# more topology

• Nov 17th 2008, 04:04 PM
EricaMae
more topology
This problem has five parts, but I think I can get two of them. So

Let B be the collection of subsets of R in the form [a,b) for all a < b, the empty set, and the whole set included.

Let T be the topology on R determined by taking all arbitrary unions of sets in B. Show the sequence {1+ (1/n), n= 1,2, ...} converges to one in the topology T.

I thought it was obvious that this sequence converges to 1 because 1/n converges to 0, but then I got confused by the next part.

Show that the sequence {1 - (1/n), n=1,2,...} does NOT converge to 1 in the topology T. I'm assuming this may have something to do with [a,b) where b is not included, but I"m not sure.

Also, considering the product topology R x R determined by T. Show that the set of points {(x,-x) | x is in R} is discrete in the relative topology.

I'm not sure what discrete in the relative topology means. Does it have anything to do with discrete metrics?

Thanks
• Nov 17th 2008, 11:35 PM
Moo
Hello,

Quote:

Let B be the collection of subsets of R in the form [a,b) for all a < b, the empty set, and the whole set included.

Let T be the topology on R determined by taking all arbitrary unions of sets in B. Show the sequence {1+ (1/n), n= 1,2, ...} converges to one in the topology T.

I thought it was obvious that this sequence converges to 1 because 1/n converges to 0, but then I got confused by the next part.

Show that the sequence {1 - (1/n), n=1,2,...} does NOT converge to 1 in the topology T. I'm assuming this may have something to do with [a,b) where b is not included, but I"m not sure.
No, it has to do with the definition of convergence in a topology (Wink)
Here is the general definition :
Let $\displaystyle (x_n)_n \in X$ be a sequence of points in X.
$\displaystyle (x_n)$ converges to $\displaystyle x$ if for any open neighbourhood $\displaystyle \mathcal{U}$ of x :
$\displaystyle \exists N_x \in \mathbb{N}, ~ \forall n \geq N, x_n \in \mathcal{U}$

So for the first one, you're approaching 1 by superior values. It is easy to prove that for any open set containing 1 (which is a neighbourhood), and particulary for the open sets in the form [1,a), a>1, the definition is verified.

For the second one, you're approaching 1 by inferior values. It is sufficient to take a counterexample :
[1,a) is an open set, but in no way there is an N such that for all n>N, $\displaystyle x_n=1-\frac 1n$ is in this open set, because it is $\displaystyle <1$.
hence it doesn't converge to 1.

For the second part, to prove a set is discrete, show that any of its element and subset is an open set in the relative topology. I'm sorry I cannot help much here as I don't understand what the set exactly is (Doh)
• Nov 18th 2008, 01:25 AM
Opalg
Quote:

Originally Posted by EricaMae
Also, considering the product topology R x R determined by T. Show that the set of points {(x,-x) | x is in R} is discrete in the relative topology.

I'm not sure what discrete in the relative topology means. Does it have anything to do with discrete metrics?

The relative topology on a subspace simply means the topology where the open sets are the open sets in the whole space, intersected with the subspace.

In $\displaystyle \mathbb{R}^2$ with the product half-open interval topology, the set $\displaystyle \{(x,y):a\leqslant x<a+1,\ -a\leqslant y<-a+1\}$ is an open set containing the point (a,-a). It's easy to check that this open set does not contain any point of the form (x,-x) apart from (a,-a). That shows that the subspace $\displaystyle \{(x,-x):x\in\mathbb{R}\}$ is discrete for the relative topology.