I am trying to solve the following exercise, but I have some questions.
Let $(X,\tau)$ be a topological space, for $p\notin X$ we define $Y=X\sqcup\{p\}$ and $\tau_{Y}=\tau\cup\{Y\}$. Study the countability and separation axioms, and the Lindelöf property in $(Y,\tau_{Y})$.
We know that $p$ is closed in $Y$, since $Y\setminus \{p\}=X$ was open in $\tau$ and that means it is open in $\tau_Y$. So, if $X$ was $T_1$, then every point of $X$ was closed and then $Y$ is $T_1$ too.
$T_1$ implies $T_0$, so $Y$ is $T_0$ too if $X$ was $T_1$. But I think maybe $Y$ is trivially $T_0$, since every other point is contained in an open set that doesn't contain $p$.
I think $Y$ is not Hausdorff, since the only open set containing $p$ is $Y$ and for any $y\in Y$ there doesn't exist an open set $U$ such that $y\in U$ and $U\cap Y = \emptyset$.
Am I right about these suppositions?
I have trouble thinking about bases in the new space to see the countability.
If $x \in X$, then $Y \setminus \{x\}$ is not open, as it is not in $\tau_X$ (as $ p\in Y \setminus \{x\}$) and the only open set that contains $p$ is $Y$.
So $Y$ is not $T_1$ if $X$ is non-empty. If $X$ was $T_0$, so is $Y$ (check this).
$Y$ is Lindelöf regardless of $X$ (also compact) because the only way to cover $p$ by open sets is to have $Y$ as a set in the cover giving a one-set subcover.
If $X$ is separable, or second countable, so is $Y$. (easy to check). Likewise for first countability.