How Does Theorem 4.29 Illustrate Continuity Issues in Inverse Functions?

[Math Amateur]

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

I am focused on Chapter 4: Limits and Continuity ... ...

I need help in order to fully understand the example given after Theorem 4.29 ... ... Theorem 4.29 (including its proof) and the following example read as follows:

View attachment 9237
View attachment 9238
In the Example above we read the following:

" ... ... However, \(\displaystyle f^{ -1 }\) is not continuous at the point \(\displaystyle f(0)\). For example, if \(\displaystyle x_n = 1 - 1/n\), the sequence \(\displaystyle \{ f(x_n) \}\) converges to \(\displaystyle f(0)\) but \(\displaystyle \{ x_n \}\) does not converge in \(\displaystyle S\). ... ... "My question is as follows:

Can someone please explain exactly how/why ... the sequence \(\displaystyle \{ f(x_n) \}\) converges to \(\displaystyle f(0)\) but \(\displaystyle \{ x_n \}\) does not converge in \(\displaystyle S\) ... ... implies that \(\displaystyle f^{ -1 }\) is not continuous at the point \(\displaystyle f(0)\) ... ... ?

-----------------------------------------------------------------------------------------------------------------------------------------------My thoughts ...

I think that the relevant theorem regarding answering my question is Apostol, Theorem 4.16 which reads as follows:View attachment 9239

If we let \(\displaystyle t_n \in f(S)\) be such that \(\displaystyle t_n = f(x_n)\) ... so the sequence \(\displaystyle \{ t_n \} = \{ f(x_n) \}\) is in the domain of \(\displaystyle f^{ -1 }\) ...

Then ... sequence \(\displaystyle \{ t_n \} = \[ f(x_n) \}\) converges to \(\displaystyle f(0) = t_0\) say ...

Then following Theorem 4.16 above ... for \(\displaystyle f^{ -1 }\) to be continuous we need \(\displaystyle \{ f^{ -1 } (t_n) \} = \{ f^{ -1 } ( f(x_n) ) \} = \{ x_n \}\) to converge in \(\displaystyle S\) ... but it does not do so ...

(mind you ... I'm not sure how to prove it doesn't converge in \(\displaystyle S\) ...)

Is that correct?

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Hope that someone can help ...

Peter

 

[castor28]

Hi Peter,

Your idea is correct. Now, the sequence $\{x_n\}$ converges to $1$ in $\mathbb{R}$, but, as $S$ does not contain $1$, that sequence does not converge in $S$.

Intuitively, what happens is that, when a point $P$ travels counterclockwise in circles in $f(S)$, the image $f^{-1}(P)$ jumps discontinuously from $(1-\varepsilon)$ to $0$ in $S$ whenever $P$ passes through $f(0)=1$.

 

[Math Amateur]

Hi Peter,

Your idea is correct. Now, the sequence $\{x_n\}$ converges to $1$ in $\mathbb{R}$, but, as $S$ does not contain $1$, that sequence does not converge in $S$.

Intuitively, what happens is that, when a point $P$ travels counterclockwise in circles in $f(S)$, the image $f^{-1}(P)$ jumps discontinuously from $(1-\varepsilon)$ to $0$ in $S$ whenever $P$ passes through $f(0)=1$.

Thanks castor28

Appreciate your help ...

Peter