Let T be the cofinite topology on the set Z of integers. Show that the sequence {1,2,3...} converges in (Z,T) to each point of Z. Describe the convergent sequences in (Z,T).

thank you (Hi)

Printable View

- Jul 6th 2009, 04:00 PMmmsConvergent sequence
Let T be the cofinite topology on the set Z of integers. Show that the sequence {1,2,3...} converges in (Z,T) to each point of Z. Describe the convergent sequences in (Z,T).

thank you (Hi) - Jul 6th 2009, 05:02 PMPlato
Have you answered this question.

If not, how do you expect to understand this current question? - Jul 6th 2009, 05:08 PMmms
umm a user gave an answer, and i understood it, although i asked something else, but it was a small doubt i had.

- Jul 7th 2009, 06:59 AMhalbard
For $\displaystyle \mathbb Z$ with the cofinite topology,

*every*sequence of*distinct*integers converges to*every*point of $\displaystyle \mathbb Z$.

Let $\displaystyle x\in\mathbb Z$ and let $\displaystyle A$ be any open set containing $\displaystyle x$. Then $\displaystyle A\neq\emptyset$, and so $\displaystyle \mathbb Z-A$ is a finite set.

If $\displaystyle \{x_n\}$ is a sequence of distinct integers then there is an integer $\displaystyle N$ such that $\displaystyle x_n\notin \mathbb Z-A$ for all $\displaystyle n\geq N$ (since $\displaystyle \mathbb Z-A$ is finite) and therefore $\displaystyle x_n\in A$ for all $\displaystyle n\geq N$. Therefore $\displaystyle x_n$ converges to $\displaystyle x$.

However, a sequence such as $\displaystyle \{1,2,1,2,1,2,\dots\}$ does not converge to any point. It does not converge to $\displaystyle 1$ since $\displaystyle \mathbb Z-\{2\}$ is an open set containing $\displaystyle 1$ but not containing all terms of the sequence from some point onwards. Similarly it does not converge to $\displaystyle 2$. Furthermore if $\displaystyle x\neq1$ and $\displaystyle x\neq 2$ then $\displaystyle \mathbb Z-\{1,2\}$ is an open set containing $\displaystyle x$ but having no points of the sequence at all, and so the sequence fails to converge to $\displaystyle x$. - Jul 7th 2009, 08:34 AMmms
thanks for your help (Surprised)(and thank for your explanation on the other topic, it helped a lot),

but could you please help me "visualize" the proof? xD

how can a sequence converge to every point of Z? - Jul 7th 2009, 11:38 AMhalbard
I feel strangely compelled to respond to your request for a "visual aid". This is certainly not what I use when thinking about convergence. With Euclidean spaces or metric spaces some kind of diagram might help with notions of distance or closeness, but even here they are of limited scope. With general topological spaces, well I'd forget about it.

You may have discovered a paucity of diagrams in most undergraduate textbooks on this subject, which may have prompted your plea. There are reasons for it. Perhaps most authors are visually illiterate, or perhaps the subject doesn't lend itself to pictorial paraphernalia. Or perhaps not.

In any case, as with most higher abstract mathematics, you have to get your head around the concept, the definition of convergence being a point in case.

"A sequence ($\displaystyle x_n$) converges to a point $\displaystyle s$ in a topological space if for any open set $\displaystyle A$ containing $\displaystyle s$ we have $\displaystyle x_n\in A$ for all sufficiently large $\displaystyle n$."

Think about it. Do you understand it? Have you got your head around it? Does that statement even make sense? What does it mean to say you understand it, or not as the case may be? Can you prove things with it?

What does this kind of convergence mean? Does it mean that the points $\displaystyle x_n$ eventually get closer to $\displaystyle s$ as $\displaystyle n$ tends to infinity? Yes, in the sense that the terms $\displaystyle x_n$ must eventually lie in every neighbourhood of $\displaystyle s$.

You may wish to think of a neighbourhood as a set of points close to something. If that helps you that's fine. But how does it help to show that the sequence $\displaystyle x_n=n$ converges to every point in $\displaystyle \mathbb Z$ with the cofinite topology? How can these points get closer to every member of $\displaystyle \mathbb Z$? Don't these points get further away, eventually, from every point? No. That's where your ordinary notions of closeness break down. That is where the visual picture starts to pixellate and fizzle out. It is not a geometrical notion.

Why does the sequence converge to every point? Because eventually the terms of the sequence are in every neighbourhood of every point. How can this be? Because neighbourhoods of points contain*every*integer*except possibly*a finite number of them. How can I visualise that? I don't know. Think of these neighbourhoods as clouds reaching up above you infinitely far and the positive integers as a vast ladder disappearing high into the billowing mass of ... no, don't bother.

Don't get me wrong. I'm not saying that no one could possibly visualise what's going on here. All I know is that I don't do it that way (eh?), I wasn't taught how to do it that way (what way?), and I don't ever find it necessary to do it that way (what's are you talking about, man?).

There is no Royal Cinema now showing topology as far as I know. - Jul 7th 2009, 04:11 PMmms
THANK YOU!

Quote:

Why does the sequence converge to every point? Because eventually the terms of the sequence are in every neighbourhood of every point. How can this be? Because neighbourhoods of points contain*every*integer*except possibly*a finite number of them. How can I visualise that? I don't know. Think of these neighbourhoods as clouds reaching up above you infinitely far and the positive integers as a vast ladder disappearing high into the billowing mass of ... no, don't bother.