Continuous Functions and Open Sets .... D&K Example 1.3.8 ....

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

I need help with an aspect of Example 1.3.8 ...

Duistermaat and Kolk"s Example 1.3.8 reads as follows:View attachment 7697In the above example we read the following:

"... ... Then \(\displaystyle f( \mathbb{R} ) = [ -1, 1]\) where \(\displaystyle \mathbb{R}\) is open ... ... "My question is as follows:

Can someone please show, formally and rigorously, that \(\displaystyle f( \mathbb{R} ) = [ -1, 1]\) ... ... ?Help will be much appreciated ... ...

Peter

 

[Greg]

Can someone please show, formally and rigorously, that \(\displaystyle f( \mathbb{R} ) = [ -1, 1]\) ... ... ?

I believe you can arrive at that by elementary calculus, i.e., by taking the derivative of $f$, setting it equal to 0 and solving for $x$ to show that $f(x)$ has a minimum of $-1$, a maximum of $1$ and since $f$ is smooth and continuous over all of $\mathbb{R}$, these values are the closed boundaries of the image of $f$.

 

[S.G. Janssens]

... and since $f$ is smooth and continuous over all of $\mathbb{R}$, these values are the closed boundaries of the image of $f$.

A little more explicitly: Once you have established that $f$ assumes its global minimum $-1$ at, say, $a$ and its global maximum $+1$ at $b > a$, you could refer to the Intermediate Value Theorem to conclude that $f$ assumes on $[a, b]$ all values in $[-1,1]$.