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  1. MHB Master
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    #1
    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 Theorem 3.16 ...


    Theorem 3.16 and its proof read as follows:








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

    " ... ... But $ \displaystyle f(b) \gt y$ implies (since $ \displaystyle f$ is continuous at $ \displaystyle b$) that there exists $ \displaystyle \delta \gt 0$ such that $ \displaystyle f(t) \gt y$ for all $ \displaystyle t$ with $ \displaystyle b - \delta \lt t \leq b$. ... ... "


    My question is as follows:

    How do we demonstrate explicitly and rigorously that since $ \displaystyle f$ is continuous at $ \displaystyle b$ and $ \displaystyle f(b) \gt y $ therefore we have that there exists $ \displaystyle \delta \gt 0$ such that $ \displaystyle f(t) \gt y$ for all $ \displaystyle t$ with $ \displaystyle b - \delta \lt t \leq b$. ... ...


    Help will be much appreciated ...

    Peter




    ***NOTE***

    The relevant definition of one-sided continuity for the above is as follows:

    $ \displaystyle f$ is continuous from the left at $ \displaystyle b$ implies that for every $ \displaystyle \epsilon \gt 0$ there exists $ \displaystyle \delta \gt 0$ such that for all $ \displaystyle x \in [a, b]$ ...

    we have that $ \displaystyle b - \delta \lt x \lt b \Longrightarrow \mid f(x) - f(b) \mid \lt \epsilon$

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  3. MHB Apprentice
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    #2
    We have $f(b) > y$ or $f(b)-y >0$. Define $g(b) = f(b)-y>0$.

    Since $f$ is continuous at $b$, $g$ is continuous at $b$.

    Thus for all $\epsilon_0 >0$, there exist $\delta_0 >0$ s.t. $t \in [a,b]$ and $|t-b|<\delta_0 $ imply $|g(t)-g(b)| < \epsilon_0$.

    In particular, for $ \epsilon := g(b)/2>0$ there exists $\delta >0$ s.t. $t \in [a,b]$ and $|t-b|<\delta $ imply $|g(t)-g(b)| < g(b)/2$.

    Therefore $-g(b)/2 <g(t)-g(b) < g(b)/2$. Hence $g(t)> g(b)-g(b)/2 = g(b)/2$.

    Rewriting $g(t) > g(b)/2$ in terms of $f$ we have $f(t) > \frac{1}{2}(f(b)+y)> \frac{1}{2}(y+y) = y$ as $f(b)>y$.

    Thus $f(t)>y$ whenever $t \in [a,b]$ and $|t-b|<\delta $; that's $t \in [a,b] \cap (b-\delta, b+\delta)$, i.e. whenever $b-\delta < t \le b.$

    This calculation can be skipped by appealing to sign-preserving property of limits/continuous functions:

    Sign-preserving property (for continuous functions): Let $f: I \to \mathbb{R}$ be continuous at $ c \in I \subseteq \mathbb{R}$.

    1. If $f(c)>0$ then there exists $M>0$ and $\delta >0$ s.t. $x \in I$ and $|x-c|< \delta$ implies $f(x) > M$.

    2. If $f(c)<0$ then there exists $N<0$ and $\delta >0$ s.t. $x \in I$ and $|x-c|< \delta$ implies $f(x) <N$.
    Last edited by MountEvariste; February 1st, 2020 at 11:04.

  4. MHB Master
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    #3 Thread Author
    Quote Originally Posted by MountEvariste View Post
    We have $f(b) > y$ or $f(b)-y >0$. Define $g(b) = f(b)-y>0$.

    Since $f$ is continuous at $b$, $g$ is continuous at $b$.

    Thus for all $\epsilon_0 >0$, there exist $\delta_0 >0$ s.t. $t \in [a,b]$ and $|t-b|<\delta_0 $ imply $|g(t)-g(b)| < \epsilon_0$.

    In particular, for $ \epsilon := g(b)/2>0$ there exists $\delta >0$ s.t. $t \in [a,b]$ and $|t-b|<\delta $ imply $|g(t)-g(b)| < g(b)/2$.

    Therefore $-g(b)/2 <g(t)-g(b) < g(b)/2$. Hence $g(t)> g(b)-g(b)/2 = g(b)/2$.

    Rewriting $g(t) > g(b)/2$ in terms of $f$ we have $f(t) > \frac{1}{2}(f(b)+y)> \frac{1}{2}(y+y) = y$ as $f(b)>y$.

    Thus $f(t)>y$ whenever $t \in [a,b]$ and $|t-b|<\delta $; that's $t \in [a,b] \cap (b-\delta, b+\delta)$, i.e. whenever $b-\delta < t \le b.$

    This calculation can be skipped by appealing to sign-preserving property of limits/continuous functions:

    Sign-preserving property (for continuous functions): Let $f: I \to \mathbb{R}$ be continuous at $ c \in I \subseteq \mathbb{R}$.

    1. If $f(c)>0$ then there exists $M>0$ and $\delta >0$ s.t. $x \in I$ and $|x-c|< \delta$ implies $f(x) > M$.

    2. If $f(c)<0$ then there exists $N<0$ and $\delta >0$ s.t. $x \in I$ and $|x-c|< \delta$ implies $f(x) <N$.


    Thanks MountEvariste for an informative and really helpful reply ...

    Now working through what you have written ...

    Thanks again ...

    Peter

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