$X$ an infinite space, and $x_0 \in X$

$\mathcal{O} = \{O \subseteq X : (x_0 \notin O) \vee (X \setminus O \text{ is finite})\}$

$\mathcal{B} = \{\{x\} : x \in X \setminus \{x_0\}\} \cup \{X \setminus F : F \text{ is finite}\}$

  1. Show that $\mathcal{O}$ is a topology.
  2. Show that $\mathcal{B}$ is a basis for $\mathcal{O}$.

I managed to solve 1 via the definition of a topology. However, I'm quite stuck on showing that that $\mathcal{B}$ is a basis for $\mathcal{O}$. I came up with two ways:

First show that $\mathcal{B}$ is a basis for a topology on $X$, then possibly show that the topology generated by $\mathcal{B}$ (write $\mathcal{T}$) equals $\mathcal{O}$. I managed to show 1) that $\mathcal{B}$ is a basis for a topology on $X$ via the definition of a basis, and 2) that $\mathcal{B} \subset \mathcal{O}$. Using the fact that $\mathcal{T}$ is the set of all unions of elements of $\mathcal{B}$ and that $\mathcal{O}$ is closed under arbitrary unions, we have that $\mathcal{T} \subset \mathcal{O}$. At this point I'm stuck, since when I take an $O \in \mathcal{O}$ such that $X \setminus O$ is finite and $x_0 \in O$, this $O$ does not seem to be able to be written in terms of arbitrary unions and finite intersections of $\mathcal{T}$ (taking unions of elements of $\{\{x\} : x \in X \setminus \{x_0\}\}$ will not contain $x_0$, and taking intersections of $\{X \setminus F : F \text{ is finite}\}$ will never be finite). Is my last conclusion here wrong?

Another possible way of showing this is using lemma 13.2 in Munkres' Topology, which states the following:

Let $X$ be a topological space. Suppose that $\mathcal{C}$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x$ in $U$, there is an element $C$ of $\mathcal{C}$ such that $x \in C \subset U$. Then $\mathcal{C}$ is a basis for the topology of $X$.

So I have showed that $\mathcal{B} \subset \mathcal{O}$ (thus $\mathcal{B}$ is a collection of open sets of X). However, I seem to be running into more or less the same problem as above (take an $O \in \mathcal{O}$ such that $X \setminus O$ is finite and $x_0 \in O$, then if we want to find a $B \in \mathcal{B}$ such that $x_0 \in B \subset \mathcal{O}$, this seems impossible).

Am I missing some other way to show that $\mathcal{B}$ is a basis for $\mathcal{O}$, is some of my logic flawed, or is $\mathcal{B}$ simply not a basis for $\mathcal{O}$?

  • $\begingroup$ Then $O = X \setminus (X \setminus O)$. $\endgroup$ – Daniel Schepler Feb 13 '18 at 22:08

You need to show that any set in $\mathcal O$ is a union of sets in $\mathcal B$. There are two kinds of open sets in your topology; such that exclude $x_0$ and such that their complements are finite.

  • Let $O\in \mathcal O$ such that $x_0\notin O$. Then $O=\cup\{\{x\}\;|\;x\in O\}$ is a union of singletons, which are all members of $\mathcal B$ because $x_0\notin O$.
  • Let $O\in \mathcal O$ such that $X\setminus O$ is finite. If $x_0\notin O$, the above case applies. So let $x_0\in O$. Now $X\setminus (X\setminus O)=O$ is a member of $\mathcal B$ since $X\setminus O$ is finite.
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