Proving There Does Not Exist a Continuous Function

[Lee33]

Homework Statement

Prove that there does not exist a continuous, bijective function $f:[0,1)\to \mathbb{R}.$

2. The attempt at a solution

I am stumped on how to do this question. What I was thinking of doing was assume that there is a function and arrive at a contradiction, in doing so I know I need to use the IVT since $f$ is assumed to be continuous, but I can't create such a contradiction.

 

[SqueeSpleen]

Theorem:
Let be f : A \subset E \longrightarrow F
The following properties are equivalent:
1) f is continuos in A
2) If T is a open set in (F,d'), then f^{-1}(T) is open set in the subspace (A,d)
3) If T is a closed set in (F,d'), then f^{-1}(T) is closed set in the subspace (A,d)
4) For all set S with S \subset A:
f(A \cap \overline{S} \subset \overline{f(S)}

I studied it in a course of "Introduction to Analysis" (It was roughly Metric Spaces Topology).

 

[Dick]

Homework Statement

Prove that there does not exist a continuous, bijective function $f:[0,1)\to \mathbb{R}.$

2. The attempt at a solution

I am stumped on how to do this question. What I was thinking of doing was assume that there is a function and arrive at a contradiction, in doing so I know I need to use the IVT since $f$ is assumed to be continuous, but I can't create such a contradiction.

Think back to one of your previous exercises and draw pictures of possibilities for the function. Using the IVT, isn't it true that since f is one-to-one then f([0,1/2])=[f(0),f(1/2)]? Now where are you going to put f((1/2,1))??

 

[Lee33]

Dick - So I can assume an $x\in [0,1/2]$ such that $f(x)\le f(0)$?

 

[Dick]

Dick - So I can assume an $x\in [0,1/2)$ such that $f(x)<f(0)$?

Well, no. I actually stated what I said a little wrong. I meant that f([0,1/2]) is equal to either [f(0),f(1/2)] or [f(1/2),f(0)] depending on whether f(0)<f(1/2) or vice versa. You remember the previous exercise, right?

 

[Lee33]

Hmm, can you elaborate a bit more? If I consider the case when $f(0) <f(1/2)$ then that must imply that $[f(0),f(1/2]$?

 

[Dick]

Hmm, can you elaborate a bit more? If I consider the case when $f(0) <f(1/2)$ then that must imply that $[f(0),f(1/2]$?

I thought we went through this yesterday. https://www.physicsforums.com/showthread.php?t=723428 A bijection is one-to-one.

 

[Lee33]

But in that proof, we considered $x\in (a,b)$ such that $f(x) <f(a)$ and you said I can't do that here.

 

[Dick]

But in that proof, we considered $x\in (a,b)$ such that $f(x) <f(a)$ and you said I can't do that here.

In that proof they assumed a<b because they wanted to write the interval [a,b] without having to say [a,b] if a<b or [b,a] if b<a. It's not important. The substance of the proof is. In this case I know 0<1/2. So I don't have to say 'assume 0<1/2'. f([0,1/2]) is either [f(0),f(1/2)] or [f(1/2),f(0)] depending. That's what I would like you to confirm that you still understand. I don't know what this x is you are speaking of here.

 

[Lee33]

So, since this is a bijection then for $0<1/2$ we have $[f(0), f(1/2)]$ and for $1/2 < 1$ we have $[f(1/2),f(1)]$.

 

[Dick]

So, since this is a bijection then for $0<1/2$ we have $[f(0), f(1/2)]$ and for $1/2 < 1$ we have $[f(1/2),f(1)]$.

No, you don't. I don't even think that means anything. f(1) isn't even necessarily defined. Can you just confirm that you understand why f([0,1/2]) is equal to either [f(0),f(1/2)] or [f(1/2),f(0)], you seemed comfortable with the proof yesterday.

 

[Lee33]

Yes, I can confirm that because we proved that yesterday.

 

[Dick]

Yes, I can confirm that because we proved that yesterday.

Alright! Let's assume for the moment that f([0,1/2])=[f(0),f(1/2)]. Then f(3/4) is either greater than f(1/2) or less than f(0), right? You should really be drawing pictures here.

 

[Lee33]

I drew a graph of a function $f(x) = x$ so it's just a diagonal line. Now for $f(3/4)$ why would it be less that $f(0)$?

 

[Dick]

I drew a graph of a function $f(x) = x$ so it just a diagonal line. Now for $f(3/4)$ why would it be less that $f(0)$?

f(x) is any bijective continuous function. It doesn't have to be f(x)=x. Draw a picture where you wiggle f(x) around anyway you want as long as it's continuous. I'm going to bed soon so I'll try and give you too many hints and see if you can figure it out. If f(3/4)>f(1/2), and since the function is onto, there must also be an x such that f(x)<f(0) and x must be greater than 1/2. Why? Try and draw such functions see why they can't be bijective. Using the IVT that would give you a contradiction with one-to-oneness. Just try it. I'll be back tomorrow.

 

[Lee33]

Okay, I will try it. Thanks!

 

[Dick]

Okay, I will try it. Thanks!

Actually trying to work your previous result into this isn't making it any simpler. Just think about f(0). Since f is onto, there must be an x in (0,1) such that f(x)=f(0)+1 and a y in (0,1) such that f(y)=f(0)-1. Just think about that.