Stuck at proving a bounded above Subsequence

[CGandC]

Summary:: x

Let $\{ a_{n} \}$ be a sequence.
Prove: If for all $N \in { \bf{N} }$ there exists $n> N$ such that $a_{n} \leq L$ , then there exists a subsequence $\{ a_{n_{k}} \}$ such that $a_{n_{k}} \leq L$

My attempt:
Suppose that for all $N \in {\bf{N}}$ there exists $n> N$, such that $a_{n} \leq L$. Since $\{ a_{n} \}$ is a sequence, there exists a subsequence $\{ a_{n_{k}} \}$ such that there exists a monotonically increasing sequence of natural numbers $\{ n_{k} \}$ such that $\forall k \in {\bf{N}}, n_{k} \geq k$ . Let $k \in {\bf{N}}$ be arbitrary so there exists $n_{k} \geq k$. Thus ( by universal instantiation of $n_{k}$ in the statement " for all $N \in {\bf{N}}$ there ... " ) there exists $n > n_{k} > k$ such that $a_{n} \leq L$

However ,I don't know if $a_{n_{k}}$ is less than or greater than $a_{n}$ . So I'm stuck, how am I supposed to proceed from here to show that $a_{n_{k}} \leq L$ ?

 

[mfb]

Thus ( by universal instantiation of $n_{k}$ in the statement " for all $N \in {\bf{N}}$ there ... " ) there exists $n > n_{k} > k$ such that $a_{n} \leq L$

That is correct but you are not finding the right sequence here.

However ,I don't know if $a_{n_{k}}$ is less than or greater than $a_{n}$

Why would that matter?

So I'm stuck, how am I supposed to proceed from here to show that $a_{n_{k}} \leq L$ ?

You can't because it won't be true for any n_k you come up with. You need to define n_k based on the sequence a. The best approach is to do this one element at a time. Let's say a_1, a_6, a_245 are three elements smaller than L. Can you find a fourth one? Once you have that, can you find a fifth one? Generalize and you get a full sequence.

 

[CGandC]

Maybe the formal generalization goes as follows :

Let there be arbitrary $N' \in { \bf{N} }$, so there exists $n > N'$ such that $a_{n} \leq L$.
We'll choose $k' \in { \bf{N} }$ so that there exists $n_{k'} \in { \bf{N} }$ ( " there exists $n_{k'} \in { \bf{N} }$ " because since $k'$ is a specific value and not arbitrary, then $n_{k'}$ is also a specific value and not arbitrary ) such that $n_{k'} = n$. So it immediately follows that $a_{n} = a_{n_{k'}} \leq L$

Is this right?

 

[Office_Shredder]

You spend a lot of , big words justifying the existence of $n_{k'}$ that both doesn't seem necessary to me and sounds very confusing.I would just say something like, given $n_1,...,n_k$ in the subsequence, we know there is some $n>n_k$ such that $a_n<L$. Let $n_{k+1}=n$. And do this repeatedly to form the sequence.

 

[CGandC]

So I can say the following?:
We know there is some $n>n_k$ such that $a_n < L$ . Let $n_{k+1} = n$ .
We know there is some $n>n_{k+1}$ such that $a_n < L$ . Let $n_{k+2} = n$ .
We know there is some $n>n_{k+2}$ such that $a_n < L$ . Let $n_{k+3} = n$ .
.
.
.
...

So that the new subsequence we form is $( a_{n_k+1} , a_{n_{k+2}} , a_{n_{k+3}} , ... )$ ?

 

[Office_Shredder]

The point of what I wrote is that I assumed we already had the first k elements of the sequence which you dropped from your final list, but your sequence also works. I would consider what you wrote a proof.

It's possible if you're taking a class they will require you to write something more formal, but

1.) Most of the mathematical community doesn't actually speak this way talking about universal instantiation etc.
2.) I think it's mostly a blocker to understanding the important point.

If you think what you wrote is not acceptable for a person grading this and you're not sure how to turn it into something acceptable we can discuss it.

 

[CGandC]

I think my difficulty comes from trying to apply the definition of a subsequence ( from the book Introduction to Real Analysis . Robert G. Bartle , Donald R. Sherbert ):

Does this definition tell me:
1. If I have a sequence $( x_n )$ and an increasing sequence of natural numbers $( n_k )$ , then there exists a new sequence $X' = ( x_{n_{k}} ) = ( x_{n_{1}} , x_{n_{2}}, ... , x_{n_{k}} , ... )$

Or

2. for all sequence $( x_n )$ and for all increasing sequence of natural numbers $( n_k )$ , then there exists a new sequence $X' = ( x_{n_{k}} ) = ( x_{n_{1}} , x_{n_{2}}, ... , x_{n_{k}} , ... )$

Or

3. for all sequence $( x_n )$ ,there exists an increasing sequence of natural numbers $( n_k )$ , then there exists a new sequence $X' = ( x_{n_{k}} ) = ( x_{n_{1}} , x_{n_{2}}, ... , x_{n_{k}} , ... )$

Which is correct ,and basically what is the logical format of the above definition?
I'm used to being able to use definitions and proven statements which start with quantifiers like " there exists " and " for all ". But in the above definition I have " Let ", so I feel like I don't exactly know how to use the definition ( not the first time I encountered such definition ).

 

[mfb]

1 and 2 are both correct.
X' depends on the sequence n_k of course.

You are overthinking this.

Here is a sequence:
$x_1, x_2, x_3, x_4, x_5, x_6, ...$
Here is a subsequence I chose:
$x_2, x_4, x_8, x_{16}, ...$, here n_k = 2,4,8,16,...
Here is another subsequence:
$x_1, x_3, x_5, x_7, ...$, here n_k = 1,3,5,7,...

n_k is just a list telling you which elements to use from the original sequence.

 

[Delta2]

Basically your problem in the OP is to give a proper and possibly recursive definition of the sequence $n_k$ and then prove that $n_k$ is strictly increasing and also prove that $a_{n_k}\leq L$. If you define properly the sequence $n_k$ then those two proofs will be immediate consequence of the proper definition.