0
$\begingroup$

Given $a ,b \in \mathbb{R}$. Let’s consider the subset of $\mathbb{R}^2$ defined as

$A_{a,b}=\{(x,y) \in \mathbb{R}^2\mid x >a,y>b\}$

and let $\mathscr{B}=\{ A_{a,b}\mid a,b \in \mathbb{R}\}$. Let $(\mathbb{R}^2, \tau )$ be the topology generated by it.

Prove or disprove that it is compact and connected

For connectedness I know I have to find a couple of disjoint open sets whose union gives me the whole space, I guess I can do it just with elements of the base. My guess is that the elements of the basis are connected but how to prove it?

For compactness, since the elements of the base are not closed nor bounded, they can't be compact because in $\mathbb{R}^n$ compactness is equivalent to closedness + boundedness, but that holds if the topology is the euclidean one, for this topology I am unsure. I also tried finding a cover that has not finite subcover, but I am unsure what that could be

Any suggestion?

$\endgroup$
1
  • $\begingroup$ I think it's not compact. Take the open cover $A_{-n,-n}$ where $n\in \mathbb{N}$ $\endgroup$ Jul 9, 2020 at 10:29

2 Answers 2

1
$\begingroup$

As I've mentioned here: Verify the topology having as base $\mathscr{B}=\{ A_{a,b}\mid a,b \in \mathbb{R}\}$ is coarser than the euclidean one any two nonempty open subsets in the topology have nonempty intersection. Thus the topology is connected.

For compactness, note that "bounded+closed" property applies to the Euclidean topology only. Not to the one you are considering. So how to solve that? Consider $\mathscr{U}=\big\{A_{-n,-n}\ |\ n\in\mathbb{N}\}$ which covers whole $\mathbb{R}^2$ and note that if $\mathscr{V}=\{A_{-n_1,-n_1},\ldots,A_{-n_k,-n_k}\}$ is a finite subcollection, then $\bigcup\mathscr{V}=A_{-m,-m}$ where $m=\max(n_1,\ldots,n_k)$. Meaning $\mathscr{V}$ does not cover $\mathbb{R}^2$. Thus the topology is not compact.

$\endgroup$
2
  • $\begingroup$ How does the nonempty intersection of the elements of the base imply connectedness? $\endgroup$
    – J.C.VegaO
    Jul 9, 2020 at 10:44
  • 1
    $\begingroup$ @J.C.VegaO In order for a topological space $X$ to be disconnected, you need two, open, nonempty and disjoint subsets $U,V$ such that $X=U\cup V$. If there are no open disjoint subsets then the space cannot be disconnected, i.e. it has to be connected. $\endgroup$
    – freakish
    Jul 9, 2020 at 10:46
0
$\begingroup$

Hint: For connectedness, show that the standard Euclidean topology is richer, and note that a topology coarser than a connected one is also connected.

For compactness, you're way off: the bounded+closed criterion for compactness is for subsets of an Euclidean space with the standard topology. This topology is quite different. To show that it is not compact, find a not-essentially-finite cover of $(0,1]$ and try to do the same here.

$\endgroup$
2
  • $\begingroup$ what do yuu mean with not-essentially-finite ? $\endgroup$
    – J.C.VegaO
    Jul 9, 2020 at 10:43
  • $\begingroup$ I mean one without a finite subcover. $\endgroup$
    – tomasz
    Jul 9, 2020 at 11:40

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.