This is a continuation of my previous post, Properties of $\tau_1 \lor \tau_2$
Let $X$ be a set equipped with any topology $\tau_1$, let $\tau_2$ be the cocountable topology (i.e. a subset $U$ of $X$ is open if either $U=\emptyset$ or $X\setminus U$ is finite or countable), and let $\tau=\tau_1\lor \tau_2$ be defined as the subsets $W\subseteq X$ such that for every $x \in W$, there exist subsets $U,V \subseteq X$ with $x \in U$, $x \in V$, $U \in \tau_1$, $V \in \tau_2$, and $U \cap V \subseteq W$.
1.) Which sequences in $X$ converge with respect to $\tau$?
My thought is that a sequence which is eventually constant will converge, but I'm not sure how to go about proving this.
2.) Prove or disprove: $(X, \tau)$ is a $T_1$-space.
My intuition is that $(X, \tau)$ is in fact a $T_1$-space, but I'm not sure if this is correct. By definition, $(X, \tau)$ is $T_1$ if for every $x,y \in X$ with $x \neq y$, there exists an open set $U \subseteq X$ such that $x \in U$ but $y \notin U$.
3.) Suppose that $(X, \tau_1)$ satisfies the local countability condition at a point $p \in X$. Show that if $p$ is a condensation point of $X$ with respect to $\tau_1$, then $(X, \tau)$ does not satisfy the local countability condition at $p$.
Since $(X, \tau_1)$ satisfies the local countability condition at $p$, there exists a sequence of subsets $\{U_i\}_{i=1}^{\infty}$ such that
1) $U_i \in \tau_1$ for all $i$,
2) $ x \in U_i$ for all $i$, and
3) If $V$ is an open subset of $X$ with $x \in V$, then $U_i \subseteq V$ for some $i$.
Since $p$ is a condensation point of $X$ with respect to $\tau_1$, $p$ is a limit point of $X$ with respect to $\tau$. Where do I go from here?
4.) Suppose again that $(X, \tau_1)$ satisfies the local countability condition at $p$. Under what condition will $(X, \tau)$ satisfy the local countability condition at $p$?
I have no idea about this part of the question, so a gentle nudge in the right direction would be appreciated.
Indeed, the only convergent sequences of $\tau:= \tau_1 \lor \tau_2$ are the eventually constant ones. This follows as this holds for the co-countable topology $\tau_2$ and $\tau_2 \subseteq \tau$ (because we can take $U =X$ from $\tau_1$ in the definition of $\tau_1 \lor \tau_2$ or because $\tau_1 \cup \tau_2$ is a subbase for $\tau$ etc.); If $x_n \to x$ in $\tau$ then a fortiori (we can take open sets in $\tau_2$ to use in the definition of convergence w.r.t. $\tau$) $x_n \to x$ in $\tau_2$ so $(x_n)$ is eventually constant. (The latter I showed in this answer).
For all $x \in X$, $X \setminus \{x\} \in \tau_2$, by definition (a singleton is at most countable) and so $X \setminus \{x\} \in \tau$, so that all $\{x\}$ are closed in $(X,\tau)$ and thus $X$ is $T_1$, regardless of $\tau_1$.
Suppose $p$ is a condensation point of $X$ in that $(X,\tau_1)$. Then $p$ is a condensation point of $(X,\tau)$: Let $O$ be open in $\tau$. Then there are open sets $U \in \tau_1$ and a co-countable $V= X \setminus N \in \tau_2$ (so $N \subseteq X$ is at most countable) such that $p \in U \cap V = U\setminus N \subseteq O$. As $p$ is a $\tau_1$-condensation point, $U$ is uncountable so surely $U \setminus N$ is infinite (uncountable even) and so, as $O$ is any open neighbourhood of $p$ in $\tau$, $p$ is a condensation point of $(X,\tau)$.
If $(X, \tau)$ were first countable at $p$ then as $p$ is a condensation point of $X$ in $\tau$ we can always define a sequence $(x_n)$ of distinct points unequal to $x$ such that $x_n \to_\tau x$. (just use a simple recursive definition, we could already do this in a $T_1$ space of which $p$ is a non-isolated point, so condensation point is overkill). But this contradicts 1) and so $p$ cannot be first countable in $(X, \tau)$.
Suppose $p$ is first countable at $(X,\tau)$, let $O_n$ be a countable local $\tau$-base at $p$. As $(X,\tau)$ is $T_1$, we can define a sequence $x_n$ of all distinct points, all $x_n \neq x$, with $x_n \to_\tau x$ iff $x$ is not an isolated point of $X$ (i.e. $X$ is a limit point of $(X,\tau)$. This is a standard argument. But we already saw such sequences cannot exist, so this means that if $p$ is $\tau$-first countable, it must be a $\tau$-isolated point, or equivalently $p$ has a neighbourhood $O \in \tau_1$ such that $O \cap (X\setminus N_p) = \{p\}$ for some $N_p \subseteq X$ at most countable. This means exactly that $p$ has a countable neighbourhood in $\tau_1$. First countability of $p$ in $(X,\tau_1)$ I don't see any effect of.
So $p$ is first countable at $(X,\tau)$ iff $p$ has a countable neighbourhood in $(X,\tau_1)$ iff $p$ is isolated in $(X,\tau)$.