Showing $\mathscr{T}$ is a Topology and is Hausdorff.

Question

I need help showing that $\mathscr{T}= \{U \subseteq \mathbb{N} \mid \mathbb{N} - U \hspace{2mm} \textrm{is finite or} \hspace{2mm} 1 \in (\mathbb{N} - U) \}$ is a topology. I understand how to show $\mathscr{T}$ is a topology for the cofinite case, but I don't understand how to show it for when $1 \in (\mathbb{N} - U)$.

Closed Under Union:

Let $\mathcal{U} $ be an arbitrary family of sets where $\mathcal{U} \in \mathscr{T}$; we aim to show this family is closed under arbitrary union. For $i \in I$ let $U_i$ denote a set in $\mathcal{U}$. So assume $ U_i = \emptyset \hspace{2mm} \forall i \in I$ then clearly $\displaystyle{ \bigcup_{\substack{i \in I }} U_i = \emptyset \in \mathscr{T}}$. So suppose $\mathbb{N} - U_i$ is finite for every $i \in I$ . Then the union $\displaystyle{ \mathbb{N} - \bigcup_{i \in I} U_i = \bigcap_{i \in I} (\mathbb{N} - U_i)}$ by De Morgan's Laws and the fact that showing the union is open is the same as showing the complement is closed. Thus in this case an arbitrary union of finite sets is again open and is in $\mathscr{T}$.

So how do I include the case when $1 \in (\mathbb{N} - U)$? Do I have to consider two different cases since it could be finite or infinite? I know this is probably really simple, but I'm not seeing it.

Closed Under Intersection

Let $\mathcal{U}$ be a finite family of sets containing $U_1, U_2, \cdots, U_n$ where $\mathcal{U} \in \mathscr{T}$. We aim to show that the family is closed under finite intersection. If any $U_i = \emptyset$ then clearly $\displaystyle{\bigcap_{i = 1}^{n} U_i = \emptyset}$. So assume that for each $U_i, \mathbb{N} - U_i$ is finite. Then $\displaystyle{\mathbb{N} - \bigg(\bigcap_{i = 1}^{n} U_i \bigg) = \bigcup_{i = 1}^{n} (\mathbb{N} - U_i)}$ again by De Morgan's Laws since showing the intersection is open is the same as showing its complement is closed. So we have have finite unions of finite sets which are finite; $\displaystyle{\bigcap_{i = 1}^{n} U_i \in \mathscr{T}}$

Similarly here how do I include the case of $1 \in (\mathbb{N} - U)$?

Lastly, can someone get me started on showing $\mathscr{T}$ is Hausdorff?

Answer 1

Hint: to show $\mathscr{T}$ is a topology, it is easier to think about the closed sets. A set is closed in $\mathscr{T}$ if it is finite or if it contains $1$. We have to show that finite unions and arbitrary intersections of closed sets are closed. For finite unions, $X_1 \cup \ldots \cup X_n$ is finite if all the $X_i$ are finite and contains $1$ if any of the $X_i$ contains $1$. For intersections, $\bigcap_i X_i$ is finite if any of the $X_i$ is finite and contains $1$ if all the $X_i$ contain $1$.

For the Hausdorff property, given $x \neq y$, assume w.l.o.g. that $y \neq 1$, then $\{y\}$ and $\Bbb{N} \setminus \{y\}$ are open sets that separate $x$ and $y$.

Answer 2

For the Hausdorf property:

Let $m,n \in \mathbb{N}$ such that $m \neq n$ and $m,n>1$

Take $U=\{m\}$ and $V=\{n\}$

$U,V$ belong to the topology.

Now if one of $m,n$ is equal to $1$,suppose $m=1$ then take

$U=\mathbb{N}$ \ $\{n\}$ and $V=\{n\}$.

In each case $U \cap V= \emptyset$