Math Is Fun Forum

sumpm1

Member

Registered: 2007-03-05

Posts: 42

Real Analysis Help

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Real Analysis Help

Last edited by JaneFairfax (2009-03-01 12:35:18)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Real Analysis Help

Avon

Member

Registered: 2007-06-28

Posts: 80

Re: Real Analysis Help

Jane, for any function

we have

so if your proof were correct then no such function could be continuous.
Clearly (0,1) is a closed subset of itself.

Here's a proof that works:

Suppose

is continuous bijection.
Let

and

.

If a < b then consider the set

.
Since [a,b] is connected, so is X. Also

so we must have X = [0,1].
This contradicts that f is a bijection.

Similarly if b < a.

PS: I feel pretty silly now. My proof is really just an application of the Intermediate Value Theorem.

Last edited by Avon (2009-03-01 11:37:57)

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Real Analysis Help

Clearly (0,1) is a closed subset of itself.

Apologies for nit-picking, but I'm a stickler for terminology.  (0,1) is a closed subset relative to itself.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."