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.


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 $M_1$.

  • $\begingroup$ 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

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