Let $\mathbb{N}=\{0,1,2,\dotso\}$ and $(\mathbb{N},<)$ with the usual ordering $<$. Let $\tau_<$ be the order topology with regards to $<$. Then $\tau_<$ is the discrete topology (topology of all subsets) on $\mathbb{N}$.
I am asked to proof this, but I doubt that this is correct...
A base of the order topology is given by: $\mathcal{O}=\{(u,v)|u,v\in X, u<v\}\cup\{(-\infty, u), (u,\infty)|u\in X\}\cup\{X\}$
That means for $V\in\tau_<$ there is for every $v\in V$ a $U\in\mathcal{O}$ such that $v\in U\subseteq V$.
We want to show, that $\tau_<=\tau_{disc}$, so every subset of $\mathbb{N}$ is open. Clearly it sufficies to show, that $\{n\}$ is open for every $n\in\mathbb{N}$.
For $n\neq 0$ this is obviously true, since $\{n\}=(n-1,n+1)$.
But I do not see, why $\{0\}$ should be an open set, with regards to $\tau_<$. The only elements of $\mathcal{O}$ which contain $0$ are the sets $(-\infty, n)$ with $n>0$. But with these we do not get equality.
So there is no chance to get $\{0\}=\bigcup \mathcal{U}$ where $\mathcal{U}\subset\mathcal{O}$. So $\{0\}$ should not be open. In fact no subset of $\mathbb{N}$ which has $0$ as an element should be open. So $\tau_<\neq\tau_{disc}$
Or am I missing something?
Thanks in advance.
I think the confusion comes from the notation of your open rays as $(n,\infty)$ and $(-\infty,n)$. I personally find the following notation less confusing:
Define ${x\!\!\uparrow}=\{y\in X\mid y>x\}$ and ${x\!\!\downarrow}=\{y\in X\mid y<x\}$, then the set $\mathcal B=\{{x\!\!\uparrow}\mid x\in X\}\cup\{{x\!\!\downarrow}\mid x\in X\}$ forms a subbase for the order topology. You could get an open interval $(a,b)$ by taking the intersection of ${a\!\!\uparrow}\cap {b\!\!\downarrow}$.
Now for $X=\mathbb N$ it should become clear that $\{0\}={1\!\!\downarrow}$.