# Is this function continuous?

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#2

(Original post by

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

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

I wouldn't call that a function on the reals, since f(1) is not defined.

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#3

**atsruser**)

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

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.

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#4

(Original post by

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.

**razzor**)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.

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#5

(Original post by

Are you sure?

**metaltron**)Are you sure?

Spoiler:

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It is clear the function is not continuous. It is not even defined at x=1.

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#6

**atsruser**)

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

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(Original post by

Yes

**razzor**)Yes

Spoiler:

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It is clear the function is not continuous. It is not even defined at x=1.

2. Now apply the "open set" definition of continuity

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#8

**razzor**)

Yes

Spoiler:

Show

It is clear the function is not continuous. It is not even defined at x=1.

(Original post by

1. What is the domain of the function?

2. Now apply the "open set" definition of continuity

**atsruser**)1. What is the domain of the function?

2. Now apply the "open set" definition of continuity

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#9

**razzor**)

Yes

Spoiler:

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It is clear the function is not continuous. It is not even defined at x=1.

then (implicitly) f is a function with domain , and f

**is**continuous on this domain.

Does also saying "Is this function on the reals..." imply that the domain should be all of ? 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.

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(Original post by

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?

**metaltron**)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?

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(Original post by

Does also saying "Is this function on the reals..." imply that the domain should be all of ? 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

**DFranklin**)Does also saying "Is this function on the reals..." imply that the domain should be all of ? 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.

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#12

(Original post by

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.

**atsruser**)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.

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.

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#13

(Original post by

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.

**atsruser**)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.

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(Original post by

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.

**DFranklin**)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.

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.

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(Original post by

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

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

How about on [0,1] intersection Q?

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#16

(Original post by

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 ?

**atsruser**)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 ?

(Original post by

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?)

**atsruser**)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?)

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(Original post by

But 0 is a rational, so something that blows up at an irrational would work, like

**Remyxomatosis**)But 0 is a rational, so something that blows up at an irrational would work, like

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.

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#18

(Original post by

Maybe I'm being especially dense, but given that there aren't any irrationals in , then can't be in the domain of the function..

**atsruser**)Maybe I'm being especially dense, but given that there aren't any irrationals in , then can't be in the domain of the function..

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#19

(Original post by

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 ?

**atsruser**)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 ?

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(Original post by

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.

**metaltron**)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.

No, in fact, I was thinking that you can find a continuous unbounded function on a non-compact subset of . Using the lack of completeness didn't occur to me due to old age and forgetfulness.

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