Question: Define the following relation on the collection of all Cauchy filters on X:

$F ∼\_R G$ ⇔ $∀U$ nbhood of the origin in $X$, $∃A ∈ F$, $∃B ∈ G$ s.t. $A − B ⊂ U$.

Show that :-

> 
> If a filter $F$ is equivalent modulo $R$ to a filter which converges to $x$,
>  then also $F$ converges to $x$.
> 
> 
> 

An Attempt:-

Let $\mathcal F$ and $\mathcal G$ be two equivalent filters such that $\mathcal G \to x$

$\implies$ $∀U$ nbhood of the origin in $X$, $∃A ∈ F$, $∃B ∈ G$ s.t. $A − B ⊂ U$. 

$\implies$ $F(x)⊂ \mathcal G$ where $F(x)$ nhood filter of $x$

update:-

let $\mathcal{F} \to t$ , so $\mathcal{F}- \mathcal{G} \to t-x $ 

but $F(0)⊂\mathcal{F}- \mathcal{G} \to t-x$, because $F ∼\_R G$

$\implies$ $ t-x=0$ 

$\implies$ $x=t$

Answer: 
When $x=\frac12$, you have $\arcsin(x)=\frac{\pi}{6}$ and $\arcsin(1-x)=\frac{\pi}{6}$ 

so $\sin (\arcsin (1-x))=\sin \left( \frac {π}{6} \right)$ and $\sin \left( \frac {π}{2} + 2\arcsin (x)\right) =\sin \left( \frac {5π}{6} \right) $ 

showing your third line would be a correct equality when $x=\frac12$ since $\sin \left( \frac {π}{6} \right)=\sin \left( \frac {5π}{6} \right) =\frac12$

but your second line would not be a correct equality when $x=\frac12$ since $\frac {π}{6}\not = \frac {5π}{6}$ 

and it is this use of $\sin$ which creates an equality which was not in the original expression. 

$\arcsin (1-x) - 2\arcsin (x) = \frac{π}{2} \Rightarrow x=0 \text{ or } x=\frac12$ is a correct statement, but checking shows $x=0 \Rightarrow \arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$ while $x=\frac12 \not \Rightarrow \arcsin (1-x) - 2\arcsin (x) = \frac{π}{2}$
$$-2\arcsin(x)=\dfrac\pi2-\arcsin(1-x)=\arccos(1-x)$$

Now using [Principal values](https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values) of $\arccos$, 

$0\le-2\arcsin(x)\le\pi\iff0\ge\arcsin(x)\ge-\dfrac\pi2\implies-1\le x\le0$

Now let $\arcsin x=y\implies x=\sin y$

and $\arcsin(1-\sin y)=\dfrac\pi2+2y$

$$1-\sin y=\sin\left(\dfrac\pi2+2y\right)=\cos2y=1-2\sin^2y$$

$$\implies \sin y=0,\dfrac12$$ but $\sin y=x\le0$