Topology: Finding open sets

84 Views Asked by At

1. The problem statement, all variables and given/known data

We define $X=\mathbb{N}^2\cup\{(0,0)\}$ and $\tau$ ( the family of open sets) like this $$U\in\tau\iff(0,0)\notin U\lor \exists N\in\mathbb{N},\ \forall n>N\implies(\{n\}\times\mathbb{N})\setminus U\text{ is finite}$$

$a)$ Show that $\tau$ satisfies that axioms for open sets

$b)$ Show that $(0,0)$ lies in the Closure of $\mathbb{N}^2$

$c)$ Describe closed sets in topology $\tau$

$d)$ show that there doesn't exists a sequence$(x_{n})_{n\in\mathbb{N}} \subset\mathbb{N}^2 \text{ for which }\lim x_{n}\xrightarrow[n->\infty]{X}(0,0)$. Conclude that X is not first-countable

2. The attempt at a solution

I'm having trouble visualizing the sets in $\tau$ I know from the first part that the every point which is not $(0,0)$ is open.

Also I know that every union of such sets will also be open. Therefore $\mathbb{N}^2$ in itself is open. However I don't know how to visualize the other condition $\exists N n\in\mathbb{N},\ \forall n>N\implies(\{n\}\times\mathbb{N})\setminus U\text{ is finite}$

I would really appreciate it if someone could explain to me how this sets look as I am unable to continue with the problem.

Thank you

1

There are 1 best solutions below

0
On BEST ANSWER

All points of $\Bbb N^2$ are isolated and a set $U\ni (0,0)$ is open iff it intersects all but finitely many verticals $\{n\}\times\Bbb N$ by all but finitely many elements each, that is complements of neighborhoods of the point $(0,0)$ are “small”.