# Rising Sun Lemma proof from Spivak.

Here's the proof outline by Spivak: I don't understand why do we need part a) (I can prove it though). Why do the following proof won't work?

Suppose $$f(a)>f(b)$$.
Since $$f$$ is continuous on $$R$$, it is continuous at $$a$$, and so $$\exists\delta>0\ \forall x \ |x-a|<\delta\implies f(x)>f(b)$$.
So, we have $$x\in(a,b)$$, and $$f(x)>f(b)$$, which contradicts the statement that all points in $$(a,b)$$ are "shadow points".
Which leaves us with only possibility that $$f(a)=f(b)$$.

• Well, technically speaking in order to reach a contradiction in your argument above you should also conclude that there is no point $y$ in $(a,b)$ such that $f(y)>f(b)$. Doing this amounts to proving that $f$ attains its maximum at $a$. – Leo163 Jan 14 at 13:04
• @Leo163 Yes, but isn't it already stated in the problem, "... all points of $(a,b)$ are shadow points..." (which is equivalent to $x\in(a,b) \implies \exists y>x$ with $f(y)>f(x)$)? – yellowcat Jan 14 at 13:26
• That implication is not immediate, at least to me: it could be that for every point $y\in (a,b)$ there is a point $z\in (y,b)$ such that $f(z)>f(y)$. The fact that this is not the case is something that requires a proof, at least in my perspective. – Leo163 Jan 14 at 14:39

Your proof is correct. It boils down to the fact that if $$f(a) > f(b)$$, continuity at $$a$$ would mean $$f$$ is still be $$> f(b)$$ in a neighborhood of $$a$$.