# Homework Help: Topology: Understanding open sets

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1. Nov 29, 2017

### nightingale123

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
$U\in\tau\iff(0,0)\notin U\lor \exists N\ni : n\in\mathbb{N},n>N\implies(\{n\}\times\mathbb{N})\backslash U\text{ is finite}$
$a)$ 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$(x_{n})_{n\in\mathbb{N}}\subset\mathbb{N}^2 \text{ for which }\lim x_{n}\xrightarrow[n->\infty]{X}(0,0)$.
Assume that X is not first-countable
$e)$
2. Relevant equations

3. 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 $\exists N\ni : n\in\mathbb{N},n>N\implies(\{n\}\times\mathbb{N})\backslash U\text{ is finite}$ 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

2. Nov 29, 2017

### Staff: Mentor

What does $\ni :$ mean? And are you sure it is an OR in the defining property? That makes every set which doesn't contain $(0,0)$ an open set. Can you draw an open set which includes the origin, e.g. are $\{(0,0),(1,1)\}$ or $\{(0,0),(1,\mathbb{N})\}$ open?

3. Nov 29, 2017

### nightingale123

https://i.gyazo.com/a809d7be047d7ed855211b4d51372367.png

This is the original picture of the problem and pretty I'm sure that I translated it correctly. The $\ni$ is the main part that is bothering me because I don't know exactly what it means

4. Nov 29, 2017

### WWGD

Well, let's see:
So the open sets $U$ are those which:
1) Do not contain $(0,0)$ or :
2) The collection of pairs $(n+k) \times \mathbb N - U ; k=0,1,2,...= \{ (n \times \mathbb N), (n+1) \times \mathbb N ,.....,(n+k) \times \mathbb N ,.... \} -U$ is finite . This means $U$ must contain EDIT all but finitely-many of the sets in each of the collections $(n+k) \times \mathbb N; k=0,1,2,...$
Does this help?

5. Nov 29, 2017

### WWGD

Best I can tell
It means something like " So that" , or " With the property that" . EDIT and so it seems like this may be:

https://en.wikipedia.org/wiki/Cofiniteness#Cofinite_topology

Last edited: Nov 29, 2017
6. Nov 29, 2017

### Staff: Mentor

Never seen it before. It probably means $\forall$. It's a bit strange and should be explained somewhere in the book.
Google translates "ali" by "but", so I assume it to mean "and" and not "or", which means that not all sets without the origin are open. Thus I read it as follows:
$$U \in \tau \Longleftrightarrow (0,0) \notin U\; \wedge \; \exists N \in \mathbb{N}\; \forall n > N \, : \,\vert \, \{n\}\times \mathbb{N} \backslash U \,\vert < \infty$$
I haven't checked whether it's a topology or not. I would feel more comfortable, if this guesswork above wouldn't had been necessary. Especially "ali" makes a major difference.

7. Nov 29, 2017

### WWGD

No, I think only those who contain $(0,0)$ and all-but-finitely-many elements in the set. Ze Risky topology you suggested in Random Thoughts. EDIT Then open sets are either those that contain all-but-finitely many elements of $\mathbb N^2$ or those that miss the origin. I think it is clear to see closedeness under union: 1) If neither set contains the origin, neither will the union; if each contains all-but-finitely-many, so will the union. 2) Clearly the whole space is there, as it contains all but finitely many ( none) . And clearly the empty set is also there. EDIT2: Still, don't b) and d) contradict each other?

Last edited: Nov 29, 2017
8. Nov 29, 2017

### Staff: Mentor

You are right and Google translate was wrong!
I checked two other translation pages and they both had "ali = OR"
Corrected condition:
$$U\in \tau \Longleftrightarrow (0,0)\notin U \vee \exists N\in \mathbb{N} \; \forall n>N\, : \, |\,\{n\}×\mathbb{N} \,\backslash \,U\,| < \infty$$

9. Nov 29, 2017

### nightingale123

Thank you both
Now I tried to continue what WWGD said. A set open if it does not contain $(0,0)$ or there exists some number $N$ so that for every number greater than N $(\{n\}\times\mathbb{N})\backslash U$ is finite.
So we need to check
$a)$ whether any number of unions of such sets is again open
$b)$ a finite number of intersection of such sets is again open
So checking for intersections:
Lets ignore the first part because if $(0,0)\notin (A\wedge B)\implies (0,0)\notin A\cap B$
Now lets say that there exist two numbers $N,M$ for sets $U,V$such that $\forall n>N\implies(\{n\}\times\mathbb{N})\backslash U\text{ is finite}$ and $\forall n>M \implies(\{n\}\times\mathbb{N})\backslash V\text{ is finite}$ if we define $k=\max\{M,N\}$ the set we again get is open because both $U$ and $V$ can only miss finite many points therefore their intersection can only miss finite many points.

I believe I understand now how this topology works and will probably be able to continue.

Thank you both very much for the help