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Let $(a_n)$ be bounded and $b > 0$ s.t. for all $n \in \mathbb N, \ |a_n| < b.$ Then $b \in \mathbb R$ with $\{n \in \mathbb N: a_n \le b\} = \mathbb N$ and $\{n \in \mathbb N: a_n \le -b\} = \varnothing.$ Thus $b \in X = \{x \in \mathbb R: \{n \in \mathbb N: a_n \le x\} \text{ is infinite}\}$. Because $X$ is bounded below by $-b, \ L= \inf X$ exists.

Now for all $k \in \mathbb N, \ H_k^+ = \displaystyle{\left\{n \in \mathbb N: a_n \le L + \frac 1k\right\}}$ is infinite and $H_k^- = \displaystyle{\left\{n \in \mathbb N: a_n \le L - \frac 1k\right\}}$ is finite so that $T_k = H_k^+ \setminus H_k^-$ infinite for all $k \in \mathbb N$.

Because $T_1$ is infinite, it's not empty and so let $n_1$ be in $T_1$. Assume $n_k$ is chosen. Then exists $n_{k+1} \in T_{k+1}$ with $n_{k+1} > n_k$. Otherwise, $n_k$ would be the largest number in $T_{k+1}$ making it finite. Thus $(a_{n_k})$ exists and for all $k \in \mathbb N$, we have $\displaystyle{|a_{n_k} - L| \le \frac 1k}$ as $n_k \in T_k$. Since $\displaystyle{\frac 1k \to 0, \ |a_{n_k} - L| \to 0 \text{ and so }\lim_{k \to \infty}a_{n_k} = L.}$

My question:

Why is $H_k^+$ is infinite and $H_k^-$ is finite?

I think $H_k^-$ contains all naturals $m$ that satisfy $\displaystyle{a_m = a_n + L \le -\frac 1k}$ which means $H_k^- = \varnothing$ implying $H_k^-$ is finite and the elements $p$ in $H_k^+$ satisfy $\displaystyle{a_p = a_n - L \le \frac 1k}$ meaning $H_k^+ = \mathbb N$ implying $H_k^+$ is infinite. Does that make sense? If not, what makes $H_k^-$ finite and $H_k^+$ infinite? Thanks.

user1008945
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1 Answers1

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Your explanation doesn't seem quite right to me because when you write $a_m=a_n+L\leq -\frac{1}{k}$ the variable $n$ is undefined. Do you mean that there exists $n$ such that this inequality is true? Or that it is true for all $n$? Or something different?

The way to justify that $H_k^+$ is infinite is to note that $L+\frac{1}{k}>L$ for every $k$, and since $L$ is the infimum of $X$ there is some $x_0\in X$ such that $x_0<L+\frac{1}{k}$ (otherwise, $L+\frac{1}{k}$ would be a lower bound of $X$ which is bigger than $L$). By definition of $X$ this means that $\{n\in\mathbb{N}:a_n\leq x_0\}$ is infinite, and since $x_0<L+\frac{1}{k}$ this means that $\{n\in\mathbb{N}:a_n\leq L+\frac{1}{k}\}$ is infinite, and this is just $H_k^+$. It should go similarly for $H_k^-$.

(I have to add that it seems to me that the notation used in this proof is likely to be quite confusing for someone who is supposed to be seeing a proof of Bolzano-Weierstrass for the first time.)

Snaw
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