# real analysis question

• Apr 20th 2008, 01:10 AM
squarerootof2
real analysis question
this is from ross's book elementary analysis: the theory of calculus section 21 #10. i tried reading the section and stuff but it seems like the problems are still hard for me. here are the problems:

show that there exist continuous functions
(a) mapping (0,1) onto [0,1]
(b) mapping (0,1) onto R
(c) mapping [0,1]U[2,3] onto [0,1]

also, can someone give me tips on how i should study for this class? this class is the 2nd quarter of analysis i'm taking, and we are 4 weeks into the class. so i'd read the book and think i understand the material, only to get stumped on the questions... the midterm exam is on next week friday, so i'm getting really anxious... help!
• Apr 20th 2008, 02:28 AM
CaptainBlack
Quote:

Originally Posted by squarerootof2
this is from ross's book elementary analysis: the theory of calculus section 21 #10. i tried reading the section and stuff but it seems like the problems are still hard for me. here are the problems:

show that there exist continuous functions
(a) mapping (0,1) onto [0,1]
(b) mapping (0,1) onto R
(c) mapping [0,1]U[2,3] onto [0,1]

(a) Look at $\sin(2\pi x)$ this takes the interval $(0,1)$ to $[-1,1]$, so can you do something to modify this function so the image of $(0,1)$ is $[0,1]$?

RonL
• Apr 20th 2008, 02:35 AM
CaptainBlack
Quote:

Originally Posted by squarerootof2
this is from ross's book elementary analysis: the theory of calculus section 21 #10. i tried reading the section and stuff but it seems like the problems are still hard for me. here are the problems:

show that there exist continuous functions
(a) mapping (0,1) onto [0,1]
(b) mapping (0,1) onto R
(c) mapping [0,1]U[2,3] onto [0,1]

also, can someone give me tips on how i should study for this class? this class is the 2nd quarter of analysis i'm taking, and we are 4 weeks into the class. so i'd read the book and think i understand the material, only to get stumped on the questions... the midterm exam is on next week friday, so i'm getting really anxious... help!

For (b) consider somthing like $\tan( \pi x -\pi/2)$

RonL
• Apr 20th 2008, 10:15 AM
squarerootof2
Quote:

Originally Posted by CaptainBlack
(a) Look at $\sin(2\pi x)$ this takes the interval $(0,1)$ to $[-1,1]$, so can you do something to modify this function so the image of $(0,1)$ is $[0,1]$?

RonL

isn't the image of the sin (2pi*x) function already [0,1]? it seems like all the pts in [0,1] are hit already by putting in (0,1) for x? i dont see how we need to modify the function. hmm...
• Apr 20th 2008, 10:24 AM
Jhevon
Quote:

Originally Posted by squarerootof2
isn't the image of the sin (2pi*x) function already [0,1]? it seems like all the pts in [0,1] are hit already by putting in (0,1) for x? i dont see how we need to modify the function. hmm...

try doing something like shrinking the period and shifting the graph
• Apr 20th 2008, 10:38 AM
Isomorphism
Quote:

Originally Posted by squarerootof2

show that there exist continuous functions
(a) mapping (0,1) onto [0,1]
(b) mapping (0,1) onto R
(c) mapping [0,1]U[2,3] onto [0,1]

Actually we can construct ugly functions to satisfy requirements....
For (c) , we can get onto mapping with $\sin(\frac{\pi x}{2})$ from [0,1] to [0,1]. Now define f([2,3]) = 1. This will make the function continuous, but the function does not look natural :(
• Apr 20th 2008, 10:41 AM
CaptainBlack
Quote:

Originally Posted by squarerootof2
isn't the image of the sin (2pi*x) function already [0,1]? it seems like all the pts in [0,1] are hit already by putting in (0,1) for x? i dont see how we need to modify the function. hmm...

$\sin(2\pi x)$ achives a maximum value of $1$ and a minimum of $-1$ in $(0,1)$ .

RonL
• Apr 20th 2008, 10:43 AM
CaptainBlack
Quote:

Originally Posted by Jhevon
try doing something like shrinking the period and shifting the graph

I would leave the period alone and shift the function upwards by 1 and shrink it by a factor of 2.

RonL
• Apr 20th 2008, 01:38 PM
squarerootof2
Quote:

Originally Posted by CaptainBlack
$\sin(2\pi x)$ achives a maximum value of $1$ and a minimum of $-1$ in $(0,1)$ .

RonL

so i've been thinking , and would it work if i take the absolute value of sin (2pi*x) (just to make the function onto [0,1]. would that work?
• Apr 20th 2008, 10:47 PM
CaptainBlack
Quote:

Originally Posted by squarerootof2
so i've been thinking , and would it work if i take the absolute value of sin (2pi*x) (just to make the function onto [0,1]. would that work?

1. It would leave the function continuous, so that's OK

2. It would map (0,1) to [0,1] so that's Ok

so yes it would

RonL