Let $X$ be an uncountable set endowed with the co-countable topology
Set $X=[0,3]$ and $A=[2,3]$
- Show that $1$ belongs to $A’$.
I have to show that all neighborhoods of $1$ intersect $A$. This means
$\forall V\in V(1), V \cap A\ne \emptyset$.
I think, any neighborhood of $1$ will be in the form $X\setminus C$ where $C$ is a countable set, and $1 \notin C$. And we have $(X\setminus C) \cap A\ne \emptyset$ (but I don’t know how can I prove that).
- Let $(x_n)$ be a sequence in $A$. Can $(x_n)$ converges to $1$?
If $(x_n)$ is a sequence in $A\setminus \{1\}$, define $B=X\setminus \{ x_n : n\in \Bbb{N} \}$, which is open and is a neighborhood of $1$ that does not contain any points of the sequence. Therefore $(x_n)$ does not converge to $1$.
Is that true, please?
Second part is correct. For the first part assume that $(X\setminus C)\cap A=\emptyset$. Then we get $A \subset C$ which contradicts the fact that $A$ is uncountable and $C$ is countable. Hence $(X\setminus C)\cap A\neq \emptyset$.