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In general, I understand the proof, however, I have a difficulty understanding the proof given by my professor. It goes like the following which is very similar to most proofs given in many textbooks.

Suppose $f$ is continuous on the interval $[a,b]$ and $f(a)<S<f(b)$, then $\exists c$ s.t. $f(c)=S$

Proof:

Define $V= \{x \in[a,b];f(x)\leq S \}$

Then $V$ is nonempty and has a least upper bound $c$.

Now, we claim that $f(c)=S$

First, suppose, $f(c)>S$ then by continuity $\exists \delta$ s.t. $f(x)$ is on the interval $(c-\delta,c]$

Then $c-\delta$ is an upper bound of $V$

Now suppose $S>f(c)$, then similarly, $\exists \delta$ s.t.$\forall x\in[c,c+\delta)$, we have $f(x)<S$

This implies that $c$ is not an upper bound of $V$

Therefore $f(c)=S$

Now what I'm especially confused is how he deduced $c-\delta$ as being an upper bound of $V$ by showing that $\exists \delta$ s.t. $f(x)$ is on the interval $(c-\delta,c]$

What is the justification behind this step?

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If $f(c)>S$, there is a $\delta>0$ such that $f(x)>S$ for each $x\in(c-\delta,c]$. So, the set $V$ contains no element of $(c-\delta,c]$, which is impossible, since $c=\sup V$, by definition.

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We have $V\subset [a,c].$

Suppose $f(c)>S.$ Then $c>a$. Let $e=f(c)-S.$ Then for some $d\in (0,c-a)$ we have $$x\in (-d+c,c]\implies |f(x)-f(c)|<e/2\implies f(x)>S.$$ That is, $(-d+c,c]$ is disjoint from $V,$ and since $V\subset [a,c],$ this implies $V\subset [a,-d+c].$ But then $\sup V\leq -d+c<c=\sup V,$ which is absurd.

Suppose $f(c)<S.$ Then $c<b$. Let $e'=S-f(c).$ So for some $d'\in (0,b-c)$ we have $$x\in [c,d'+c)\implies |f(x)-f(c)|<e'/2\implies f(x)<S$$ implying $[c, c+d')\subset V$. But then $\sup V\geq c+d'>c=\sup V.$ (Or at this point we could also say that $[c,c+d')\subset V$ contradicts $V\subset [a,c]).$

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