4
$\begingroup$

(I am aware this question may be considered subjective and somewhat nitpicky. I hope it is not too inappropriate.)

When doing proofs in analysis, for example, one may encounter some convergent sequence $(x_n)_{n\in\mathbb{N}}$ in some metric space $(X,d)$. Symbolically, one may write: $\forall \varepsilon >0 \exists N\in\mathbb{N} \forall n\geqslant N:d(x,x_n)<\varepsilon $, where $x\in X$ is the limit of $(x_n)_{n\in\mathbb{N}}$. Some argument using the convergence of $(x_n)_{n\in\mathbb{N}}$ may start like this:

"Let $\varepsilon >0$. Since $(x_n)_{n\in\mathbb{N}}$ converges to $x$, we can find $N\in\mathbb{N}$ such that for all $n\geqslant N$ we have $d(x,x_n)<\varepsilon$."

It is possible that the natural number $N$ (such that $\forall n\geqslant N: d(x,x_n)<\varepsilon$) is needed in some further part of the argument, so we want to fix the letter $N$ as being that specific natural number. However the above line does not explicitly specify that the letter $N$ should represent this natural number. It really just says that such a number exists.

When writing proofs, is it considered good custom to add another line explicitly specyfing $N$? For example by adding a line such as

"Let $N\in\mathbb{N}$ be such that $\forall n\geqslant N: d(x,x_n)<\varepsilon$."

or

"Let $N\in\mathbb{N}$ such that the above holds."

Or, is it more customary to just leave the specification implicit? (That is, write "$\ldots$we can find $N\in\mathbb{N}$ such that$\ldots$", and just use the letter $N$ assuming that it has been appropriately specified.) We could also write:

"Let $\varepsilon >0$. Let $N\in\mathbb{N}$ such that $\forall n \geqslant N: d(x,x_n)<\varepsilon$, which we can find because $(x_n)_{n\in\mathbb{N}}$ converges.

This somewhat combines the existence and specification of $N$ into one line. Is there any other way of handling this nicely?

Thank you for your time.

$\endgroup$
2
  • 1
    $\begingroup$ The only way to "formally" disambigaute such cases is with the scope of the quantifiers; if we write: $∀ε > 0 \ ∃N ∈ \mathbb N \ ∀n ⩾ N \ d(x,x_n)<ε$, we have no problem. If we have a longer semi-formal expression like e.g.: "$∀ε > 0 \ ∃N ∈ \mathbb N \ ∀n ⩾ N(d(x,x_n)<ε$ and $N > 5$", we may have the doubt that the last occurrence of $N$ is outside the scope of the initial quantifier. In cases like this, we have to add parentheses. $\endgroup$ Sep 2, 2016 at 11:37
  • 1
    $\begingroup$ Existential instantiation is definitely a thing. The tricky matter is when you have existentials which are not at top level, such as your $N $. It can be helpful to Skolemize in this case, which in your example means introducing a function $N (\varepsilon) $. In my class we did this for understanding uniform continuity and uniform convergence. Skolemization can have bad effects with respect to obfuscation foundations sometimes; for example it can make usage of the axiom of choice less explicit. $\endgroup$
    – Ian
    Sep 2, 2016 at 11:42

1 Answer 1

2
$\begingroup$

It's very rare for people to add a whole extra sentence just to introduce formally something whose existence was already proven. Here are some more constructions that can be used instead:

Since $(x_n)_{n\in\mathbb{N}}$ converges to $x$, we can find $N\in\mathbb{N}$ such that for all $n\geqslant N$ we have $d(x,x_n)<\varepsilon$. Given such an $N$, ...

Since $(x_n)_{n\in\mathbb{N}}$ converges to $x$, we can (and do) choose $N\in\mathbb{N}$ such that for all $n\geqslant N$ we have $d(x,x_n)<\varepsilon$.

Since $(x_n)_{n\in\mathbb{N}}$ converges to $x$, pick $N\in\mathbb{N}$ such that for all $n\geqslant N$ we have $d(x,x_n)<\varepsilon$.

The last is the briefest, although perhaps a bit ungrammatical. Alternatively it is acceptable to, as you say, "leave the specification implicit".

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .