Find accumulation points (real analysis)

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  • #1
quasar987
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We must find the accumulation point for the set

[tex]E = \left\{ \frac{n^2 + 3n + 5}{n^2 + 2} \vert n \in \mathbb{N} \right\} [/tex]

Now that is easy, we first rearange the terms so we see what happens in this mess when n varies. I did the following thing..

[tex]E = \left\{ \frac{n^2 + 2 + 3n + 3}{n^2 + 2} | n \in \mathbb{N} \right\} [/tex]

[tex]E = \left\{ \frac{n^2 + 2}{n^2 + 2} + \frac{3n + 3}{n^2 + 2} | n \in \mathbb{N} \right\} [/tex]

[tex]E = \left\{ 1 + \frac{3n}{n^2 + 2} + \frac {3}{n^2 + 2} | n \in \mathbb{N} \right\} [/tex]

[tex]E = \left\{ 1 + \frac{3}{n + \frac{2}{n}} + \frac {3}{n^2 + 2} | n \in \mathbb{N} \right\} [/tex]

Now it is clear that 1 is the only accumulation point and it happens when n is arbritrarily humongeous. But we must prove it! Usually, for sets that look like

[tex]E = \left\{ \frac{1}{n} | n \in \mathbb{N} \right\} [/tex]
or
[tex]E = \left\{ \frac{1}{2^n} | n \in \mathbb{N} \right\} [/tex],

we can prove by use of the Archimedean property (given [tex]\delta[/tex] element of real such that [tex]\delta >0[/tex], there exist an n element of the positive intergers such that for any y element of real, [tex]n \delta > y[/tex]), that

[tex] \forall \delta>0, V'(1,\delta) \cap E \neq \emptyset [/tex] (the definition of 1 being an accumulation point)

Now I can't see how we can use the Archimedean property here. Anyone sees a way to do this?

Thanks for your inputs!
 

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  • #2
Hurkyl
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Well, let's see. One thing you would like to do is to show that [itex]3/(n^2+2) < \epsilon[/itex] for some [itex]\epsilon[/itex] that you have chosen, right?

Well, that's equivalent to showing that [itex]n^2 > (3 / \epsilon) - 2[/itex], right? Any ideas on using the archmedian property to show this?
 
  • #3
quasar987
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So you're saying that it would be okay, in order to prove that E has 1 as an accumulation point, to show that the terms,

[tex]\frac{3}{n + \frac{2}{n}} [/tex]

and

[tex]\frac {3}{n^2 + 2}[/tex]

have an accumulation point at 0. It crossed my mind but I didn't have any justification for it.


To answer your question: How about this way? We chose y to be [itex]3 - 2\delta[/itex] and since [itex]\delta>0[/itex], there exist [itex]n[/itex] such that

[tex]\delta n>3 - 2\delta[/tex]

divide both side by [itex]\delta[/itex] and argue that if there exist such a n, then [itex]n^2[/itex] satisfy the inequality just has much

[tex]n^2 > \frac{3}{\delta} - 2[/tex]

And now we can go back in time and find what we wanted, that is, [itex]\forall\delta>0[/itex],

[tex]\delta > \frac{3}{n^2 + 2} > 0[/tex]

Awesome! Ok, for [tex]\frac {3}{n + \frac{2}{n}}[/tex] now. Using your trick, we see that this is the same as showing that there exists n such that

[tex]n + \frac{2}{n} > \frac{3}{\delta}[/tex]

and that this is the same as showing that there exist n such that

[tex]n > \frac{3}{\delta}[/tex]

because if this is true, then for any, n, since [itex]n + \frac{2}{n}>n[/itex], our inequality is true too.

But proving that there exist n such that [itex]n > \frac{3}{\delta}[/itex] is just basic Archemede with y = 3, so we're done there too.


Is this OK? Did you have something else in mind?
 
  • #4
Hurkyl
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Yep; it was this basic idea I wanted to motivate. This is the basic principle behind many arguments: looking at each contribution to the "error", and show they vanish.
 

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