All subsets of $\mathbb{N}$ are open in $\mathbb{N}$? Why precisely is this true? I've always sort of overlooked this and taken it for granted, but why exactly are all subsets of $\mathbb{N}$ open in $\mathbb{N}$?
 A: All subsets of $\Bbb N$ are open in $\Bbb N$ if and only if $\Bbb N$ is given the discrete topology, and one can certainly endow $\Bbb N$ with other topologies. However, the topology that $\Bbb N$ inherits from $\Bbb R$ is the discrete topology, and that’s probably the one that you have in mind. To see why this is the discrete topology, let $n\in\Bbb N$; then $(n-1,n+1)$ is an open set in $\Bbb R$, and $$(n-1,n+1)\cap\Bbb N=\{n\}\;,$$ so $\{n\}$ is an open set in the subspace topology on $\Bbb N$. Arbitrary unions of open sets are open, and every non-empty subset of $\Bbb N$ is a union of sets of the form $\{n\}$, so every subset of $\Bbb N$ is open in the topology inherited from $\Bbb R$.
A: The topology that $\mathbb N$ has here is the topology coming from $\mathbb R$: the open subsets of $\mathbb N$ are those of the form $U \cap {\mathbb N}$ with $U$ open in $\mathbb R$. So, in particular, $\{n\}$ is open in $\mathbb N$ for every $n \in \mathbb N$, as $\{ n \} = (n - 1/2, n + 1/2) \cap {\mathbb N}$. Consequently, since a union of open sets is open, every subset of ${\mathbb N}$ is open.
A: If we equip $\Bbb N$ with the topology inherited from $\Bbb R$, then a singleton $\{n\}$ is equal to the intersection $(n-1/2,n+1/2)\cap\Bbb N$, thus it is open by definition of the subspace topology.
On the other hand, $\Bbb N$ is a linearly ordered space. That means it has an order "$<$" which induces a topology, by taking as basic sets the intervals $$(a,b)=\{x\mid a<x<b\}\\(-\infty,b)=\{x\mid x<b\},\\ (a,\infty)=\{x\mid x>a\}$$ where $a,b\in\Bbb N$. But then again, since $\{n\}=(n-1,n+1)$, we get the discrete topology.
