Give N the cofinite topology. Is N connected in the cofinite topology?

Question

This is how I showed it, but I am not sure whether this is correct or wrong. Hence, it would be great if someone verifies this proof. also, kindly let me know on notation and better structural proof because I dont think this proof is in good shape.

Appreciate your help.

Answer 1

Suppose $\Bbb N$ in the cofinite topology is not connected. Then $\Bbb N = A \cup B$ where $A$, $B$ are disjoint non-empty open subsets of $\Bbb N$. It follows (as $A$ is $B$'s complement and vice versa) that $A$ and $B$ are also closed and by assumption $A,B\neq \Bbb N$ (or the other one is empty) and so by definition of the cofinite topology $A$ and $B$ must be finite. This contradicts that $\Bbb N$ is infinite.

So $\Bbb N$ is connected in the cofinite topology.

Answer 2

There's something wrong with your argument. You assume that $U=\emptyset$ which is not ok. One way is to show that if both $U\neq\emptyset$, $V\neq\emptyset$ and $U\cup V=\mathbb{N}$ then $U\cap V\neq\emptyset$ as well.

And this follows because if $X$ is an infnite set with cofinite topology, $U,V\subseteq X$ are open and nonempty, then there is $x\in X$ such that $x\not\in X\backslash U$ and $x\not\in X\backslash V$. That's because both $X\backslash U$ and $X\backslash V$ are finite while $X$ is infnite. And so $x\in U\cap V$, meaning no two open nonempty subsets of $X$ are disjoint (regardles whether their union is $X$ or not)

Answer 3

Since $\Bbb N$ is infinite, there does not exist a separation of $\Bbb N$. That is, you can't write $\Bbb N$ as a union of two disjoint cofinite sets. For $U$ and $V$ would both then be finite, which would make the union finite (note: unions of cofinite sets are also cofinite, so all the open sets in this topology are cofinite).