Given that $f$ is a continuous mapping of $I = [0,1]$ into $I$, I want to prove that $f(x) = x$ for at least one $x \in I.$ I have a feeling that I need to build this by using some form of continuous pigeonhole principle, but I am not sure that is reasonable given that I have tended to use that principle only in discrete problems. Is it possible that if $\forall x \in I. f(x) \neq x$ would contradict the definition of $f$ being some continuous mapping? It is possible that I may be able to prove that the preimage of an open set will not be open in this case, but I am having difficulty formalizing this idea.
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Try the intermediate value theorem, applied to a strategically chosen function. – Eric Auld Dec 09 '15 at 04:05
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Hint: you want to show that there is some $x$ such that $f(x)=x$, which can be rewritten as $f(x)-x=0$. Define a new function $g(x):=f(x)-x$. The problem then becomes whether or not this function attains a $0$ value on $I$. Can you show this using the Intermediate Value Theorem?
J.G
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