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Prove that any topological vector space gives rise to a uniform structure via: $$\Phi:=\uparrow\{U_N:N\in\mathcal{N}_0\}\text{ with }U_N:=\{(x,y):y-x\in N\}$$ where $C\in\uparrow\mathcal{A}$ iff $A\subseteq C$ for some $A\in\mathcal{A}$.
Moreover prove that the uniform topology coincides with the original topology: $$\mathcal{N}\mapsto\Phi\mapsto\mathcal{N}$$ where the precise definition of the uniform topology is given in here.
Uniform Structure
Firstly, $\Phi$ is upward closed by construction.
Next, note that $U_M\cap U_N=U_{M\cap N}$.
Thus if $A,B\in\Phi$ then so also $A\cap B\in\Phi$.
Now, $\Phi$ is nonempty since so is $\mathcal{N}_0$
Also, $\Delta\subseteq U$ for $U\in\Phi$ since $0\in N$ for any $N\in\mathcal{N}_0$.
Moreover, inversion (dilation by the factor minus one) is a homeomorphism.
Thus $U^{-1}\in\Phi$ for $U\in\Phi$ since $U_N^{-1}=U_{-N}$.
Finally, since addition is continuous there is $N'\in\mathcal{N}_0$ for any $N\in\mathcal{N}_0$ s.t. $N_0+N_0\subseteq N$.
Thus for any $U\in\Phi$ there is $V\in\Phi$ with $V^2\subseteq U$ since $U_M^2=U_{M+M}$.
Induced Topology
Since $N+z=U_N[z]$ we have $\mathcal{N}_\Phi=\mathcal{N}$.