MHB Metric Spaces & Compactness - Apostol Theorem 4.28

[Math Amateur]

I need help with the proof of Theorem 4.28 in Tom Apostol's book: Mathematical Analysis (2nd Edition).

Theorem 4.28 reads as follows:View attachment 3855In the proof of the above theorem, Apostol writes:

" ... ... Let $$m = \text{ inf } f(X)$$. Then $$m$$ is adherent to $$f(X)$$ ... ... "

Can someone please explain to me exactly why $$m = \text{ inf } f(X)$$ implies that $$m$$ is adherent to $$f(X)$$?

Help will be appreciated ... ...NOTE: In the above proof Apostol makes mention of an adherent point, so I am providing Apostol's definition of an adherent point together with (for good measure) his definition of an accumulation point ... ...View attachment 3856

 

[Fallen Angel]

Hi Peter,

$m=inf \ f(X)$ implies that there exists a sequence $\{x_{n}\}_{n\in \Bbb{N}}\subseteq X$ such that $\underset{n \to +\infty}{lim}f(x_{n})=m$,
so $m$ is in the closure of $f(X)$

 

[Math Amateur]

Hi Peter,

$m=inf \ f(X)$ implies that there exists a sequence $\{x_{n}\}_{n\in \Bbb{N}}\subseteq X$ such that $\underset{n \to +\infty}{lim}f(x_{n})=m$,
so $m$ is in the closure of $f(X)$

Thanks for the help Fallen Angel ...

I guess for the case of finite sets such as $$A \subset \mathbb{R}$$ where $$A = \{1,2,3 \}$$ and $$inf A = 1$$, the sequence $$\{x_n \}$$ would be $$1,1,1, \ ... \ ... \ $$

Is that correct?

Peter

 

[Fallen Angel]

Hi Peter,

Yes, it's correct, but this sequences are not unique, $\{2,2,3,2,3,1,1,1,1,\ldots\}$ it's another one, for example.