I have a lemma which states the following:
The closure of $A, \overline{A}$, is the set
$\overline{A}:=\{x\in T: U\cap A \neq \emptyset \text{ for every open set U that contains x}\}$
I would like to give an example to know if I'm do anything wrong:
Let $T=[0,2]$, let the topology on $T$ be $\mathcal{T}=\{\emptyset,T,(0.3,0.4)\}$. Let $A=(1,1.5)$.
By the definition, the closure of $\overline{A}$ of a set $A\subset T$ is the intersection of all closed sets that contain $A$.
So, the closed sets are $T, \emptyset, [0,0.3]\cup [0.4,2]$. $A$ is in $T$ and $[0,0.3]\cup [0.4,2]$. The intersection of these is $[0,0.3]\cup [0.4,2]$. So $\overline{A}=[0,0.3]\cup [0.4,2]$. Is this correct?
By the lemma, $\overline{A}:=\{x\in T: U\cap A \neq \emptyset \text{ for every open set U that contains x}\} = T$ since $T$ is the only open set that contains $A$
What am I doing wrong?
Best Answer
Your very last equality is wrong. If $a\in(0.3,0.4)$, then we can take $U=(0.3,0.4)$ which does not intersect $A$. Hence, $a\notin\{x\in T:U\cap A\neq\emptyset\text{ for every open }U\text{ containing }x\}$. After fixing this, you see that the closure is indeed $[0,0.3]\cup[0.4,2]$.