Let $(X,\tau)$ be a $T_1$ space, which is not compact, $x_0 \in X$, and $\mathcal{F}$ an ultrafilter........ Let $(X,\tau)$ be a $T_1$ space, which is not compact, $x_0 \in X$, and $\mathcal{F}$
an ultrafilter in $X$, such that $x_0$ is not a  $\tau$-limit of $\mathcal{F}$.
Suppose a new topology $\sigma$ is defined on $X$, such that $U$ is a  $\sigma$-open set if it is $\tau$-open and satisfies one of the following conditions:
(i) $x_0 \in U$ and $U \in \mathcal{F}$
(ii) $x_0 \not\in U$

Is $\sigma$ a $T_1$-topology, and is $\sigma$ strictly weaker than $\tau$?

 A: Since $x_0$ is not a $\tau$-limit of $\mathscr{F}$, $x_0$ has an open nbhd $U_0$ such that $U_0\notin\mathscr{F}$; by definition $U_0\in\tau\setminus\sigma$, so $\sigma$ is strictly weaker (coarser) than $\tau$. 
If $\mathscr{F}$ is non-principal, then $\langle X,\sigma\rangle$ is $T_1$. $\langle X,\tau\rangle$ is $T_1$, so if $x\in X\setminus\{x_0\}$ and $y\in X\setminus\{x\}$, there is a $U\in\tau$ such $x\in U$ and $\{y,x_0\}\cap U=\varnothing$; $x_0\notin U$, so $U\in\sigma$. Thus, $\langle X,\sigma\rangle$ is $T_1$ except possibly at $x_0$. Suppose that $x\in X\setminus\{x_0\}$, and let $U=X\setminus\{x\}$; $\langle X,\tau\rangle$ is $T_1$, so $U\in\tau$, and certainly $x_0\in U$. Finally, $X\setminus U$ is finite, and $\mathscr{F}$ is non-principal, so $U\in\mathscr{F}$. Thus, $U\in\sigma$, and $\langle X,\sigma\rangle$ is $T_1$.
If $\mathscr{F}$ is the principal ultrafilter at some point $y\in X\setminus\{x_0\}$, then $\langle X,\sigma\rangle$ is not $T_1$: every $\sigma$-nbhd of $x_0$ contains $y$. Finally, $\mathscr{F}$ cannot be the principal ultrafilter at $x_0$, because that converges to $x_0$.
