Can someone help a brainlet out? How do I show that [math]\mathscr{T}=

>>9163804

1) a topology
Just check that intersection of 2 open sets is open and any union of open sets is open. Which of these is not obvious?
2) is Hausdorff
For each n \in \N holds {n} \in T. So each 2 point can be divided.

>>9163804
>those tits
I would help you if I was a mathead

>>9164303

I'm getting held up in the case of [math]1 \in (\mathbb{N} - U [/math] here's what I have

Closed Under Union:

First we aim to show that [math]\emptyset[/math] and [math]\mathbb{N}[/math] are elements of [math]\mathscr{T}[/math]. Observe that the empty set is in [math]\mathscr{T}[/math] since [math]\mathbb{N} \subseteq \mathbb{N}[/math] and let [math]U = \mathbb{N}[/math] then [math]\mathbb{N} - U = \mathbb{N} - \mathbb{N} = \emptyset \in \mathscr{T}[/math] Next observe that [math]\mathbb{N} \in \mathscr{T}[/math] since [math]\emptyset \subseteq \mathbb{N}[/math] and let [math]U = \emptyset[/math] then [math]\mathbb{N} - \emptyset = \mathbb{N} \in \mathscr{T}[/math] This satisfies condition (1) for being a topology.

How do I include the case(s) when [math]1 \in (\mathbb{N} - U)[/math]? Since it could be finite or infinite am I looking at the complement again?

>>9163804
I propose that we create a new intentions section on sci in order for smart faggots to collaborate and create cool shit for the rest of us.

>>9164334

I copied wrong code

Closed Under Union:

Let [math]\mathcal{U}[/math] be an arbitrary family of sets where [math]\mathcal{U} \in \mathscr{T}[/math]; we aim to show this family is closed under arbitrary union. For [math]i \in I[/math] let [math]U_i[/math] denote a set in [math]\mathcal{U}[/math]. So assume [math] U_i = \emptyset \hspace{2mm} \forall i \in I[/math] then clearly [math]\displaystyle{ \bigcup_{\substack{i \in I }} U_i = \emptyset \in \mathscr{T}}[/math]. So suppose [math]\mathbb{N} - U_i[/math] is finite for every [math]i \in I[/math] . Then the union [math]\displaystyle{ \mathbb{N} - \bigcup_{i \in I} U_i = \bigcap_{i \in I} (\mathbb{N} - U_i)}[/math] by De Morgan's Laws and the fact that showing the union is open is the same as showing the complement is closed. Thus in this case an arbitrary union of finite sets is again open and is in [math]\mathscr{T}[/math].

>>9164340
Its hard to read the code. If at list one of the sets has finite complement then the whole union also has finite complement, so it is open. If all the sets do not contain 1, then the union also does not, so it is open.

>>9164356
Fuck, at least*. How will I pass TOEFL?

>>9163804
I can have my men onto it but first I need some source on your pic

>>9163804
Clearly [math] \emptyset, \mathbb{N} \in T[/math]. Let [math] F\subset T[/math]. If [math] 1\in\mathbb{N}-f\quad\forall f\in F[/math] then clearly [eqn]1\in\bigcap_{f\in F}(\mathbb{N}-f)=\mathbb{N}-\bigcup_{f\in F}f[/eqn]. If 1 is in none of em' then we must have [math] \mathbb{N}-f [/math] is finite for each [math] f\in F[/math] so that [eqn] \mathbb{N}-\bigcup_{f\in F}f=\bigcap_{f\in F}(\mathbb{N}-f) [/eqn] where the last thing is the intersection of finite sets which is finite or empty. In each case the union is in [math] T[/math]. In a similar manner, if we have some finite collection [math] f_i\in T,\quad i=1,...,n[/math], then either [math] 1\in f_i [/math] for some index so that [math] 1\in\bigcup_{i=1}^n(\mathbb{N}-f_i)=\mathbb{N}-\bigcap_{i=1}^nf_i[/math]. Else, each [math] \mathbb{N}-f_i [math] is finite so through similar things we get a finite union of finite sets, which is finite. Thus we got ourselves a topology. Given two elements [math] m,n\mathbb{N} [/math] not 1, we clearly have [math] 1\in\mathbb{N}-\{m\},\quad 1\in\mathbb{N}-\{n\} [/math] so that the desired disjoint open sets are the singletons of [math] m,n [/math]. In the case where [math] m=1 [/math], take the neighborhood [math] \{n\} [/math] of [math] n [/math], and [math] \{1\}\cup\{n+1,n+2,...\} [/math] of 1. These are clearly disjoint and the latter is in our topology since its complement is finite. God I hope this formatting looks good. Cheers!

>>9164635
Goddamnit. The fucked stuff says:

Else, each [math] \mathbb{N}-f_i [/math] is finite so through similar reasoning we get a finite union of finite sets, which is finite. Thus we got ourselves a topology. Given two elements [math] m,n\in\mathbb{N} [/math] ...

>>9164635
Also, when I concern myself with the finite intersections, i meant to have [math] 1\in\mathbb{N}-f_i [/math] for some index. Damn it's hard not seeing this stuff as you type.

>>9163804

>the thumbnail

i'm only here because I thought she had a boogie