5
$\begingroup$

I've personally found things are a lot easier to prove using the sequences rather than the $\epsilon-\delta$ method. I do understand the $\epsilon-\delta$ is more intuitive, but it's quite difficult to prove a function is continuous on $x_0 \in D$ since it implies finding a function that takes $\epsilon$ and $x_0$ as arguments and outputs some $\delta>0$, or at least proving such a function exists. This issue is even more prevalent when proving a function is uniformly continuous. On all of my homework problems involving uniform continuity, I always tried to use the $\epsilon-\delta$ definition, yet I always failed. When searching for the solution, the proofs are quite confusing and complex. I would then switch to the sequence definition and would succeed in solving the problem quickly. My professor says that the mathematics community nonetheless refers the $\epsilon-\delta$ definition, so I am curious why when it's a more difficult path.

Here are the definitions.

For a function $f:D \rightarrow \mathbb{R}$,

Continuity:$f$ is continuous at $x_0 \in D$ if for any sequence $(x_n)$ in $D$ where $\lim_{n \rightarrow \infty} x_n=x_0$, it follows $\lim_{n \rightarrow \infty} f(x_n)=f(x_0)$.

Uniform continuity: $f$ is uniformly continuous if for any sequences $(u_n),(v_n)$ in $D$ where $\lim_{n \rightarrow \infty} (u_n-v_n)=0$, it follows $\lim_{n \rightarrow \infty}(f(u_n)-f(v_n))=0$ .

An instance of this is proving the function $f:[0,1) \rightarrow \infty$ where $f(x)=\frac{1}{1-x}$ is not uniformly continuous. I found it very difficult to prove this using the $\epsilon-\delta$ definition, and the proofs I found online were quite confusing and there's definitely no way I would've come across them by myself. The sequential definition however was very simple. I simply used the sequences $u_n=1-\frac{1}{n^2}$ and $v_n=1-\frac{1}{n}$. The differences of the sequences converge to 0, and the differences of the image of the sequences diverge to $\infty$. $0 \not = \infty$ and therefore $f$ is not uniformly continuous.

Another example is showing that any continuous function whose domain is a closed interval is uniformly continuous. There's no way I could come up with a way to prove this using the $\epsilon-\delta$ definition, but I definitely could using the sequence definition.

I may be wrong, but I believe I read once that for some metric spaces, continuity and sequential continuity are not equivalent. I don't know much about that matter though, but would love to learn about it.

$\endgroup$
  • $\begingroup$ Is there a sequential definition of uniform continuity? $\endgroup$ – Aweygan Jun 7 at 23:11
  • $\begingroup$ Are you able to provide an example? $\endgroup$ – zugzug Jun 7 at 23:14
  • $\begingroup$ @Aweygan I edited my post and included the definition. It's interesting, my professor told me that the sequence definition is rarely taught so many do not know about it. $\endgroup$ – Spencer Kraisler Jun 7 at 23:35
  • 1
    $\begingroup$ There are more general topological spaces that are not metric where this equivalence does not hold. Usually first contact with them is when defining spaces of distributions. The analogon to sequences there are what is called nets, some kind of directed graph with points as nodes. $\endgroup$ – LutzL Jun 7 at 23:42
  • 1
    $\begingroup$ Indeed I asked a related question in Math educator recently, see here $\endgroup$ – Arctic Char Jun 8 at 7:46
5
$\begingroup$

Just to show that it is not at all hard or confusing to show that $f \colon [0..1) → ℝ,~x ↦ \frac 1 {1 - x}$ is not uniformly continuous using the $ε$-$δ$-definition:

Since $(0..1] → [0..1),~x ↦ 1 - x$ is uniformly continuous and uniform continuity is stable under composition, it suffices to show that $f \colon (0..1] → ℝ,~x ↦ \frac 1 x$ is not uniformly continuous.

For this, fix $ε = 1$. Then for any $δ > 0$ with $δ < 1$, let $x = δ$ and $x' = \frac δ 2$. Then clearly $|x - x'| < δ$, whereas $|1/x - 1/x'| = 1/δ ≥ 1 = ε$. For $δ ≥ 1$, choose $x = 1$ and $x' = \frac 1 2$.


Now regarding your question. To show the equivalence of the $ε$-$δ$-definition and the sequential definition of continuity for metric spaces, one needs the Axiom of Countable Choice. (And to show the equivalence of the generalized neighbourhood definition and the net definition, one needs the full axiom of choice, I think.)

That is to say, the equivalence is non-constructive. This already makes it clear that neither approach is in itself more useful for all cases, but it’s the equivalence of these approaches that is really useful: Thinking non-constructively is hard and having two languages to talk about the same phenomena from two constructively separated places that have already been linked by a bridge saves one the effort of building a new bridge every time it seems handy to switch perspectives.

Philosophically, I’d say that the $ε$-$δ$-perspective of continuity and uniform continuity capture these notions more directly, whereas the perspective on these from sequences and net is more indirect, as if to say: “There is no counterexample of sequences such that they are converging here, but not there. Such a wild thing does not exist!” This makes this perspective perfect for the realm of counterexamples. Proofs using sequences tend to be indirect proofs, using contradiction. It’s often about supposing something isn’t true and then, often non-constructively, choosing a counterexample sequence.

The $ε$-$δ$-perspective, however, not only generalises more easily to more abstract spaces, I also find it to be easier to think about conceptually. For example, I did have an absurdly hard time thinking about how to prove that continuous maps are uniformly continuous on compact sets when thinking about this sequentially, whereas it’s crystal clear from the $ε$-$δ$-perspective. Maybe it’s a matter of socialisation, but it’s easier for me to see things from this perspective. Another example would be that open subgroups of compact topological groups (which are uniform spaces) are exactly closed subgroups of finite index. It’s obvious from this perspective, not so much from the other.

In short: Topological spaces can be really weird, making set-theoretic topology a study of weird things. If you want to do neat topology, then it might become a study about how some weird things are not happening. That makes the sequential approach a bit more powerful, albeit less conceptual.

And whenever you encounter an extension of this equivalence such as the equivalence of compactness and sequential compactness, expect proofs to be non-constructive.

$\endgroup$
  • $\begingroup$ Thanks for the answer, this opened my eyes on some things. $\endgroup$ – Spencer Kraisler Jun 8 at 5:22
3
$\begingroup$

Well, you're just substituting the sequential definition of a limit versus a neighborhood approach, and then applying that to continuity. As for why that isn't done in general, I shall quote Anthony E. Labarre, Jr.'s Intermediate Mathematical Analysis, page 67:

On the other hand, we emphasize that the neighborhood approach is often more convenient in theoretical investigations. As a matter of fact, in the abstract setting of a topological space, the "sequential limit" and the "neighborhood limit" are not equivalent concepts and, moreover, the concept of "sequential limit" proves to be inadequate.

The comments provide a few examples of topological spaces where they fail to be equivalent.

$\endgroup$
  • $\begingroup$ I see, so it seems the only reason the $\epsilon-\delta$ def. is preferred is simply because the sequential def. fails for some topological spaces. Thank you. $\endgroup$ – Spencer Kraisler Jun 8 at 0:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.