Is this function continuous?

Is this function on the reals continuous? If so, why? If not, where is it discontinuous?

$f(x) = \begin{cases} x & \text{if } x < 1 \\ x+1 & \text{if } x > 1 \end{cases}$

Is this function on the reals continuous? If so, why? If not, where is it discontinuous?

$f(x) = \begin{cases} x & \text{if } x < 1 \\ x+1 & \text{if } x > 1 \end{cases}$

First, see if you can sketch the graph.
Compute the limit as x tends to 1 from the left and the limit as x tends to 1 from the right.
If both of the limits do not match up, then it is not continuous.

Are you sure?

Is this function on the reals continuous? If so, why? If not, where is it discontinuous?$f(x) = \begin{cases} x & \text{if } x < 1 \\ x+1 & \text{if } x > 1 \end{cases}$

Yes

Yes

1. What is the domain of the function?
2. Now apply the "open set" definition of continuity

Yes

The function is continuous on its domain as poorform has said.


Ha, was this a test of us then, or did you need help with some part of it?

Does also saying "Is this function on the reals..." imply that the domain should be all of $\mathbb{R}$? Probably not, since (as you point out yourself), it isn't defined on that domain, therefore not even a function on that domain.

My guess would be that the purpose of the question as set is precisely to get people to realise that a function can be "obviously" discontinuous at a point, but if that point isn't in the domain it doesn't matter. But if that was the intent it could certainly have been worded better.

I had to choose the wording carefully, so as not to point out that the domain had a hole in it; "on the reals" was the best I could think of, but I admit it may be a bit ambiguous.

I'll come clean: I knew the answer, but I wanted to shake the forum to its very foundations, moribund as it is in the dog days of August, with a result that confused me as I lay in bed last night.

If you want to be tricky but not actively misleading, I'd just leave it out, the (correct) domain is implicit from the ranges on which you define the function.

If you want an "opposite" possibly counter-intuitive function:

Over (all of the!) reals, define

f(x) = x (if x rational),
f(x) = 0 (if x irrational).

Then f is continuous at exactly one point.

A similar question: Is there an unbounded continuous function on [0,1]?

How about on [0,1] intersection Q?

Without cheating (or thinking), I don't know, but I will guess yes, and I'd guess that it's possible due to failure of compactness. So maybe something as easy as $f(x)=1/x$?

That's nice. I've probably seen it before though. There's also some function that's continuous on all rational x, discontinuous elsewhere, though I can't recall the details (Thomae function?)

But 0 is a rational, so something that blows up at an irrational would work, like $\dfrac{1}{x - \pi/4}$

The thing about Thomae's function (it's continuous at all irrational, discontinuous at all rational) is that it's even Riemann-integrable which blew my mind when I first learnt it, but of course now it's obvious from Lebesgue's criterion.

Maybe I'm being especially dense, but given that there aren't any irrationals in $[0,1] \cap \mathbb{Q}$, then $\pi/4$ can't be in the domain of the function..

Unless you're being sneaky with your wording, then no - this is a standard result.



Without cheating (or thinking), I don't know, but I will guess yes, and I'd guess that it's possible due to failure of compactness. So maybe something as easy as $f(x)=1/x$?

I take it you mean completeness rather than compactness (at least completeness is more explicity useful, though a metric space is compact iff it is complete and totally bounded); and yeah something like Remyxomatosis' function will work.