How to show that the sets $U$ that satisfy $\lim \inf_{n \to \infty} \frac{|U \cap \{1,...,n\}|}{n} = 1$ form a topology?

Question

I want to construct a topology on $\mathbb{N}$ where $\emptyset$ is open and all of the sets $U$ that satisfy $\lim \inf_{n \to \infty} \frac{|U \cap \{1,...,n\}|}{n} = 1$ are open. I have that $\emptyset$ and $\mathbb{N}$ are open pretty trivially, and it is also pretty easy to see that the union of open sets will be open. However, I'm having a lot of trouble showing that finite intersections of open sets are open. Also, once I show that this is indeed a topology, how would I prove or disprove that it is Hausdorff? My intuition is that this topology is not Hausdorff, partially because it reminds me of the co-finite topology.

Answer 1

For the finite intersections, consider two open sets $U$ and $V$.

Then let us prove that $\liminf \frac1n |U\cap V \cap \{1,\ldots, n\}|=1$.

Let $\epsilon>0$. Since $U$ is open, for any large enough $n\in \mathbb{N}$, $$ (1-\epsilon)n \leq |U\cap \{1,\ldots ,n\}|. $$ Since $V$ is open, for any large enough $n\in \mathbb{N}$, $$ (1-\epsilon)n\leq |V\cap \{1,\ldots , n\}|. $$ From $|A\cap B| = |A|+|B|-|A\cup B|$ for any sets $A$ and $B$, we have $$ |U\cap V\cap \{1,\ldots n\}|=|U\cap\{1,\ldots,n\}|+|V\cap\{1,\ldots,n\}|-|(U\cup V)\cap \{1,\ldots,n\}| $$

this is $$\geq |U\cap\{1,\ldots,n\}|+|V\cap\{1,\ldots,n\}|-n= 2(1-\epsilon)n-n = (1-2\epsilon)n .$$ Therefore $\liminf\frac1n|U\cap V\cap \{1,\ldots,n\}| \geq 1-2\epsilon.$ Since $\epsilon>0$ is arbitrary, we have the result $$ \liminf\frac1n|U\cap V\cap \{1,\ldots,n\}| =1. $$

For the remark on Hausdorff, let $a, b$ be distinct. We need to find disjoint nonempty open sets $U$ and $V$ such that $a\in U$ and $b\in V$. From what we proved, intersections of nonempty open sets cannot be empty. Thus, the topology is not Hausdorff.