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Let $Z$ be the set of positive integers. For each positive integer n, let $O_{n}=\{ n,n+1,n+2,\dots \}$.Let $λ=\{\emptyset,O_{1},O_{2}, \dots, O_{n}.\dots\}$. Then $(Z,λ)$ is a topological space but not metrizable.

My proof is: Let $d(1,2)=\varepsilon_{1,2}$ , if $\forall x \in Z$, $d(1,2) \leq d(1,x)$ then choose $\delta < \varepsilon_{1,2}$, $B(1,\delta)$ only contains 1, and there does not exist a subset $\{1\}$ in λ,

if $\exists x\in Z$ ,st $d(1,x)<d(1,2)$ then $B(1,\delta)$ does not contain 2 and 1 is in the set, and there is not a such set.

so $(Z,λ)$ is not metrizable.

Is this proof correct?Is the metrizable topological space arise from the metric space have the same open set? I read a book about topology but it does not explain it and I got confused

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4  
Hint: Show that your space is not Hausdorff. –  user38268 Mar 16 '13 at 14:39
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You can even show that not all points are closed. In fact there is only one closed point in this space. –  Dune Mar 16 '13 at 14:46
    
You do not need to distinguish two cases. If $d(1,2)=\epsilon$ Then every $B(1,\delta)$ for $\delta<\epsilon$ does not contain $2$. But every set containing $1$ also contains $2$. –  Stefan Hamcke Mar 16 '13 at 22:52

2 Answers 2

up vote 2 down vote accepted

A metrizable topological space is in particular a $T_1$-space, i.e. for all $n,m \in \mathbb{N}$, $n \not=m$, there have to exist open sets $U,V \in \lambda$ such that $n \in U$, $m \in V$, $U \cap V = \emptyset$.

But this is clearly not the case: Consider $n := 1$ and $m \in \mathbb{N}$, $m>1$ arbritrary. The set $O_1 = \{n \in \mathbb{N}; n \geq 1\}=\mathbb{N}$ is the only open set containing $1$, so $U=O_1=\mathbb{N}$. But since this implies $m \in U$, we have $$m \in U \cap V$$ i.e $U \cap V \not= \emptyset$ for all $V \in \lambda$ such that $m \in V$.


Concerning your own proof: The idea is correct. Assume that the topology is generated by some metric $d$. Then $B(1,r)$ is a open set for all $r>0$, so there has to exist in particular a set $O \in \lambda$ such that $O=B(1,\varepsilon)$ where $\varepsilon:=d(1,2)$. By definition, we have $2 \notin O$. But since $O=O_1 = \mathbb{N}$ is the only set in $\lambda$ containing $1$, this is clearly a contradiction.

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Does the metrizable topological space arise from the metric space have the same open set? –  Jebei Mar 16 '13 at 14:50
    
@frame99 Well, in this case, there is no metric space. In general, a topological space is metrizable iff there exists a metric such that the open sets (with respect to the topology) are open with respect to the metric and vica versa. –  saz Mar 16 '13 at 14:55
    
@frame99 I added a remark concerning your proof. –  saz Mar 16 '13 at 15:07
    
Thanks,Is it a metrizable space a T4 space? –  Jebei Mar 16 '13 at 15:31
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@frame99 A metrizable space is a $T_4$-space, yes. –  saz Mar 16 '13 at 15:37

If this topology is coming from a metric on $Z$, then it has to be Hausdorff. But you can easily show that this topology is not Hausdorff. For example take two points 1 and 2 in $Z$. every open set containing $1$ contains 2 as well.

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