#### Archived

This topic is now archived and is closed to further replies.

# Bijection from open interval to closed

This topic is 5734 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hi, I came across this problem and out of curiosity I''ve been trying to figure out a way to make a bijection between the open interval (0,1) and the closed interval [0,1]. No luck so far, because I said, if you start with the identity function, then you can''t get 0 or 1 as an output of the function. But then say you assign some 2 numbers in the open interval to map to 0 to 1. Say, f(.3) = 0 and f(.5) = 1. (The choice of the 2 numbers doesn''t matter). Anyways, but now you have nothing in (0,1) which maps to .3 or .5! Anyways if anyone''s curious and wants to try solving this, let me know if you figure it out! =) Thanks, Raj

##### Share on other sites
My initial response is that a bijection isn''t possible. First, just to satisfy the subjection requirement, we need to find a function which outputs everything in the interval [0,1] given the input (0,1). The identity function doesn''t even satisfy this. Next, we have to make sure our function is injective, which basically means that the inverse is also a function. Since we can''t think of a function that satisfies the first condition, we''re already in a tight spot!

Tell me if you find such a function

##### Share on other sites
I don''t think its possible either.
The function couldn''t be continuous, because we would have to achieve the min (0) and max (1) in the interval, and then local neighbourhoods of the maxima / minima would have overlapping image sets, disallowing a bijection.

Although I am a bit rusty...

"Most people think, great God will come from the sky, take away everything, and make everybody feel high" - Bob Marley

##### Share on other sites
I believe that one exists, although I'm afraid I can't give you an example just yet. However, consider the following:
f: (0,1) -> [0,1]f(x) = (x+1/2)/2

Then (clearly?) f is 1-1. Also, a function g defined in the same way, with the domain and range swapped round:
g: [0,1] -> (0,1)g(x) = (x+1/2)/2

is also 1-1, and both f and g map into the appropriate sets. I think that's all fine. Now comes the slightly dodgy bit (as I've only just learnt it in lectures). By the Schröder-Bernstein theorem, as there is a 1-1 mapping from (0,1) to [0,1] and also a 1-1 mapping from [0,1] to (0,1), there exists a bijection between (0,1) and [0,1].

I've only just been shown the proof, and someone's borrowed my lecture notes, but I think it may be possible to give the bijective function explicitly.

I hope all I've said is true...

Miles

[EDITx1001: sorry, messed up the tags, etc.]

[edited by - Miles Clapham on November 7, 2002 12:37:42 PM]

##### Share on other sites
Hi there!

There is no continuous bijection between open and closed interval, acording to Weierstrass theorem ( continuous image of closed interval is always closed interval ).

Discontinuous bijection exist, but do not expect some nice and simple formula for it.

eg.:
f : [0,1] --> [0,1)
f(1)=1/2
f(1/2)=1/4
f(1/4)=1/8
f(1/8)=1/16
...
and f(x)=x if x != 1, 1/2, 1/4, ...
===>f IS bijection (discontinuous)

Hope that helps, and apologize for my English
Kruno

##### Share on other sites
Hi there!

There is no continuous bijection between open and closed interval, acording to Weierstrass theorem ( continuous image of closed interval is always closed interval ).

Discontinuous bijection exist, but do not expect some nice and simple formula for it.

eg.:
f : [0,1] --> [0,1)
f(1)=1/2
f(1/2)=1/4
f(1/4)=1/8
f(1/8)=1/16
...
and f(x)=x if x != 1, 1/2, 1/4, ...
===>f IS bijection (discontinuous)

Hope that helps, and apologize for my English
Kruno

##### Share on other sites
Hi there!

There is no continuous bijection between open and closed interval, acording to Weierstrass theorem ( continuous image of closed interval is always closed interval ).

Discontinuous bijection exist, but do not expect some nice and simple formula for it.

eg.:
f : [0,1] --> [0,1)
f(1)=1/2
f(1/2)=1/4
f(1/4)=1/8
f(1/8)=1/16
...
and f(x)=x if x != 1, 1/2, 1/4, ...
===>f IS bijection (discontinuous)

Hope that helps, and apologize for my English
Kruno

##### Share on other sites
Hi there!

There is no continuous bijection between open and closed interval, acording to Weierstrass theorem ( continuous image of closed interval is always closed interval ).

Discontinuous bijection exist, but do not expect some nice and simple formula for it.

eg.:
f : [0,1] --> [0,1)
f(1)=1/2
f(1/2)=1/4
f(1/4)=1/8
f(1/8)=1/16
...
and f(x)=x if x != 1, 1/2, 1/4, ...
===>f IS bijection (discontinuous)

Hope that helps, and apologize for my English
Kruno

##### Share on other sites
Hi there!

There is no continuous bijection between open and closed interval, acording to Weierstrass theorem ( continuous image of closed interval is always closed interval ).

Discontinuous bijection exist, but do not expect some nice and simple formula for it.

eg.:
f : [0,1] --> [0,1)
f(1)=1/2
f(1/2)=1/4
f(1/4)=1/8
f(1/8)=1/16
...
and f(x)=x if x != 1, 1/2, 1/4, ...
===>f IS bijection (discontinuous)

Hope that helps, and apologize for my English
Kruno

1. 1
Rutin
25
2. 2
3. 3
4. 4
JoeJ
18
5. 5

• 14
• 14
• 11
• 11
• 9
• ### Forum Statistics

• Total Topics
631757
• Total Posts
3002131
×