MHB Continuity and Compact Sets - Bolzano's Theorem

[Math Amateur]

I am reading Tom Apostol's book: Mathematical Analysis (Second Edition).

I am currently studying Chapter 4: Limits and Continuity.

I am having trouble in fully understanding the proof of Bolzano's Theorem (Apostol Theorem 4.32).

Bolzano's Theorem and its proof reads as follows:
https://www.physicsforums.com/attachments/3863
In the above proof, Apostol writes the following:" ... ... If $$f(c) \lt 0$$, then $$c - \delta/2$$ is an upper bound for A, again contradicting the definition of $$c$$. ... ... "
Can someone please explain why $$f(c) \lt 0$$ implies that $$c - \delta/2$$ is an upper bound for A?

Help will be appreciated ... ...

Peter

 

[Fallen Angel]

Hi Peter,

$f$ is continuous and $f(c)<0$ so there exists a 1-ball $B(c,\delta)$ (that's an open interval) where $f$ is negative, so $f(c-\delta/2)<0$ and $c-\delta/2<c=sup(A)$ which is a contradiction.

 

[Math Amateur]

Hi Peter,

$f$ is continuous and $f(c)<0$ so there exists a 1-ball $B(c,\delta)$ (that's an open interval) where $f$ is negative, so $f(c-\delta/2)<0$ and $c-\delta/2<c=sup(A)$ which is a contradiction.

Hello Fallen Angel,

Thanks for your help!

... BUT ... still reflecting on what you have written ... and need further help ...

... I can see that $$f$$ is negative on $$B( c; \delta)$$ and thus, that $$f ( c - \delta/2 ) \lt 0$$ ... ... but fail to fully understand your contradiction ...I suspect the argument may be something like this ... ...

$$f ( c - \delta/2 ) \lt 0$$ from the argument of yours above ...

But then ... ...

$$c = \text{ sup } A$$ ... ... and therefore the point $$c - \delta/2 \in A$$ ... ... [ BUT ... ! I cannot justify this step rigorously! ...]

But $$c - \delta/2 \in A$$ implies that $$f( c - \delta/2) \ge 0$$ ... ... which contradicts what we established above ... ...Can you comment on my analysis ...

Hopefully you can clarify the situation for me ...

Thanks again for you help with the above issue ... ...

Peter

 

[Fallen Angel]

Hi Peter,

Yes, you can justify it like you have done, the fact that $c-\delta/2 \in A$ implies $f(c-\delta/2)\geq 0$ it's just by definition of $A$.

If $c-\delta/2\notin A$ then you have an upper bound for $A$ that is strictly less than the supremum, which is another contradiction.