Real Analysis - Continuity
a. Give an example of a function defined everywhere on the interval $[0,1]$, which does not achieve its maximum.
b. Give an example of a function defined on $mathbb{R}$, that is nowhere continuous.
c. Give an example of a continuous function defined on a bounded set, which is not uniformly continuous.
My answers are
- $f(x) = x^2$
$f(x) = {1$, if $x$ is rational; $-1$, if $x$ is irrational$}$
- $f(x) = 1/x$
My workings are attached. Please help verify if the working is correct.Solutions
real-analysis continuity maxima-minima uniform-continuity
asked Dec 1 at 22:36
Ty Johnson
283
add a comment |
a. Give an example of a function defined everywhere on the interval $[0,1]$, which does not achieve its maximum.
b. Give an example of a function defined on $mathbb{R}$, that is nowhere continuous.
c. Give an example of a continuous function defined on a bounded set, which is not uniformly continuous.
My answers are
- $f(x) = x^2$
$f(x) = {1$, if $x$ is rational; $-1$, if $x$ is irrational$}$
- $f(x) = 1/x$
My workings are attached. Please help verify if the working is correct.Solutions
real-analysis continuity maxima-minima uniform-continuity
asked Dec 1 at 22:36
Ty Johnson
283
4
2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41
add a comment |
a. Give an example of a function defined everywhere on the interval $[0,1]$, which does not achieve its maximum.
b. Give an example of a function defined on $mathbb{R}$, that is nowhere continuous.
c. Give an example of a continuous function defined on a bounded set, which is not uniformly continuous.
My answers are
- $f(x) = x^2$
$f(x) = {1$, if $x$ is rational; $-1$, if $x$ is irrational$}$
- $f(x) = 1/x$
My workings are attached. Please help verify if the working is correct.Solutions
real-analysis continuity maxima-minima uniform-continuity
asked Dec 1 at 22:36
Ty Johnson
283
a. Give an example of a function defined everywhere on the interval $[0,1]$, which does not achieve its maximum.
b. Give an example of a function defined on $mathbb{R}$, that is nowhere continuous.
c. Give an example of a continuous function defined on a bounded set, which is not uniformly continuous.
My answers are
- $f(x) = x^2$
$f(x) = {1$, if $x$ is rational; $-1$, if $x$ is irrational$}$
- $f(x) = 1/x$
My workings are attached. Please help verify if the working is correct.Solutions
real-analysis continuity maxima-minima uniform-continuity
real-analysis continuity maxima-minima uniform-continuity
asked Dec 1 at 22:36
Ty Johnson
283
asked Dec 1 at 22:36
Ty Johnson
283
asked Dec 1 at 22:36
Ty Johnson
283
asked Dec 1 at 22:36
Ty Johnson
283
asked Dec 1 at 22:36
Ty Johnson
283
283
4
2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41
add a comment |
4
2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41
4
4
2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41
2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41
add a comment |
3 Answers
3
active
oldest
votes
$a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.
$b)$ The function you had is a good one.
$c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.
answered Dec 1 at 22:50
DeepSea
70.9k54487
add a comment |
- Wrong: the maximum is $1$, which is $f(1)$.
- Right.
- Your answer is incomplete, at best, since you did not state what is the domain of your function.
answered Dec 1 at 22:44
José Carlos Santos
150k22120221
add a comment |
1)
Extreme value theorem says that every continuous function over a closed interval has a maximum.
You need a function that is not continuous.
$f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$
answered Dec 1 at 23:27
Doug M
43.9k31854
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.
$b)$ The function you had is a good one.
$c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.
answered Dec 1 at 22:50
DeepSea
70.9k54487
add a comment |
$a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.
$b)$ The function you had is a good one.
$c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.
answered Dec 1 at 22:50
DeepSea
70.9k54487
add a comment |
$a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.
$b)$ The function you had is a good one.
$c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.
answered Dec 1 at 22:50
DeepSea
70.9k54487
$a)$ Consider $f(x)$ on $[0,1]$ such that $f(0) = 0, f(1) = 0, f(x) = dfrac{1}{x}, 0 <x < 1$.
$b)$ The function you had is a good one.
$c)$ Consider $f(x)$ on $(0,1)$ and $f(x) = dfrac{1}{x}$.
answered Dec 1 at 22:50
DeepSea
70.9k54487
answered Dec 1 at 22:50
DeepSea
70.9k54487
answered Dec 1 at 22:50
DeepSea
70.9k54487
answered Dec 1 at 22:50
DeepSea
70.9k54487
70.9k54487
add a comment |
add a comment |
- Wrong: the maximum is $1$, which is $f(1)$.
- Right.
- Your answer is incomplete, at best, since you did not state what is the domain of your function.
answered Dec 1 at 22:44
José Carlos Santos
150k22120221
add a comment |
- Wrong: the maximum is $1$, which is $f(1)$.
- Right.
- Your answer is incomplete, at best, since you did not state what is the domain of your function.
answered Dec 1 at 22:44
José Carlos Santos
150k22120221
add a comment |
- Wrong: the maximum is $1$, which is $f(1)$.
- Right.
- Your answer is incomplete, at best, since you did not state what is the domain of your function.
answered Dec 1 at 22:44
José Carlos Santos
150k22120221
- Wrong: the maximum is $1$, which is $f(1)$.
- Right.
- Your answer is incomplete, at best, since you did not state what is the domain of your function.
answered Dec 1 at 22:44
José Carlos Santos
150k22120221
answered Dec 1 at 22:44
José Carlos Santos
150k22120221
answered Dec 1 at 22:44
José Carlos Santos
150k22120221
answered Dec 1 at 22:44
José Carlos Santos
150k22120221
150k22120221
add a comment |
add a comment |
1)
Extreme value theorem says that every continuous function over a closed interval has a maximum.
You need a function that is not continuous.
$f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$
answered Dec 1 at 23:27
Doug M
43.9k31854
add a comment |
1)
Extreme value theorem says that every continuous function over a closed interval has a maximum.
You need a function that is not continuous.
$f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$
answered Dec 1 at 23:27
Doug M
43.9k31854
add a comment |
1)
Extreme value theorem says that every continuous function over a closed interval has a maximum.
You need a function that is not continuous.
$f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$
answered Dec 1 at 23:27
Doug M
43.9k31854
1)
Extreme value theorem says that every continuous function over a closed interval has a maximum.
You need a function that is not continuous.
$f(x) = begin{cases} f(x) = x & x<1\f(x) = 0 & x=1 end{cases}$
answered Dec 1 at 23:27
Doug M
43.9k31854
answered Dec 1 at 23:27
Doug M
43.9k31854
answered Dec 1 at 23:27
Doug M
43.9k31854
answered Dec 1 at 23:27
Doug M
43.9k31854
43.9k31854
add a comment |
add a comment |
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4
2. is correct. For 3, you need to specify the bounded domain. Your answer to 1. is incorrect, as the function certainly achieves its maximum on the interval $[0,1]$.
– Ted Shifrin
Dec 1 at 22:41