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: 
The sine function is not one-to-one. When you applied that function you also got

$\sin(\arcsin(1-x))=\sin(\frac{\pi}{2}-2\arcsin(x))$,

as this RHS equals the one you intended. 
Note the sign change on the right side. 

Then $x=\frac{1}{2}$ satisfies

$\arcsin(1-x)+2\arcsin(x)=\frac{\pi}{2}$.
None of your implication steps are wrong
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Let's review the concept of **implication**. In mathematics, an **implication** is represented by the symbol $\implies$ and when we write $A \implies B$, it means **precisely** that **if A is true, then B is true**.

Here is a quote of your second implication:

> 
> $$\arcsin(1-x) = \frac{\pi}{2} + 2\arcsin (x) \implies \sin (\arcsin (1-x)) = \sin \left( \frac {\pi}{2} + 2\arcsin (x)\right)$$
> 
> 
> 

This is perfectly correct, since the form of that statement is $A = B \implies \sin(A) = \sin(B)$, and clearly if A = B, they are the same thing, and their sine will also be equal.

Well then where is my mistake? Surely there is a mistake somewhere
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Let's summarize what you've proved:

$$\arcsin (1-x) - 2\arcsin (x) = \frac{\pi}{2} \implies x = 0 \text{ or } x = \dfrac{1}{2}$$

This is **correct!** (bear with me, I'll get there).

Assume my car is blue. Then I tell you the following sentence: "My car is blue or my car is red." That sentence is **correct!**

See what's happening here?

**When we solve equations, we should be using "equivalence" steps, not "implication" steps**, because we want to find all the values of $x$ that satisfy the equation, no more no less.

I find it enlightening to look at it this way because we don't lose the benefit of formal math which IMO makes much easier to find mistakes. By understanding well how implications work, it is clear that there is no mistake in the sequence of implications you created.

Of course, the other answers are correct as well, but I'd like to provide this formal insight because I find it very useful.

If you replace the implication symbols ($\implies$) in your attempt by equivalence symbols ($\iff$), you'll have a mistake on your second step, for the reasons other people already mentioned.