3
$\begingroup$

Consider $\mathbb{N}$ as natural number with following topology $T$ :

$U\subset \mathbb{N}$ is nonempty, $U\in T$ if and only if $U$ has the property that natural number $n$ belongs to $U$ only if every devisor $k\in \mathbb{N}$ of $n$ belongs to $U$.

a) Is $(\mathbb{N},T)$ a $T_1$-space?

b) Is $(\mathbb{N},T)$ a $T_0$-space?

c)Is $(\mathbb{N},T)$ a second countable space?

d)Is $(\mathbb{N},T)$ a first countable space?

My solution:

Part a) Based on the definition of $T_1$-space that said $(\mathbb{N},T)$ a $T_1$-space if for all $x,y\in \mathbb{N}$ , $\exists$ open $U,V$ such that $x\in U$ , $y\in V$, $x\notin V$ , $y\notin U$. so as $1$ divide every element in $\mathbb{N}$, $1$ belong to every $U$. So if we let $x=1$ , $y$ we cannot found any open set that is not contain $1$.

I will appreciate any every help to correct or improve my way and my writing mathematically.

Part b) A topological apace $(\mathbb{N},T)$ is $T_1$-space if for all distinct point $x,y\in \mathbb{N}$ , $\exists$ open $U$, such that $x\in U$, $y\notin U$. If we let $x=1$ and $y\in \mathbb{N}$, there always exist open set such that $x\in U$ and $y\notin U$.

I also have same problem, I do not know it is correct and enough?

Part c) $(\mathbb{N},T)$ is a second countable space if there exist a countable basis for the topology on $\mathbb{N}$. I can prove it by following example but i need more precise way.

let $B=${$U_1 ,U_2 , ...$} such that $U_1 =${$1$} , $u_2 =${$1,2$} , $U_3 =${$1,3$} and ets. First, $ \forall x\in \mathbb{N}$, $\exists U_i \in B$ such that $x\in U_i$. Second, $\forall U_i , U_j \in B$ for $x\in U_i \cap U_j$ $\exists U_k \in B$ , $x\in U_k \subset U_j\cap U_i$. for example $U_1\cap U_3 =${$1$}, $\exists$ {$1$} such that $1\in$ {$1$}$\subset${$1$}.

Part d) $(\mathbb{N},T)$ is a first countable space if $ \forall x\in \mathbb{N}$ has countable local basis. I mean, $\forall x\in\mathbb{N}$ , $B=${$U_i, 1\le i \le n$}, $B\subset \mathbb{N}$,

$\forall U \in T$, $x\in U$, $\exists U_i \in B$ such that $U_i \subset U$

$\forall x \in mathbb{N}$ , let $x=2$ and $B=${$U_1 , U_2 ,...$} such that $u_1=${$1$} , $U_2 =${$1,2$} , $U_3=${$1,3$} if $U=${$1,2,4$}, we see $2\in U$ then $\exists U_2 \in B$ such that $U_2 \subset U$ .

But I need some help to prove it without example.

$\endgroup$
2
$\begingroup$

a) Your answer is perfect. The definition I knew involved only one open set around $x$ which doesn't contain $y$, but it is the same by symmetry reasons.

b) If you wanted to prove that it is $T_0$, then you cannot choose freely values for $x,y$ but assume that $x,y$ are given. Now, this space is indeed $T_0$. But $T_0$ means that for any $x,y$ there is an open set $U$ which contains exactly one of $x,y$ (but it's not said which one). So, if neither $x,y$ is $1$, and e.g. $x<y$ then $x\in G_x\not\ni y$.
If $x=1$ and $y> 1$ then consider $\{1\}=G_1$ around $x$, this doesn't contain $y$.

c) For each $n\in\Bbb N$ consider the set $G_n:=\{m\in\Bbb N\,:\,m\,|\,n\}$, then it is open ($G_n\in T$), and, the definition of $T$ asserts just that $U\in T\iff U$ is a union of $G_n$'s (whenever $n\in U$ we have $G_n\subset U$).

d) If a space is $M_2$ then it implies that it is also $M_1$.

$\endgroup$
  • $\begingroup$ Thank you so much for your answer. in the part a I proved that defined topology is not $T_1$ so we can not imply that is $T_0$. So, I proved that this toplogy is $T_0$. $\endgroup$ – Zeezee Nov 26 '13 at 23:43
  • $\begingroup$ I have a little question, $M_2$ means second countable space and $M_1$ means fierst countable space? $\endgroup$ – Zeezee Nov 26 '13 at 23:45
  • $\begingroup$ Yes, you're right. I correct it. Yes, $M_2$ means second and $M_1$ first compatible. $\endgroup$ – Berci Nov 26 '13 at 23:56

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.