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

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

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 \(\displaystyle A \subset \mathbb{R}\) where \(\displaystyle A = \{1,2,3 \}\) and \(\displaystyle inf A = 1\), the sequence \(\displaystyle \{x_n \}\) would be \(\displaystyle 1,1,1, \ ... \ ... \ \)

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.

 

[blue_raver22]

An adherent point of a set A is a point x such that every neighborhood of x contains at least one point of A.

An accumulation point of a set A is a point x such that every neighborhood of x contains infinitely many points of A.

To understand why m = inf f(X) implies that m is adherent to f(X), we first need to understand what inf f(X) represents. In this case, inf f(X) represents the infimum (greatest lower bound) of the set of values of f(X). This means that m is the smallest value that f(X) can take on.

Now, let's consider what it means for m to be adherent to f(X). This means that every neighborhood of m contains at least one point of f(X).

Since m is the smallest value that f(X) can take on, this means that every neighborhood of m contains at least one point of f(X). In other words, every neighborhood of m contains a value of f(X) that is smaller than or equal to m.

But according to the definition of infimum, m is the smallest value that f(X) can take on. This means that there are no values of f(X) that are smaller than m. Therefore, every neighborhood of m must contain at least one point of f(X) that is equal to m.

Thus, m = inf f(X) implies that m is adherent to f(X) because every neighborhood of m contains at least one point of f(X), and there are no values of f(X) that are smaller than m.