Difficulty with accumulations points

[sitia]

Homework Statement

Hi guys,

I'm having real difficulty with understanding accumulation points. I don' really know why that is since others seem to understand the concept fine but I'm very lost.
For example, I'm practicing some questions and one of the is :
If S is the set of rational numbers with 1<x<2, then is √2 is an accumulation point?

I am completely lost in how to go about figuring this out.
Examples help, so if you have any good examples that could make this concept a little clearer, I would truly appreciate it.

Thanks!

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

What, exactly, is your understanding of the definition of "accumulation point"? One that is commonly used is "p is an accumulation point of set A if and only if there exist a sequence of points in A (not including p) that converges to p. Another is that every neighborhood of p contains at least one point of A (other than p).

\sqrt{2}= 1.41421..., right? So given any \delta&gt; 0, there exist a power of 10 such that 10^n&lt; \delta. Cutting that number off after n decimal places gives a rational number closer to \sqrt{2} than \delta.

More generally, given any real number, there exist a sequence of rational numbers converging to it.

 

[sitia]

I struggle with how/what process I need to follow to find accumulation pts of a set or determining of a given value is an accumulation pt.

 

[HallsofIvy]

Do you know what the definition of "accumulation point" is? It sounds like you are saying you don't.

 

[sitia]

You are right, I have the definition but that's when I'm struggling...truly understanding it and applying it.

 

[haruspex]

Does it help to put it like this: p is an accumulation point of S if you can get arbitrarily close to it by picking points of S-{p}.
There's any number of sequences of rationals in [1,2] that converge to sqrt(2). Halls gave you a very easy and obvious one. Another is to start with x = 1 and generate a sequence of rationals by iterating x' = 1/(1+x) + 1. (You can get that formula by writing y² - 1 = 1, so (y-1) = 1/(y+1).)