Lindelöf and co-countable topology

Question

Excuse me i have a question i can not fully got it Let Y be a non-empty set . Endow Y with the co-countable topology , so that a subset A of Y is closed if and only if A=Y or A is countable . Prove that Y is $$\text{Lindelöf}$$

i know that i must take open cover of Y and got to it has countable $$\text{subcover}$$ but i could not do it can you help me please ?

Answer 1

Let $\{U_i, i \in I\}$ be an open cover of $Y$.

Let $U_{i_0}$ be any fixed open set from the cover. Either $U_{i_0} = Y$ and we have a finite subcover $\{U_{i_0}\}$, or $U_{i_0} \neq Y$ and this means that $A:=Y\setminus U_{i_0}$ is non-empty and closed and so $A$ must be countable by the definition of the co-countable topology.

Now for each $a \in A$ pick $U_{i(a)}$ from our cover so that $a \in U_{i(a)}$.

Finally, $\{U_{i(a)} \mid a \in A\} \cup \{U_{i_0}\}$ is a countable subcover of $\{U_i, i \in I\}$. So $Y$ is Lindelöf.

It is a simple adaptation of the proof that the cofinite topology on any set is compact.

Answer 2

Let $\mathscr{U}$ be an open cover of $Y$. Let $U \in \mathscr{U}$. Since $U$ is open, then $U^c = \{x_1,x_2, x_3, \dots\}$, i.e. $U^c$ is countable. Now, note that $Y = U \cup U^c$. Since $\mathscr{U}$ covers $Y$, then for each $x_i \in U^c$, there is some $U_i \in \mathscr{U}$ so that $x_i \in U_i$. Now, $\{U, U_1, U_2, U_3, \dots\}$ is a countable subcover of $\mathscr{U}$. Hence, $Y$ is Lindelof.