Question On the Proof of The boundedness Theorem

let $f:[a,b]\rightarrow\mathbb{R}$, f continuous on $[a,b]$. I shall prove that $\exists A,B\in\mathbb{R}, \forall x\in[a,b], A\le f(x)\le B$.

Proof: Let's define $g(x)=|f(x)|$, we need to prove now that $\exists A\in\mathbb{R}, \forall x\in[a,b], g(x)\le A$. let's suppose that this claim is false, therefore we get:

$\forall A, \exists x\in[a,b], g(x)> A$.

$ \forall n\in\mathbb{N}$ we define $A=n$ therefore $\exists x_n\in[a,b], g(x_n)>n$.

We know that $a\le x_n \le b$ and from B.W. we know that $\exists x_{n_k}$ that converges to $y\in\mathbb{R}$. furthermore, we also know that $a\le y \le b$.

$g(x_n)>n$ therefore $g(x_{n_k})\ge n_k$

Because $g$ is continuous on $[a,b]$ w know that $\lim_{k\to \infty}g(x_{n_k})=g(y)\mathbb\in{R}$.

We also know that $\lim_{k\to \infty}n_k=\infty$.

Up to this point I understand each sub-claim in the proof. However, at this point in the proof it's deduced that we can compare the limits and say that because the first sequence converges to a number in $\mathbb{R}$ and the second converges to $\infty$ and there is a $ \ge $ sign between them there is a contradiction.

Could you please make this comparison more formal and explain why I can deduce this? To be more clear: Why when given two sequences that converge I can apply $\ge$ on their limits and what does it mean when a sequence that converges to a number in $\mathbb{R}$ is greater or small than $\infty$.


2022-07-25 20:47:02
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