Theorem: Let $f:[0,1] \to [0,1]$. Assume $f$ is continuous. Then there exists $c \in [0, 1]$ such that $f(c) = c$.
For any function $f$, let $g(x) = f(x) − x$. If the theorem is true, what does that say about $g(c)$?
Write a proof of the theorem. Do it by applying the Intermediate Value Theorem in the appropriate way to the function $g(x) = f(x) − x$ with $a = 0$ and $b = 1$. Clearly state why the hypotheses of the IVT are satisfied here.
Well if the theorem is true, then there exists $c$ such that $f(c) = c$. Then $f(c) - c = 0$. Then $g(c) = 0$.
Intermediate Value Theorem: Let $I$ be an interval and $f$ a function whose domain contains $I$. If $f$ is continuous, then for all $a, b \in I$ with $a < b$ and all real numbers $k$, if $k$ is strictly between $f (a)$ and $f (b)$, then there exists $c$ such that $a < c < b$ and $f (c) = k$.
Proof Attempt: Take $a = 0$ and $b = 1$. Then $a,b \in[0,1]$ and $a < b$. Since $f:[0,1] \to [0,1]$, it follows that for all real numbers $k$ such that $f(a) < k < f(b)$, then there exists $c$ such that $a < c < b$ so $0 < c < 1$ and $f(c) = k$. Then $f(c) - k = 0$. Thus, since $g(x)=f(x)-x$, it follows that $g(c) = 0$.
Looking to see if I did this correctly and/or if there is a more elegant way to prove problem 2. I am a little confused why the first theorem is true, but I went with it.