MHB Understanding Browder Proposition 3.14: Increasing Function & Discontinuities

[Math Amateur]

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 3.14 ...Proposition 3.14 and its proof read as follows:
View attachment 9536
In the above proof by Browder we read the following:

" ... ... For any $$d \in I, d$$ not an endpoint of $$I$$ we know (Proposition 3.7) that $$f(d-)$$ and $$f(d+)$$ exist with $$f(d-) \leq f(d) \leq f(d+)$$, so $$d \in D$$ if and only if $$f(d-) \lt f(d+)$$. ... ... "My question is as follows:

Can someone please demonstrate (rigorously) exactly why/how it follows that $$d \in D$$ if and only if $$f(d-) \lt f(d+)$$. ... ... Help will be much appreciated ...

Peter
=======================================================================================The above post mentions Browder Proposition 3.7 ... so I am providing the text of that proposition ... as follows:
View attachment 9537
Hope that helps ...

Peter

 

[Math Amateur]

Re: Increasing Function and Discontinuitiesl ... Browder, Proposition 3.14 ... ...

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 3.14 ...Proposition 3.14 and its proof read as follows:

In the above proof by Browder we read the following:

" ... ... For any $$d \in I, d$$ not an endpoint of $$I$$ we know (Proposition 3.7) that $$f(d-)$$ and $$f(d+)$$ exist with $$f(d-) \leq f(d) \leq f(d+)$$, so $$d \in D$$ if and only if $$f(d-) \lt f(d+)$$. ... ... "My question is as follows:

Can someone please demonstrate (rigorously) exactly why/how it follows that $$d \in D$$ if and only if $$f(d-) \lt f(d+)$$. ... ... Help will be much appreciated ...

Peter
=======================================================================================The above post mentions Browder Proposition 3.7 ... so I am providing the text of that proposition ... as follows:

Hope that helps ...

Peter

After a little reflection I have realized that $$f$$ is continuous at any point $$x$$ if and only if $$f(x-) = f(x+)$$ ... so if f is discontinuous at $$d$$ then $$f(d-) \neq f(d+)$$ ... but ... we have that $$f(d-) \leq f(d+)$$ ... so therefore $$f(d-) \lt f(d+)$$ ...Is that correct?

Peter

 

[soloenergy]

Dear Peter,

Thank you for your question. I can understand your confusion with Proposition 3.14 and its proof. Let me try to explain it in a more detailed manner.

Proposition 3.7 states that for any point d in the interval I, which is not an endpoint of I, the left and right limits of a function f exist and are bounded by the value of f at d. In other words, we can say that the function f is continuous at d.

Now, in Proposition 3.14, we are looking at a specific set D, which is defined as the set of points in the interval I where the left and right limits of f are not equal. In other words, at these points, the function is not continuous.

In the proof of Proposition 3.14, Browder is showing that if d is in D, then the left and right limits of f at d are not equal, which is the definition of D. On the other hand, if d is not in D, then the left and right limits of f at d are equal, and therefore the function is continuous at d.

Now, to answer your question, we need to understand why d is in D if and only if f(d-) < f(d+). This is because if d is in D, it means that the left and right limits of f at d are not equal, and therefore one is strictly less than the other. On the other hand, if f(d-) < f(d+), it means that the left and right limits of f at d are not equal, and therefore d is in D.

I hope this explanation helps you understand Proposition 3.14 better. If you have any further questions, please feel free to ask.