform structure:
$$\mathcal{U}:=\uparrow\{B_{\mu\varepsilon}:\mu\in\mathcal{S},\varepsilon>0\}$$
Isomorphic spaces:
$$\Omega\cong\Omega':\iff\Phi:\Omega\leftrightarrow\Omega':\quad\Phi(\mathcal{S})=\mathcal{S}'$$
$$\Phi(x+y)=\Phi(x)+'\Phi(y)\quad((x,y)\in\Omega\times\Omega)$$
$$\Phi(\lambda\cdot x)=\lambda\cdot'\Phi(x)\quad((x,\lambda)\in\Omega\times\mathbb{C})$$
Suppose one finds:
$$\text{Countable Base}$$
Induced Metric:
$$d(x,y):=\sum_{k=1}^\infty\frac{1}{2^k}\frac{\sigma_k(y-x)}{1+\sigma(y-x)}$$
Going next step up:
Metrizable Vector Space
Induced seminorm:
$$\mu(x):=d(x,0)=d(0,x)\geq0$$
Lipschitz maps:
$$\mathcal{C}_L(\Omega,\Omega'):=\{\Phi:\Omega\to\Omega':d(\Phi\cdot,\Phi\cdot)'\leq L_\Phi d(\cdot,\cdot)\}$$
It holds equality:
$$\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_L(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_U(\Omega,\Omega')$$
Isometric maps:
$$\mathcal{I}(\Omega,\Omega'):=\{\Phi:\Omega\to\Omega':d(\Phi\cdot\Phi\cdot)'=d(\cdot,\cdot)\}$$
Homomorphisms:
$$\mathrm{Hom}(\Omega,\Omega'):=\mathcal{L}(\Omega,\Omega')\cap\mathcal{I}(\Omega,\Omega')$$
Isomorphic spaces:
$$\Omega\cong\Omega':\iff\Phi:\Omega\leftrightarrow\Omega':\quad d(\Phi\cdot,\Phi\cdot)'=d(\cdot,\cdot)$$
$$\Phi(x+y)=\Phi(x)+'\Phi(y)\quad((x,y)\in\Omega\times\Omega)$$
$$\Phi(\lambda\cdot x)=\lambda\cdot'\Phi(x)\quad((x,\lambda)\in\Omega\times\mathbb{C})$$
By construction:
$$d(x+a,y+a)=d(x,y)\quad(a\in\Omega)$$
Suppose one has:
$$d(\lambda x,\lambda y)=|\lambda|d(x,y)\quad(\lambda\in\mathbb{C})$$
Induced norm:
$$\|x\|:=d(x,0)\geq0$$
Going next step up:
Normed Spaces
Induced metric:
$$d(x,y):=\|y-x\|\geq0$$
Lipschitz maps:
$$\mathcal{B}(\Omega,\Omega'):=\{\Phi:\Omega\to\Omega':\|\Phi\cdot\|'\leq\|\Phi\|\cdot\|\cdot\|\}$$
It holds equality:
$$\mathcal{L}(\Omega,\Omega')\cap\mathcal{B}(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{C}_L(\Omega,\Omega')$$
Isomorphic spaces:
$$\Omega\cong\Omega':\iff\Phi:\Omega\leftrightarrow\Omega':\quad\|\Phi(\cdot)\|'=\|\cdot\|$$
$$\Phi(x+y)=\Phi(x)+'\Phi(y)\quad((x,y)\in\Omega\times\Omega)$$
$$\Phi(\lambda\cdot x)=\lambda\cdot'\Phi(x)\quad((x,\lambda)\in\Omega\times\mathbb{C})$$
Suppose one has:
$$\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2$$
Induced scalar product:
$$\langle x,y\rangle:=\frac{1}{4}\sum_{\alpha=0\ldots3}i^\alpha\|x+i^\alpha y\|$$
Going final step up:
Hilbert Spaces
Induced norm:
$$\|x\|^2:=\langle x,x\rangle\geq0$$
Orthogonal maps:
$$\mathcal{O}(\Omega,\Omega'):=\{\Phi:\Omega\to\Omega':\langle\Phi\cdot,\Phi\cdot\rangle'=\langle\cdot,\cdot\rangle\}$$
It holds equality:
$$\mathcal{L}(\Omega,\Omega')\cap\mathcal{O}(\Omega,\Omega')=\mathcal{L}(\Omega,\Omega')\cap\mathcal{I}(\Omega,\Omega')$$
Homomorphisms:
$$\mathrm{Hom}(\Omega,\Omega'):=\mathcal{L}(\Omega,\Omega')\cap\mathcal{O}(\Omega,\Omega')$$
Isomorphic spaces:
$$\Omega\cong\Omega':\iff\Phi:\Omega\leftrightarrow\Omega':\quad \langle\Phi\cdot,\Phi\cdot\rangle'=\langle\cdot,\cdot\rangle$$
$$\Phi(x+y)=\Phi(x)+'\Phi(y)\quad((x,y)\in\Omega\times\Omega)$$
$$\Phi(\lambda\cdot x)=\lambda\cdot'\Phi(x)\quad((x,\lambda)\in\Omega\times\mathbb{C})$$
Concluding categories.