Corollary to Bolzano weierstrass theorem

Question

Let ($a_n$) be a sequence in $R$. If ($a_n$) is bounded above and $a_n \nrightarrow -\infty$, then ($a_n$) has a convergent subsequence.

Proof: The statement $a_n \nrightarrow -\infty$ means there is $\beta \in R$ such that for every $n_0 \in N$, there is $n \in N$ with $n>n_0$ and $a_n \geq \beta$. Hence there are $n_1<n_2<...$ In $N$ such that $a_{n_k} \geq \beta$ for each $k \in N$. The subsequence ($a_{n_k}$) in $R$ is thus bounded above as well as bounded below. So by Bolzano weierstrass theorem, ($a_{n_k}$) has a convergent subsequence. Finally, we now that a subsequence of ($a_{n_k}$) is a subsequence of ($a_n$) itself.

Here I'm not able to understand the first sentence. Is the $\beta$ fixed or changes for each $n_k$? That is for given $n_0$ there exist $\beta \in R$ such that $n>n_0$ and $a_n \geq \beta$ for some $n$. Or is the $\beta$ fixed? Could some explain precisely?

Answer 1

What does $a_n \rightarrow -\infty$ ? It means:

for every $ \beta \in \mathbb R$ there is $n_0=n_0(\beta) \in \mathbb N$ such that

$a_n < \beta$ for all $n>n_0$.

Hence, if $a_n \nrightarrow -\infty$, then there is $ \beta \in \mathbb R$ such that for every $n_0$ there is $n \in \mathbb N$ with $n>n_0$ and $a_n \ge \beta$

Answer 2

$\beta$ is fixed but for each $n_0$ $\beta$ fixes or find some $n>n_0$ such that $a_n>\beta$. This is because if $a_n\rightarrow -\infty$ you could arbitrary get close to $-\infty$ since some time, so as it doesn't, you find a precision such that you can't find some time since all terms are nearer that this precision.

Answer 3

This sentence is the negation of the statement that $a_n\to -\infty$. If $a_n\to -\infty$, then for every $\beta\in\mathbb R$, there exists $n_0\in\mathbb N$ such that for all $n>n_0$ we have $a_n<\beta$. But if $a_n$ does not converge to $-\infty$, the converse should be true, which is spelled out in your proof. So yes, $\beta$ is fixed.

Answer 4

$\beta$ there is fixed. Think of it this way: $a_n\to\infty$ means that $\forall M\in\Bbb{R},\exists N,\forall n\geq N: a_n<M$, so that given any number $M$ all but finitely many terms of $\{a_n\}_n$ will be smaller than $M$. Then $a_n\not\to -\infty$ means $\exists M_0,\forall N,\exists n\geq N: M\leq a_n$, so that there is some number $M_0$ ($\beta$ in your case) that we are guaranteed the existence of infinitely many terms of the sequence not less than $M_0$.

For example you can compare $\{0,-1,-2,...,-n,...\}_n$ and $\{0,1,-1,1,-2,1,...,1,-n,1,...\}$.

Answer 5

If for all $\beta\in\mathbb{R}$, there exists an $n_0\in \mathbb{N}$ such that $a_n< \beta$ for all $n>n_0$, then $a_n\rightarrow -\infty$. The contrapositive of this implication is the statement of the first line in the proof.

Namely: if $a_n\not\rightarrow -\infty$, then it is not the case that for all $\beta\in\mathbb{R}$, there exists an $n_0\in \mathbb{N}$ such that $a_n< \beta$ for all $n>n_0$.

This is equivalent to the statement: if $a_n\not\rightarrow -\infty$, then for at least one (fixed) $\beta\in\mathbb{R}$ and all $n_0\in \mathbb{N}$, $a_n\geq \beta$ For at least one $n>n_0$.

Finally, this last statement is clearly equivalent to e onerosa in the given proof. In other words, the first sentence in the proof is the contrapositive of the fact that $a_n\rightarrow -\infty$ if for all $\beta\in\mathbb{R}$, there exists an $n_0\in \mathbb{N}$ such that $a_n< \beta$ for all $n>n_0$.