# Proving Equivalence between conditions on sequences

• Feb 26th 2011, 07:49 PM
dfanforlife
Proving Equivalence between conditions on sequences
Let E be a set of real numbers. Then p is an accumulation point of E iff the following condition holds:

there is a sequence (x_n) of members of E, each different from p, such that (xn) converges to p.

I have to prove that the above condition is equivalent to:
there is a sequence (x_n) of DISTINCT members of E converging to p.

At first I thought I just had to show the second condition with E-{p}. I'm not sure how to prove equivalence between these two conditions. If someone could help me out that'd be great!
• Feb 26th 2011, 08:35 PM
Tinyboss
It is clear that the second condition implies the first, right? So, assuming the first holds, construct a sequence satisfying the second. Suppose (x_n) converges to p, and each x_n is not equal to p. So let y_1 = x_1, and since x_n converges to p, there is some x_k with k>1 and |x_k-p| < |x_1-p|/2. So let y_2 = x_k. Then just repeat the process to get a subsequence of distinct elements that converge to p.
• Feb 27th 2011, 10:25 AM
dfanforlife
awesome! I wasn't sure if I could just say that the condition 2 implies condition 1. However, what if I now wanted to prove that condtion (2) (there is a sequence (x_n) of DISTINCT members of E converging to p) is equivalent to the following new condition:

there is a strictly monotone sequence (xn) of members of E converging to p.

Now, I know that a strictly monotone sequence is a sequence in which the members are distinct. so does that mean that I can say that Condtion (we will call it 3) (3) implies condition (2)? And then how do I prove it going the other way?
• Feb 27th 2011, 10:57 AM
DrSteve
monotone means increasing or decreasing. So strictly monotone means strictly increasing or strictly decreasing (the strictly means that the elements are distinct). Thus one direction is trivial. For the other direction, assume there is no strictly decreasing subsequence. Then you should be able to construct a strictly increasing subsequence.
• Feb 27th 2011, 11:50 AM
dfanforlife
Quote:

Originally Posted by DrSteve
monotone means increasing or decreasing. So strictly monotone means strictly increasing or strictly decreasing (the strictly means that the elements are distinct). Thus one direction is trivial. For the other direction, assume there is no strictly decreasing subsequence. Then you should be able to construct a strictly increasing subsequence.

Right because we have to show it converging. However, I'm just confused when you said the one direction is trival (which i thought too until i read the rest of the post) So just to clarify is it showing condition (3) equal to condtion (2) or just the fact of an strictly increasing sequence?
• Feb 27th 2011, 12:50 PM
DrSteve
Yes - (3) implies (2) I consider trivial. Here is the argument:

If there is a strictly monotone sequence of members of E converging to p, then the elements of the sequence are distinct (because of "strictly"). Thus there is a sequence of distinct elements of E converging to p.
• Feb 28th 2011, 08:02 AM
dfanforlife
Quote:

Originally Posted by DrSteve
monotone means increasing or decreasing. So strictly monotone means strictly increasing or strictly decreasing (the strictly means that the elements are distinct). Thus one direction is trivial. For the other direction, assume there is no strictly decreasing subsequence. Then you should be able to construct a strictly increasing subsequence.

Quick question for this part. To show the increasing subsequence wouldn't that just be stating that's its strictly montone? I'm not sure how I would show this.
• Feb 28th 2011, 08:40 AM
DrSteve
I'm not sure that I understand your question. My suggestion is to assume there is no strictly decreasing subsequence, and to use this fact to produce a strictly increasing one. Here is a hint to get you started:

Since the sequence has no strictly decreasing subsequence, the sequence has a least element \$\displaystyle x_n_1\$. Now look at the sequence \$\displaystyle \{x_k|k>n_1\}\$ This sequence also has a least element,...etc.
• Feb 28th 2011, 08:51 AM
dfanforlife
Quote:

Originally Posted by DrSteve
I'm not sure that I understand your question. My suggestion is to assume there is no strictly decreasing subsequence, and to use this fact to produce a strictly increasing one. Here is a hint to get you started:

Since the sequence has no strictly decreasing subsequence, the sequence has a least element \$\displaystyle x_n_1\$. Now look at the sequence \$\displaystyle \{x_k|k>n_1\}\$ This sequence also has a least element,...etc.

I figured it would be best to copy and paste my original question to clear up confusion.
prove that condtion (2) (there is a sequence (x_n) of DISTINCT members of E converging to p) is equivalent to the following new condition:

there is a strictly monotone sequence (xn) of members of E converging to p.

When you say:Now look at the sequence \$\displaystyle \{x_k|k>n_1\}\$ This sequence also has a least element,...etc.[/QUOTE]
Am I just repeating that process?
Sorry for all the edits my computer is giving me problems
• Feb 28th 2011, 09:16 AM
DrSteve
Quote:

Originally Posted by dfanforlife
I figured it would be best to copy and paste my original question to clear up confusion.
prove that condtion (2) (there is a sequence (x_n) of DISTINCT members of E converging to p) is equivalent to the following new condition:

there is a strictly monotone sequence (xn) of members of E converging to p.

When you say:Now look at the sequence \$\displaystyle \{x_k|k>n_1\}\$ This sequence also has a least element,...etc.

Am I just repeating that process?
Sorry for all the edits my computer is giving me problems[/QUOTE]

Yes - you just keep repeating this process forming a subsequence \$\displaystyle \{x_{n_k}\}\$ of the original sequence which is strictly increasing. This subsequence is strictly monotone and converges to p.