- #1

Eclair_de_XII

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- TL;DR Summary
- Let ##\{a_n\}\subset \mathbb{R}##. Let ##A_k\equiv \sup\{a_n:n\geq k\}## and suppose that ##A_k\rightarrow \lambda##. Prove:

(a) There is a sequence ##B_n## with the properties:

(i) ##B_n<A_n## for all ##n##

(ii) ##B_n## is an increasing sequence

(iii) ##B_n\rightarrow \lambda##

(b) There are infinitely many ##k## such that ##a_n\in [B_n,A_n]##

(c) There exists a subsequence ##\{a_{n_k}\}## that converges to ##\lambda##

(d) The limit of a subsequence ##\{a_{n_k}\}## is at most ##\lambda##

First, it must be justified that ##A_n## is decreasing and is bounded from below by the point it converges to. See the other topic

(1) ##A_n## is decreasing

By definition, ##A_n## is the supremum of the set containing all ##a_k## where ##k\geq n+1## and the set containing ##a_n##. Hence, it must be an upper bound for ##A_{n+1}##. By definition, ##A_n\geq A_{n+1}##.

(2) ##A_n\geq \lambda##

If not, then choose ##k## such that ##A_k<\lambda-\epsilon## for some ##\epsilon>0##. Since ##A_n## is decreasing, it follows that for ##n\geq k##, no matter how large the interval of convergence is, ##A_n## must lie outside of it. This contradicts the fact that it converges. Hence, ##A_n\geq \lambda##.

Note: This is just a rephrasing of Mr. Office Shredder's post from that topic I mentioned.

===(a)===

Define ##B=2\lambda-A_n##.

==(i)

This follows from the fact that ##A_n## is bounded from below by ##\lambda##

##A_n\geq \lambda##

##-A_n\leq -\lambda##

##2\lambda-A_n\leq 2\lambda-\lambda##

##2\lambda-A_n\leq \lambda##

##B_n\leq \lambda\leq A_n##

==(ii)

This follows from the fact that ##A_n## is decreasing.

##A_{n+1}\leq A_n##

##-A_{n+1}\geq -A_n##

##2\lambda-A_{n+1}\geq 2\lambda-A_n##

##B_{n+1}\geq B_n##

==(iii)

Finally, if ##A_n\rightarrow \lambda##, then for any ##\epsilon>0##, there is some integer ##N>0## such that if ##n\geq N##:

##|A_n-\lambda|<\epsilon##

##-\epsilon<\lambda-A_n<\epsilon##

##\lambda-\epsilon<2\lambda-A_n<\lambda+\epsilon##

##\lambda-\epsilon<B_n<\lambda+\epsilon##

##-\epsilon<B_n-\lambda<\epsilon##

##|B_n-\lambda|<\epsilon##

===(b)===

Let ##\epsilon>0## and let ##A_n## be fixed. By definition, ##A_n\geq a_k## whenever ##k\geq n##. Hence, there are infinitely many points of this form below ##A_n##. If we can show that ##a_k\geq B_n##, for ##k\geq n##, we will have our result.

Suppose that it is not. Then ##B_n## must be an upper bound for ##\{a_k:k\geq n\}##. Since ##B_n<A_n##, this contradicts the fact that ##A_n## is the least upper bound. Hence, ##a_k\geq B## for ##k\geq n##.

===(c)===

Let ##\epsilon>0##. Then there is ##N\in \mathbb{N}## such that if ##n\geq N##:

##-\epsilon+\lambda<A_n<\epsilon+\lambda##

Choose ##\epsilon_1\in (0,\epsilon)##. Then there is ##n_1\geq n## such that ##a_{n_1}\in (A_n-\epsilon_1,A_n]##. Now choose an ##\epsilon_2>0## smaller than ##\epsilon_1## such that ##a_{n_1}\notin (A_n-\epsilon_2,A_n]##. For example, choose ##\epsilon_2\in (0,A_n-a_{n_1})##. Then there must be some ##n_2\geq n## such that ##a_{n_2}\in (A_n-\epsilon_2,A_n]##.

