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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 $\infty$.

Thanks.

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The continuous image of a compact set is compact. –  Stefan Smith May 17 '12 at 22:55
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2 Answers

Your analysis is fine.

You have constructed a sequence so that $g(x_{n_k}) \rightarrow \infty$, which contradicts the fact that $g(x_{n_k}) \rightarrow g(y)$. It is a contradiction because a sequence cannot converge to two limits (or diverge properly to $\infty$ and converge to a real number at the same time).

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Because $g(x_{n_k}) \to g(y)$, there is some $K$ such that if $k > K$, then $|g(x_{n_k}) - g(y)| < 1$, or equivalently, if $k > K$ one has $$g(y) - 1 < g(x_{n_k}) < g(y) + 1$$ But you also know that $g(x_{n_k}) > n_k \geq k$. So if $k > K$ and also $k > g(y) + 1$ then you get a contradiction; on the one hand since $k > K$ you have $g(x_{n_k}) < g(y) + 1$, and on the other hand since $k > g(y) + 1$ you have $g(x_{n_k}) > k > g(y) + 1$.

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