atsruser
Badges: 11
Rep:
?
#1
Report Thread starter 5 years ago
#1
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}
0
reply
metaltron
Badges: 11
Rep:
?
#2
Report 5 years ago
#2
(Original post by atsruser)
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}
For all a>0 and for each x in (-infinity,1)U(1,infinity) can you find a b>0 such that |x-y| < b, implies |f(x)-f(y)| < a? Also make sure you've sketched the graph to help you visualise what you need to do.

I wouldn't call that a function on the reals, since f(1) is not defined.
0
reply
razzor
Badges: 8
Rep:
?
#3
Report 5 years ago
#3
(Original post by atsruser)
Is this function on the reals continuous? If so, why? If not, where is it discontinuous?

f(x) = \begin{cases} x & \text{if } x &lt; 1 \\ x+1 & \text{if } x &gt; 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.
0
reply
metaltron
Badges: 11
Rep:
?
#4
Report 5 years ago
#4
(Original post by 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.
Are you sure?
0
reply
razzor
Badges: 8
Rep:
?
#5
Report 5 years ago
#5
(Original post by metaltron)
Are you sure?
Yes
Spoiler:
Show
It is clear the function is not continuous. It is not even defined at x=1.
2
reply
poorform
Badges: 10
Rep:
?
#6
Report 5 years ago
#6
(Original post by atsruser)
Is this function on the reals continuous? If so, why? If not, where is it discontinuous?f(x) = \begin{cases} x & \text{if } x &lt; 1 \\ x+1 & \text{if } x &gt; 1 \end{cases}
Most definitions of continuity say that a function can only be continuous on points of its domain. Since 1 is not in the domain here I would say that this function is continuous on its domain. Looking at the domain \displaystyle \mathbb{R}-\{1\} it should be fairly obvious and easy to prove that this function is continuous on its domain.
0
reply
atsruser
Badges: 11
Rep:
?
#7
Report Thread starter 5 years ago
#7
(Original post by razzor)
Yes
Spoiler:
Show
It is clear the function is not continuous. It is not even defined at x=1.
1. What is the domain of the function?
2. Now apply the "open set" definition of continuity
1
reply
metaltron
Badges: 11
Rep:
?
#8
Report 5 years ago
#8
(Original post by razzor)
Yes
Spoiler:
Show
It is clear the function is not continuous. It is not even defined at x=1.
The function is continuous on its domain as poorform has said.
(Original post by atsruser)
1. What is the domain of the function?
2. Now apply the "open set" definition of continuity
Ha, was this a test of us then, or did you need help with some part of it?
0
reply
DFranklin
Badges: 18
Rep:
?
#9
Report 5 years ago
#9
(Original post by razzor)
Yes
Spoiler:
Show
It is clear the function is not continuous. It is not even defined at x=1.
Well, this is where the ambiguity comes. If you write:

f(x) = \begin{cases} x & \text{if } x &lt; 1 \\ x+1 & \text{if } x &gt; 1 \end{cases}

then (implicitly) f is a function with domain \mathbb{R} \backslash \{1\}, and f is continuous on this domain.

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.
1
reply
atsruser
Badges: 11
Rep:
?
#10
Report Thread starter 5 years ago
#10
(Original post by 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?
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.
2
reply
atsruser
Badges: 11
Rep:
?
#11
Report Thread starter 5 years ago
#11
(Original post by DFranklin)
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.
Right. Dfranklin, perspicacious as ever. In fact, the question is pretty boring unless you complicate it by defining the function with a weird domain. But I think that it's a fairly surprising result, and not at all obvious even to people reasonable familiar with the formal definition of continuity. It certainly confused me late last night, and it was only this morning that I realised the problem was the domain, not the function (it's obvious by considering open sets that it's continuous, but I couldn't see why at first).

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.
0
reply
DFranklin
Badges: 18
Rep:
?
#12
Report 5 years ago
#12
(Original post by 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 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.
0
reply
metaltron
Badges: 11
Rep:
?
#13
Report 5 years ago
#13
(Original post by 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.
A similar question: Is there an unbounded continuous function on [0,1]? How about on [0,1] intersection Q?
0
reply
atsruser
Badges: 11
Rep:
?
#14
Report Thread starter 5 years ago
#14
(Original post by 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.
But then people would say "Is x \in \mathbb{N}, \mathbb{Q}, \mathbb{R}, \nathbb{Z}?" or whatever.

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.
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?)
0
reply
atsruser
Badges: 11
Rep:
?
#15
Report Thread starter 5 years ago
#15
(Original post by metaltron)
A similar question: Is there an unbounded continuous function on [0,1]?
Unless you're being sneaky with your wording, then no - this is a standard result.

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?
0
reply
Remyxomatosis
Badges: 1
Rep:
?
#16
Report 5 years ago
#16
(Original post by 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 f(x)=1/x?
But 0 is a rational, so something that blows up at an irrational would work, like  \dfrac{1}{x - \pi/4}

(Original post by 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?)
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.
0
reply
atsruser
Badges: 11
Rep:
?
#17
Report Thread starter 5 years ago
#17
(Original post by Remyxomatosis)
But 0 is a rational, so something that blows up at an irrational would work, like  \dfrac{1}{x - \pi/4}
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.

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.
Yes. Lebesgue's criterion removes a lot of the awe and mystery from the integration of the weird functions.
0
reply
Remyxomatosis
Badges: 1
Rep:
?
#18
Report 5 years ago
#18
(Original post by atsruser)
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..
I should have been more clear. I was thinking  \dfrac{1}{x - \pi/4} restricted to  \mathbb{Q}. As  \dfrac{1}{x - \pi/4} is continuous on  \mathbb{R} \setminus \{ \pi/4 \} and restriction of continuous functions to subspaces is again continuous.
1
reply
metaltron
Badges: 11
Rep:
?
#19
Report 5 years ago
#19
(Original post by 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 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.
0
reply
atsruser
Badges: 11
Rep:
?
#20
Report Thread starter 5 years ago
#20
(Original post by 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.
And after a ludicrous delay ...

No, in fact, I was thinking that you can find a continuous unbounded function on a non-compact subset of \mathbb{R}. Using the lack of completeness didn't occur to me due to old age and forgetfulness.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Are you tempted to change your firm university choice on A-level results day?

Yes, I'll try and go to a uni higher up the league tables (35)
28.46%
Yes, there is a uni that I prefer and I'll fit in better (11)
8.94%
No I am happy with my choice (68)
55.28%
I'm using Clearing when I have my exam results (9)
7.32%

Watched Threads

View All