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Suppose that $f \colon [a, b] \to [a, b]$ is continuous. (Note that the range of $f$ is a subset of $[a, b]$)

$a) $Prove that there exists at least one point $x \in [a, b]$ such that $f(x) = x$. A point with this property is known as a fixed point of $f$.

$b)$ what happen if $X$ equals $[0,1) $or $(0,1)$ ?

my answer :

for $ a)$ :If $f(a)=a$ or $f(b)=b$ then we are done, suppose $f(a)\ne a, f(b) \ne b$

Now what can you say about $g$ below? does it vanish somewhere in $(a,b)$?

$g(x)=f(x)-x$

For $b)$ i don't know ,,,pliz help me...

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closed as off-topic by Shaun, user99914, Math1000, Wouter, choco_addicted Apr 8 '18 at 5:52

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  • $\begingroup$ Is the question meant to read, what is $f(X)$ if $X=[0,1)$ or $(0,1)$? $\endgroup$ – Alex Clark Apr 7 '18 at 11:56
  • $\begingroup$ Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. $\endgroup$ – Shaun Apr 7 '18 at 12:46
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The range is $[a,b]$, not a subset of $[a,b]$.

a) Let $g(x)=f(x)-x$. Then $g(a)\geqslant 0\geqslant g(b)$. Therefore, $g(c)=0$ for some $c\in[a,b]$, and this means that $f(x)=c$.

b) Let $f(x)=\frac12(x+1)$. The $f\bigl([0,1)\bigr)\subset[0,1)$, but the equation $f(x)=x$ has no solutions in $[0,1)$. Or in $(0,1)$.

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  • $\begingroup$ Why is the range the codomain? $\endgroup$ – William Elliot Apr 7 '18 at 13:59

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