Suppose that $(X,\tau)$ is a $T_0$-topological space without isolated point. As it is well-known, all cofinite subsets together with the empty-set forms a topology on $X$ which is a $T_1$ topology on $X$ and we call it the cofinite topology and denote it by $(X,\tau_f)$. Now $(X,\tau\vee\tau_f)$ is another topological space. Can someone help me to prove our disprove the following statement?

If $U\in \tau\vee\tau_f$ but $U\notin \tau$, then $U$ has finite complement.

The topology $\tau\vee\tau_f$ is defined to be the smallest topology which contains both $\tau$ and $\tau_f$.


1 Answer 1


Here's a counter-example.
Define the topology $\tau$ on $\mathbb N$ with the basis: $$\{\{n,n+2,n+4,\ldots\}:n \in \mathbb N\}.$$ It is clear that this is a basis since the intersection of any two such sets is yet another one of these (or empty) and $\mathbb N = \{0,2,4,\ldots\}\cup\{1,3,5,\ldots\}$.
There are no isolated points in $\tau$: if $n \in U$ for some $U \in \tau$, then $n+2 \in U$.
The topology is $T_0$: if $n<m$, then $n \notin \{m,m+2,m+4,\ldots\}$.
The set $E$ of even numbers is open: $E=\{0,0+2,0+4,\ldots\}$.

In $\tau_f$, the set $\mathbb N\setminus\{10\}$ is open.
Let $E_{10}=(\mathbb N\setminus\{10\}) \cap E$. Then $E_{10} \in\tau\vee\tau_f$.
But $E_{10}$ is not co-finite, and so $E_{10} \notin \tau_f$.
And $E_{10} \notin \tau$ either, because $0 \in E_{10}$ and for $U \in \tau$, if $0 \in U$ then $10 \in U$.


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