With each step, we choose a point that is closer to ##A_n## than the point chosen before it. Hence, for ##n_k\geq n##:

##\lambda-\epsilon\geq A_n-\epsilon<A_n-\epsilon_k<a_{n_k}\leq A_n<\lambda+\epsilon##

##\lambda-\epsilon<a_{n_k}<\lambda+\epsilon##

##-\epsilon<a_{n_k}-\lambda<\epsilon##

##|a_{n_k}-\lambda|<\epsilon##

===(d)===

Let some subsequence ##a_{n_k}## converge to ##\gamma\in\mathbb{R}##. Suppose that ##\gamma>\lambda##. Choose ##m\in\mathbb{N}## such that ##\gamma>A_m>\lambda##. Set ##\epsilon=\gamma-A_m##. Let ##N\in\mathbb{N}##. Then whenever ##n\geq N##, specifically, when ##n\geq m##:

\begin{align*}

|\gamma-a_n|&=&\gamma-a_n\\

&\geq&\gamma-A_m\\

&=&\epsilon

\end{align*}

(1) ##A_n## is decreasing

By definition, ##A_n## is the supremum of the set containing all ##a_k## where ##k\geq n+1## and the set containing ##a_n##. Hence, it must be an upper bound for ##A_{n+1}##. By definition, ##A_n\geq A_{n+1}##.

(2) ##A_n\geq \lambda##

If not, then choose ##k## such that ##A_k<\lambda-\epsilon## for some ##\epsilon>0##. Since ##A_n## is decreasing, it follows that for ##n\geq k##, no matter how large the interval of convergence is, ##A_n## must lie outside of it. This contradicts the fact that it converges. Hence, ##A_n\geq \lambda##.

Note: This is just a rephrasing of Mr. Office Shredder's post from that topic I mentioned.

===(a)===

Define ##B=2\lambda-A_n##.

==(i)

This follows from the fact that ##A_n## is bounded from below by ##\lambda##

##A_n\geq \lambda##

##-A_n\leq -\lambda##

##2\lambda-A_n\leq 2\lambda-\lambda##

##2\lambda-A_n\leq \lambda##

##B_n\leq \lambda\leq A_n##

==(ii)

This follows from the fact that ##A_n## is decreasing.

##A_{n+1}\leq A_n##

##-A_{n+1}\geq -A_n##

##2\lambda-A_{n+1}\geq 2\lambda-A_n##

##B_{n+1}\geq B_n##

==(iii)

Finally, if ##A_n\rightarrow \lambda##, then for any ##\epsilon>0##, there is some integer ##N>0## such that if ##n\geq N##:

##|A_n-\lambda|<\epsilon##

##-\epsilon<\lambda-A_n<\epsilon##

##\lambda-\epsilon<2\lambda-A_n<\lambda+\epsilon##

##\lambda-\epsilon<B_n<\lambda+\epsilon##

##-\epsilon<B_n-\lambda<\epsilon##

##|B_n-\lambda|<\epsilon##

===(b)===

Let ##\epsilon>0## and let ##A_n## be fixed. By definition, ##A_n\geq a_k## whenever ##k\geq n##. Hence, there are infinitely many points of this form below ##A_n##. If we can show that ##a_k\geq B_n##, for ##k\geq n##, we will have our result.

Suppose that it is not. Then ##B_n## must be an upper bound for ##\{a_k:k\geq n\}##. Since ##B_n<A_n##, this contradicts the fact that ##A_n## is the least upper bound. Hence, ##a_k\geq B## for ##k\geq n##.

===(c)===

Let ##\epsilon>0##. Then there is ##N\in \mathbb{N}## such that if ##n\geq N##:

##-\epsilon+\lambda<A_n<\epsilon+\lambda##

Choose ##\epsilon_1\in (0,\epsilon)##. Then there is ##n_1\geq n## such that ##a_{n_1}\in (A_n-\epsilon_1,A_n]##. Now choose an ##\epsilon_2>0## smaller than ##\epsilon_1## such that ##a_{n_1}\notin (A_n-\epsilon_2,A_n]##. For example, choose ##\epsilon_2\in (0,A_n-a_{n_1})##. Then there must be some ##n_2\geq n## such that ##a_{n_2}\in (A_n-\epsilon_2,A_n]##.

With each step, we choose a point that is closer to ##A_n## than the point chosen before it. Hence, for ##n_k\geq n##:

##\lambda-\epsilon\geq A_n-\epsilon<A_n-\epsilon_k<a_{n_k}\leq A_n<\lambda+\epsilon##

##\lambda-\epsilon<a_{n_k}<\lambda+\epsilon##

##-\epsilon<a_{n_k}-\lambda<\epsilon##

##|a_{n_k}-\lambda|<\epsilon##

===(d)===

Let some subsequence ##a_{n_k}## converge to ##\gamma\in\mathbb{R}##. Suppose that ##\gamma>\lambda##. Choose ##m\in\mathbb{N}## such that ##\gamma>A_m>\lambda##. Set ##\epsilon=\gamma-A_m##. Let ##N\in\mathbb{N}##. Then whenever ##n\geq N##, specifically, when ##n\geq m##:

\begin{align*}

|\gamma-a_n|&=&\gamma-a_n\\

&\geq&\gamma-A_m\\

&=&\epsilon

\end{align*}

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