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I want to prove that if we have a continuous function from the closed interval [0,1] to the closed interval [0,1], that there exists a value y in [0,1] such that $f(y)=y$.

I have an idea of a theorem that I want to use: Let f is a continuous real function on the interval [a,b]. If f(a) < f(b) and if c is a number such that f(a)< c < f(b), then there exists a point x in (a,b) such that f(x)=c.

So I know that f(0) < f(1). So according to this theorem, for some number in between 0 and 1, there exists a point in the domain such that f(x)=this number. But I'm having trouble coming up with the argument

mayra
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As @sciona has already pointed out, try using $g(x) = f(x)-x$ and the Intermediate Value Theorem (which you stated). I wanted to add one very subtle point though: The IVT you stated is for points $c$ greater/less than but not equal to $f(a)$ and $f(b)$. For example, you couldn't find your desired $y$ for $f(x) = 0$ for all $x\in[0,1]$ by just checking points on the open interval $(0,1)$. That means we need to check the end points separately, and then use the IVT for the points in between.

Glare
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    Okay thanks.just to make sure I understand; if the intervals were open rather than closed,I wouldn't have to worry about that? – mayra Jan 21 '15 at 23:55
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    If your function was $f:[0,1]\to (0,1)$, then yes, that would be correct. You'd have $f(0)>0$ and $f(1)<1$ by the definition of the range, so $g(0)>0$ and $g(1)<0$. The theorem works from there. It's really just the cases $f(0) = 0$ and $f(1)=1$ in your problem that you need to worry about. – Glare Jan 22 '15 at 00:02