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Increasing Function on an Interval ... Browder, Proposition 3.7 ... ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
I am reading Andrew Browder's book: "Mathematical Analysis: An Introduction" ... ...

I am currently reading Chapter 3: Continuous Functions on Intervals and am currently focused on Section 3.1 Limits and Continuity ... ...

I need some help in understanding the proof of Proposition 3.7 ...


Proposition 3.7 and its proof read as follows:


Browder - Proposition 3.7 ... .png



In the above proof by Andrew Browder we read the following:

" ... ... Clearly \(\displaystyle A\leq f(t) \leq B\) since \(\displaystyle f\) is increasing ... ... "



Can someone demonstrate, formally and rigorously that \(\displaystyle A\leq f(t) \leq B\) ... ...


Note: Although it seems highly plausible, given the definitions of \(\displaystyle A\) and \(\displaystyle B\) and given also that \(\displaystyle f\) is increasing, that \(\displaystyle A\leq f(t) \leq B\) .. I am unable to prove it rigorously ...


Hope someone can help ...

Peter
 

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
260
Hi Peter ,

Since $f$ is increasing we know $x_{1}<x_{2}\,\Longrightarrow\, f(x_{1})\leq f(x_{2}).$ This means that $f(t)$ is an upper bound for the set $\{f(s)\,\vert\, s < t\}$ and a lower bound for $\{f(s)\,\vert\, t<s\}$. Does that help resolve the issue?
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,891
Hi Peter ,

Since $f$ is increasing we know $x_{1}<x_{2}\,\Longrightarrow\, f(x_{1})\leq f(x_{2}).$ This means that $f(t)$ is an upper bound for the set $\{f(s)\,\vert\, s < t\}$ and a lower bound for $\{f(s)\,\vert\, t<s\}$. Does that help resolve the issue?


Thanks so so much for the help GJA ...

Reflecting on what you have said ...

Beginning to suspect that I'm overthinking this issue ...

Thanks again for the help ...

Peter
 

GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
260
Hi Peter ,

Always happy to help in any way that I can. I had a tough time too when it came to understanding supremums and infimums. The words "least upper" and "greatest lower" don't hit the ear right initially.