How to formalize this argument?
Sebastian Wright 
I have an informal argument for a proof, that is so obvious that I normally see people just skim over it. However, I am trying to find out how to make this argument formally.
I want to go from
$$\exists D_{>0}\forall \delta_{>0}\exists N_{>0}:n>N\to x_n<D\cdot\delta$$ to $$\forall \delta_{>0}\exists N_{>0}:n>N\to x_n<\delta$$
Intuitively, I would simply say: for any $\hat\delta>0$ we can take the $D$ and find any $\delta$ such that $D\cdot\delta=\hat\delta$, which is a property of the completeness of real numbers. Therefore we can remove the $D$ from the formula.
How do we make this proof formally?
$\endgroup$1 Answer
$\begingroup$Your proof should go like this:
We know that $\exists D>0$ so that $\forall\delta>0\,\exists N>0$ so that if $n>N$ then $x<D\delta$. Then given $\hat{\delta}>0$, choose $D$ as we know exists from the first statement. Then we let $\delta=\dfrac{\hat{\delta}}{D}$, and since we know the first statement to be true, we can find $N>0$ so that if $n>N$, then $x_n<D\cdot\delta=\hat{\delta}$. In conclusion, we showed that given $\hat{\delta}$, we can find $N>0$ so that if $n>N$ then $x_n<\hat{\delta}$. QED.
Of course $\hat{\delta}$ is a dummy variable, so we can replace it in our conclusion with $\delta$, and the statement becomes syntactically exactly what we wanted to show. And it matters not how we find $N$, but only that we are able to, with our current knowledge (which includes knowing that the first statement is true).
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