Can A Function Be Continuous At Only One Point?

We find a function which is continuous at a single point, and discontinuous everywhere else. The first half of the video covers intuition for continuity, plus its formal definition.

00:00 Intuition
02:00 Refined intuition
04:08 Formal definition
06:39 Constructing an example
08:15 Proving continuity
10:22 Proving discontinuity
12:28 Further examples

[flyingbicycles]

So a function can be continuous at any number of finite points, countably infinite points, or uncountably infinite points. Very nice.

[bjornfeuerbacher]

As far as I know, it's usual to say that f(x) = 1/x is neither continuous nor discontinuous at x = 0, since it only makes sense to talk about continuity for values of x where a function actually is defined.

[ExplosiveBrohoof]

My favorite example of a function with degenerate continuity properties is this: If x is irrational, let f(x) = 0. If x is rational and nonzero, write x = p/q, where p and q are coprime and q is positive. Then set f(x) = 1/q. Finally, set f(0)=1. This function ends up being continuous at every irrational point, and discontinuous at every rational point.

[oli_dev]

What a brilliant solution. It seems so obvious in hindsight too!

[yakovify]

Nice video as always. Many interesting generalizations of this question exist, one of my favourites is that the set of continuity points of any function into R is always a G_delta set, a countable intersection of open sets.

[mikenorman]

Having studied analysis as an undergraduate (many decades ago) with the formal delta/epsilon definition, my gut feeling was no of course you can't have a function that is only continuous at a single point. But then the fact that the question was even posed in the first place led me to think the answer must in fact somehow be yes which is what I went for. And so I was right! However if you asked me to define such a function I would have failed. Bah! Maths eh?

[isomeme]

I love a good paradox, and the concept of a function being continuous only at a single point is a *great* paradox. I'll be turning this one over in my head for quite a while. Thanks for this video!

[koenth]

Dr Barker keeps impressing with great insights and new information. Really great channel!

[XahhaTheCrimson]

Even the answer is pretty trivial, I haven't thought about that and I think it's really worth it.

[MichaelRothwell]

I enjoyed this video, but I'm afraid I have a small nit to pick.
Right near the start, I didn't feel very comfortable about the choice of f(x)=1/x as an example of a discontinuous function. Whilst (as you said) it's not continuous everywhere, it is continuous on its domain. Later at 3:10, when you attempted to illustrate the discontinuity of f(x)=1/x at a point on its domain, I really cringed!
It would have been better (in my opinion) to give a step function as a simple example of a discontinuous function.

[chonkycat]

Always a pleasure watching your videos. More analysis !!

[landsgevaer]

So, for f(x) =
{ 0 if x is rational
{ cos(1/x) if x is irrational
we even have *infinitely many* continuous points in any open interval that contains zero, yet not continuity in *any* open interval as a whole.
Funny...

[theimmux]

short answer: yes. F.ex. f: R -> R x -> x if x is rational and otherwise x -> -x is only continuous at x = 0.

[alexwarner]

Inb4
Continuously differentiable ⊂ Lipschitz continuous ⊂
α-Hölder continuous ⊂ uniformly continuous = continuous

A fun pathological example is the Weierstrass function. Continuous everywhere, but nowhere differentiable 🤯 did challenge notions at the time it was published.

[tombratcher]

Saw the picture, thought, "huh, nice example". I assume you explained it well too :-)

[jakobr_]

My favorite way to think about continuity is to say that a function is continuous at an input if that output is what we “expect” it to be by looking at nearby outputs.

This works really well as a plain english version of “continuous means the output at an input and the limit of outputs as the inputs approach that input are equal”

[arthurkassis]

I entered in the video already knowing the answer, but the explanation was really good.

[Dr.Cassio_Esteves]

Great explanation! Every thing was so clear!
I would love if you did a topology playlist.

[MihaiNicaMath]

The hardest bonus problem in my 1st analysis class was about proving that not all sets are possible as the set of discontinuity points of a function. It turns out the Baire category theorem limits what is possible! For example you can't make a function that is only continuous on the irrational numbers.

[flockofwingeddoors]

Amazing watch at usual. You do incredibly clear explanations, and you always pick such interesting topics to talk about. Bravo!