2
$\begingroup$

Let $A = \{1 - \frac 1n : n \in \mathbb Z ^+\}$ is closed under certain topologies on $\mathbb R$.

I am supposed to figure out if this set is closed under certain topologies. I know that means I need to show the complement of the set is open. I am having a lot of difficulty with two things: 1. figuring out what this set actually is 2. figuring out the differences between the different topologies.

I know for it to be false I need to just show one counterexample. For it to be true I am going to have to show that all three conditions of a topology are met: the set and the complement are in the topology, the union of any collection of sets is in the topology and the finite intersection of sets is in the topology.

I am supposed to look at the three following topologies for this set all in $\mathbb R$

The usual topology: is just the three conditions listed above. The half open interval topology: The collection of $\mathfrak H$ of all subsets of $U$ such that either $U = \emptyset$ or for each $ x \in U$ there is an interval of the form $[a, b) \subseteq U$ is a topology for $\mathbb R$ The countable complement topology $ \{ U: \mathbb R - U$ is countable$\} \cup \{ \emptyset, \mathbb R\}$

I would appreciate help in just being able to wrap my brain around the different topologies and what this set might actually look like.

$\endgroup$
  • 1
    $\begingroup$ Do you correctly define "half" open interval topology? $\endgroup$ – Salomo May 3 '15 at 15:34
  • 1
    $\begingroup$ Under the usual topology it is clearly not closed: the sequence $1- \frac 1 n$ tends to $1$, which does not belong to $A$. $\endgroup$ – Alex M. May 3 '15 at 15:36
  • $\begingroup$ @Salomo I did not. I will edit. $\endgroup$ – user219081 May 3 '15 at 15:43
  • 1
    $\begingroup$ You should check what is meant by "the usual topology." The three conditions you mention are satisfied by any topology, not just the usual one. $\endgroup$ – AMPerrine May 3 '15 at 15:47
  • 1
    $\begingroup$ @AlexM. is that another way of saying 1 is a limit point and therefore the set is not closed since the set does not contain the limit point? $\endgroup$ – user219081 May 8 '15 at 19:34
1
$\begingroup$

As $A$ is countable (that is there is bijection $f$ from $\mathbb{N}$ to $A$, precisely, $f(n)=1-\frac{1}{n}$), it is obviously a closed set in the countable complement topology.

$\endgroup$
  • 1
    $\begingroup$ Why is A obviously countable? Sorry new learner here! $\endgroup$ – user219081 May 3 '15 at 15:45
  • 1
    $\begingroup$ @AlyssaWallace Edited with more explanation. $\endgroup$ – Salomo May 3 '15 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy