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Let $f : [a, b]\to R$ be a continuous function such that $[a,b] \subset [f(a), f(b)]$. Prove that there exists $x\in [a,b]$ such that $f(x) = x$.

My attempt: I said let there be a $\delta > 0 $and defined $c$ and $d$ to be $x + \delta$ and $x-\delta$ respectively. From here since $f$ is continuous $[f(c), f(d)]\subset [f(a), f(b)]$. Then I assumed by definition $[c,d]$ is also a subset of $[f(c), f(d)]$. Then I claimed $\delta$ can be arbitrarily small so that $f(c) = f(d) = f(x)$.

Is this correct or is there a better approach?

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    It's not necessarily true that $[f(c),f(d)]$ is a subset of $[f(a),f(b)]$. Consider a function where $f(0) = 0$, $f(1) = (1)$ where $f(x)$ starts decreasing at $0$.2012-11-16

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Your approach is not correct, since you are assuming what you are supposed to show. You cannot define $c$ and $d$ to be something which depends on $x^*$ before you have shown that there is such a number as $x^*$.

A better approach would be to consider the function $g(x)=f(x)-x$. Argue that $g$ is continuous, that $g(a)\leq 0$, and that $g(b)\geq 0$. Then there should be a result available that you can use.

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    Did you draw the two intervals with one being contained in the other? Try to find$a$numerical example where $[a,b]\subset [f(a),f(b)]$ and a. (Hint: It can't be done.)2012-11-16
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In $\mathbb R^2$ draw the straight line $y=x$ and consider any interval at the positive part of $x$-axis, you'll get some intuition. Your problem is saying that under some conditions the graph of your function is going to intersept the straight line $y=x$ in some point. Think about the conditions for it to happen..