Proving that a function can't take exactly same value twice

[Alpharup]

Homework Statement

Prove that there does not exist a continuous function f, defined on R which takes on every value exactly twice.

Homework Equations

It uses this property:
1... If f is continuous on [a,b], then there exists some y in [a,b], such that f(y)≥f(x), for all x in [a,b]

The Attempt at a Solution

This problem was taken from Spivak-calculus.
He did give a hint in the problem: He asks us to consider f(a) for some 'a' in R. By definition, the function takes the same value at some b in R. So, f(b)=f(a).
I am considering the case when x is in [a,b] and f(x)≥f(a). For all other x excluding this interval, f(x)<f(a)
f is continuous on R, which means that it must be continuous on [a,b]. By using the above property,,there exists some y in [a,b], such that f(y)≥f(x), for all x in [a,b]. But by definition of f, there is a y1 in [a,b] such that f(y)=f(y1). Here y1<y or y1>y...let us denote set D=[y,y1] or [y1,y] depending on which is greater(either y or y1). Here, let D=[y,y1]
Let us consider D. f, is continuous on D. so, there exists an k in D such that f(k)≤f(m) for all m in D.
There exists no h in D such that f(h)=f(y). Orelse, the function takes three values...a clear contradiction.
But k is in [a,b], so this means that f(k)≥f(a). But clearly f(k) is not equal to f(a).
So, f(k)>f(a) but less than f(y).
since f(y)>f(k)>f(a) and f is continuous on [a,y], there exists some p in [a,y] such that f(p)=f(k)'
Also in [y1,b], we have f(y1)>f(k)>f(b) and f is continuous. so there exists z in [y1,b] such that f(z)=f(k1)
We have three points(z,p,k) where the value of the function is same, a contradiction to the assumption of our f. Hence, there exists no f which takes the same value twice in R. Hence proof. Is my proof right?

 

[stevendaryl]

Homework Statement

Prove that there does not exist a continuous function f, defined on R which takes on every value exactly twice.

Homework Equations

It uses this property:
1... If f is continuous on [a,b], then there exists some y in [a,b], such that f(y)≥f(x), for all x in [a,b]

The Attempt at a Solution

This problem was taken from Spivak-calculus.
He did give a hint in the problem: He asks us to consider f(a) for some 'a' in R. By definition, the function takes the same value at some b in R. So, f(b)=f(a).
I am considering the case when x is in [a,b] and f(x)≥f(a). For all other x excluding this interval, f(x)<f(a)
f is continuous on R, which means that it must be continuous on [a,b]. By using the above property,,there exists some y in [a,b], such that f(y)≥f(x), for all x in [a,b]. But by definition of f, there is a y1 in [a,b] such that f(y)=f(y1).

That doesn't seem right. You have that $f(a) = f(b)$. So somewhere between $a$ and $b$, the function has its maximum value for that interval. So you're assuming that that is at the point $y$. So $a \leq x \leq b \Rightarrow f(x) \leq f(y)$. Now, you say:

"...by definition of $f$, there is a $y_1$ in $[a,b]$ such that $f(y) = f(y_1)$"

That's not right. All you know is that there is some $y_1$ such that $f(y) = f(y_1)$. You don't know that $y_1$ is in the interval $[a,b]$

 

[Alpharup]

That's not right. All you know is that there is some $y_1$ such that $f(y) = f(y_1)$. You don't know that $y_1$ is in the interval $[a,b]$

We have taken f such that
f <f(a), for all x<a and x >b.
So, y1 should lie this interval, if my assumption is not wrong.

 

[stevendaryl]

We have taken f such that
f <f(a), for all x<a and x >b.
So, y1 should lie this interval, if my assumption is not wrong.

Oh. I thought the only thing you were assuming was that $f(a) = f(b)$. How do you know you can assume that $f(x) < f(a)$ for $x < a$?

 

[Alpharup]

Let f(x)≥f(a) for x<a and x>b,
Consider the case when x<a. There is a point j present such that j<a,
The function f is contiunous at [j,a].
Here f(x)≥f(a), for all x in [j,a]
Thus the minimum value of f in this interval is f(a).
By definition of f, again there is f(b)=f(a) for some b.
for the interval [a,b], again f has some x in [a.b] such that f(x)≥f(a).
Considering the case x<b, we can easily prove that f takes 4 values for some points x in R. Hence a contradiction.

 

[stevendaryl]

Let f(x)≥f(a) for x<a and x>b

But why can you assume that?

I'm not saying you're wrong, only that you're missing a step in the reasoning.

Here's the step that you're missing (or it seems that way to me): If you look at a graph of $f(x)$ versus $x$, and draw a horizontal line through the point $x=a, y=f(a)$, you know that the graph can only cross that line twice, at $x=a$ and $x=b$. That means that the graph never crosses the line in the region $x < a$. So that means either

1. $f(x) > f(a)$ throughout the whole region, or
2. $f(x) < f(a)$ throughout the whole region.

(You can reason similarly for the region $x > b$).

So you can do a case split between the two possibilities for $x < a$.

 

[Alpharup]

Yeah, this broader intuition is more comprehendable.

 

[Alpharup]

Is the proof right?

 

[stevendaryl]

Is the proof right?

I think that under your assumption that $f(x) < f(a)$ in the regions $x < a$ and $x > b$, your proof is correct, but you need to prove it without making that assumption.

 

[Alpharup]

So, is there any way of proving it?

 

[stevendaryl]

So, is there any way of proving it?

You sort of have the right idea. But first you have to consider the possibilities outside the range $[a,b]$.

You can show that

1. either $f(x) < f(a)$, for all $x < a$, or $f(x) > f(a)$, for all $x < a$
2. either $f(x) < f(a)$, for all $x > b$, or $f(x) > f(a)$, for all $x > b$.

So there are 4 cases:

1. $f(x) < f(a)$ for all $x < a$, and $f(x) < f(a)$ for all $x > b$
2. $f(x) < f(a)$ for all $x < a$, and $f(x) > f(a)$ for all $x > b$
3. $f(x) > f(a)$ for all $x < a$, and $f(x) < f(a)$ for all $x > b$
4. $f(x) > f(a)$ for all $x < a$, and $f(x) > f(a)$ for all $x > b$

So you have to show that each of these leads to a contradiction. You have a proof of case 1, but you have 3 other cases.

 

[Alpharup]

Now I get it. You want me to consider all the possible cases. Thank you.

 

[PeroK]

Now I get it. You want me to consider all the possible cases. Thank you.

Choosing $a, b$ such that $f(a) = f(b)$ was a good start. I would then have drawn a graph and used the logic of what sort of graph you can and can't draw to guide your proof.