Be $\tau_{1}$ and $\tau_{2}$ topologies over $\mathbb{N}$ Be $\tau_{1}$ and $\tau_{2}$ topologies over $\mathbb{N}$ defined by:
$\tau_{1}=\{\{m\in \mathbb{N}:m<n\}:n\in \mathbb{N}\}\cup \{\mathbb{N}\}\}$ and $\tau_{2}=\{A\subseteq \mathbb{N}: 0\in A\}\cup \{\emptyset\}$.
Check who are these topologies and prove which is the finest.

The topology consisting of all subsets containing a fixed point $x$ is called the particular/included point topology. In your case $x=0$. You can simply say that $\tau_1\subset \tau_2$ i.e. $\tau_2$ is strictly finer than $\tau_1$ since:

*

*For all subsets $S\ne\emptyset\in\tau_1,0\in S\subseteq\Bbb N$ and thus $S\in\tau_2$, so $\tau_1\subseteq \tau_2$.


*$\{0,2\}\in\tau_2$ but does not belong to $\tau_1$. Hence $\tau_1\ne\tau_2$.
