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Disclaimer: This thread is meant informative and therefore written in Q&A style. Of course everybody is encouraged to give an answer as well!


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.

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    $\begingroup$ Why are people voting to close for "off-topic for missing context etc"? This is usually for when the OP is asking a question without showing any work, but it's the complete opposite here! $\endgroup$ – Najib Idrissi Jun 24 '14 at 11:59
  • $\begingroup$ Could somebody explain why this is not about math (off-topic)?!?! I start believing that people follow my questions just to vote to close it. Please clarify and ask whenever things seem not appropriate before downvoting, voting to close and etc - be constructive!!! $\endgroup$ – C-Star-W-Star Jun 24 '14 at 15:39
  • $\begingroup$ Here two examples how things should rather work: Question wasn't clear so Spencer did ask for (math.stackexchange.com/q/842801/79762) or form of typesetting answer was hard to read so Did pointed it out (math.stackexchange.com/a/842802/79762). Thanks to both! $\endgroup$ – C-Star-W-Star Jun 24 '14 at 15:48
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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 $\mathcal{N}_\Phi=\mathcal{N}$.

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