Elementary proof of fixed point theorem.

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SUMMARY

The discussion centers on the elementary proof of the fixed point theorem, specifically the statement that for a continuous function f: [0,1] → [0,1], there exists an x in [0,1] such that f(x) = x. The original proof attempt is deemed flawed due to misinterpretations of limits and continuity. Key issues include the incorrect handling of limits involving constants and the failure to establish the necessary conditions for the conclusion. The feedback emphasizes the need for a more rigorous approach to proving the theorem.

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ArcanaNoir
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I'd like someone to check this proof out for violations of math law. It seems like hackery to me, but then so does a lot of what my professor says, so maybe it's not. If it's flawed, just tell me why. Please don't suggest other ways to prove this, that's my job this weekend.

Thanks :)

p.s. This is my first post using [itex]\LaTeX[/itex], since my discovery that it was there all along (used to see the raw text).

Homework Statement


[itex]f: [0,1] \to [0,1][/itex] and [itex]f[/itex] is continuous [itex]\Rightarrow \exists x \in [0,1] : f(x)=x[/itex]

Homework Equations


Anything elementary calculus. Trying to keep it as simple as possible.

The Attempt at a Solution



let [itex]g(x)=x[/itex]
Then [itex]g(0) \le f(x) \le g(1)[/itex]
or, [itex]g(a) \le f(x) \le g(b)[/itex] where [itex][a,b]=[0,1][/itex]

[tex]\lim_{a\to x^-}{g(a)} = \lim_{b\to x^+}{g(b)} = x[/tex]
therefore
[tex]\lim_{x\to x}{f(x)} = x[/tex]

Since [itex]f(x)[/itex] takes on the same value as [itex]g(x)[/itex] at some point [itex]x[/itex], [itex]f(x)=g(x)=x[/itex] at this point.

So, for some [itex]x \in [0,1], f(x)=x[/itex]
 
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ArcanaNoir said:
I'd like someone to check this proof out for violations of math law. It seems like hackery to me, but then so does a lot of what my professor says, so maybe it's not. If it's flawed, just tell me why. Please don't suggest other ways to prove this, that's my job this weekend.

Thanks :)

p.s. This is my first post using [itex]\LaTeX[/itex], since my discovery that it was there all along (used to see the raw text).

Homework Statement


[itex]f: [0,1] \to [0,1][/itex] and [itex]f[/itex] is continuous [itex]\Rightarrow \exists x \in [0,1] : f(x)=x[/itex]


Homework Equations


Anything elementary calculus. Trying to keep it as simple as possible.


The Attempt at a Solution


This "proof" is flawed. Comments on each line below.

let [itex]g(x)=x[/itex]
Then [itex]g(0) \le f(x) \le g(1)[/itex]
No point renaming x to something else. This just says 0<f(x)<1
or, [itex]g(a) \le f(x) \le g(b)[/itex] where [itex][a,b]=[0,1][/itex]

And this just says the same thing.

[tex]\lim_{a\to x^-}{g(a)} = \lim_{b\to x^+}{g(b)} = x[/tex]
But a and b are constants. But anyway, x is presumably known to be continuous.

therefore
[tex]\lim_{x\to x}{f(x)} = x[/tex]
Strange way to write a limit and f(x) = x doesn't follow from anything you have done.
Since [itex]f(x)[/itex] takes on the same value as [itex]g(x)[/itex] at some point [itex]x[/itex], [itex]f(x)=g(x)=x[/itex] at this point.

So, for some [itex]x \in [0,1], f(x)=x[/itex]

You aren't close. You said no suggestions. Do you want a hint?
 
No thanks on the hint. I guess it's back to the drawing board. Thanks for the input :)
 

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