Month: August 2022

Unifying ε-N and ε-δ arguments

~ David Roberts ~ 3 Comments

I’m teaching an intro analysis topic at the moment, and so of course there’s the whole ordeal of introducing ε-δ arguments. However, when we say such a thing, we usually also have in mind the type of proofs that are used for convergence of sequences, which are not usually ε-δ, which is to do with continuity of a function, but involve finding some large N past which the sequence is close to a limit, or else some Cauchy-type condition: hence either | a_n - L| < \varepsilon or | a_n - a_m| < \varepsilon, for all n \geq N.

However, it is possible to present a convergence proof as a continuity proof, using \delta. This is not a massive secret, but it’s cute, so I thought I’d write it up.

Let us fix some data: a sequence (a_n) in \mathbb{R}. Such a thing is a function a\colon \mathbb{N}\to \mathbb{R}. We say that (a_n) converges to L\in \mathbb{R} if:

\forall \varepsilon > 0,\ \exists N\in \mathbb{N},\ \forall n> N,\ |a_N - L| < \varepsilon

Another way to think about is is if we know that the function a extends to a function a'\colon \mathbb{N}_\infty = \mathbb{N} \cup \{\infty\}\to \mathbb{R}, satisfying a special property, then we have convergence. If we have convergence, we can clearly define such an extension where a'(\infty) = L. So what is this special property? It’s nothing other than continuity of a', where we have to put a particular topology on \mathbb{N}_\infty! From the point of view of pure point-set topology, we specify the neigbourhoods of \infty to be the cofinite subsets of \mathbb{N}_\infty containing \infty—that is, the subsets only missing finitely many elements of \mathbb{N}\subset \mathbb{N}_\infty—and otherwise for every n\in \mathbb{N}, \{n\} itself is a neighbourhood. Thus \mathbb{N} is a discrete subset, and only the point \infty “has nontrivial topology”, as it were. This at least means continuity of a' makes sense.

But the ε-δ definition of continuity is a metric definition, namely it’s treating \mathbb{R} as a metric space. So how do we make \mathbb{N}_\infty a metric space, so that the topology just defined is the metric topology? Here’s where we see the link we seek. Recall the fundamental sequence \frac{1}{n+1} \to 0 in the reals, whose convergence characterises Archimedean ordered fields. We can think of the set \mathcal{N}_0 := \{\frac{1}{n+1}\in \mathbb{R}\mid n\in \mathbb{N}\}\cup \{0\} = \mathcal{N}\cup \{0\} as inheriting the subspace topology from the reals, and in fact this gives something homeomorphic to \mathbb{N}_\infty. And, since \mathbb{R} is a metric space, we can give the subset a metric inducing this topology: we have d(\frac{1}{n},\frac{1}{m}) = |\frac{1}{n} - \frac{1}{m}| and d(0\frac{1}{n}) = \frac{1}{n}

However, this metric is slightly awkward, but it at least can inspire a slightly nicer metric d' on \mathbb{N}_\infty, namely d'(\frac{1}{n},\frac{1}{m}) = 1 and d(\infty,\frac{1}{n}) = \frac{1}{n}. There is a bijective short map \mathbb{N}_\infty\to \mathcal{N}_0 (n\mapsto \frac{1}{n},\ \infty \mapsto 0, but which is not invertible as as short map (nor even as a Lipschitz map, or a uniformly continuous function), but which is still a homeomorphism. As far as mere continuity goes, either version of the space is ok, but we will use \mathbb{N}_\infty with the metric d' and the resulting topology. So here we see where our very patient \delta is going to come in. A function \mathbb{N}_\infty\to \mathbb{R} (where both of these are considered as metric spaces) is continuous, precisely if

\forall \varepsilon > 0,\ \exists \frac{1}{N},\ \forall \frac{1}{n}< \frac{1}{N},\ |a(n) - a(\infty)| < \varepsilon

So we could in principle dispense with the N, and only consider positive \delta, namely

\forall \varepsilon > 0,\ \exists \delta > 0,\ \forall \frac{1}{n}< \delta,\ |a_n - L| < \varepsilon

where we have set L:=a(\infty), and a_n = a(n), as usual. This is what happens if we know (a_n) has a limit. If we were to ask whether it has as limit, we should ask instead whether there is any continuous extension of \mathbb{N}\to \mathbb{R} along the inclusion \mathbb{N}\hookrightarrow \mathbb{N}_\infty. The usual proof of uniqueness of limits of sequences in metric space can be adapted to show that if such a continuous extension existed, then there is exactly one of them.

A similar descriptions can be made for the Cauchy property, except in this instance, one really does need to use the metric space (\mathcal{N},d) (not the space \mathcal{N}_0!), so that the function \mathbb{N}\to \mathcal{N} is a Cauchy sequence. Notice here that we do not have the limit point, since Cauchyness doesn’t refer to any actual limiting value, even assuming one exists. If we include \{0\}, hence consider the metric space \mathcal{N}_0, then we are dealing with a Cauchy sequence known to be convergent (in the reals, this is all Cauchy sequences, but one can consider all this in the rationals, for instance, where only some Cauchy sequences have a limit).

Thus there are four different things happening here. And this is where I put on my category-theorists hat: we have 1) a generic sequence, 2) a generic convergent sequence, 3) a generic Cauchy sequence, and 4) a generic convergent Cauchy sequence. There are maps from the generic sequence to the generic Cauchy sequence, from the generic sequence to the generic convergent sequence, from the generic convergent sequence to the generic convergent Cauchy sequence, and from the generic Cauchy sequence to the generic convergent Cauchy sequence. Since a X-sequence (say in \mathbb{R}, but it works in any metric space) is given by a continuous function from the generic X-sequence, precomposing with these maps just described forget properties of the sequence (for instance, take a convergent sequence together with its limit, and then forget what the limit is, or take a Cauchy sequence and forget this fact). I’ve been a bit sloppy as to what category all this is happening in, it should at least be metric spaces and continuous maps, but one could see if the screws could be tightened, and something like this work in eg uniformly continuous maps. I leave this as an exercise for the reader.