# Closure of a set $\overline{A}:=\{x\in T: U\cap A \neq \emptyset \text{ for every open set U that contains x}\}$

I have a lemma in my textbook which I don't intuitively understand completely. Here's the lemma:

The closure of $$A, \overline{A}$$, is the set

$$\overline{A}:=\{x\in T: U\cap A \neq \emptyset \text{ for every open set U that contains x}\}$$

An example with which I hope I can gain intuition on why this is true:

Let $$T=[0,4]$$ $$A=(1,2)$$ and a topology on $$T$$, $$\mathcal{T}=\{\emptyset,T,(0.5,1.5),(1.6,3),(0.5,1.5)\cup (1.6,3)\}$$

I don't understand how $$\overline{A}=[1,2]=\{x\in T: T\cap A=\emptyset\}$$

Isn't $$T\cap A= A= (1,2)$$? Since any $$x\in A^c = [0,1] \cup [2,4]$$ which intersects with $$A$$ is not possible i.e if $$x=\{1\}$$ and $$y=\{2\}$$ then $$x\cap A = \emptyset$$ and $$y\cap A = \emptyset$$.

What am I doing wrong?

• As shown in Andrew's answer, the closure of any nonempty set in the indiscrete (Trivial) topology is always the whole space.
– Alan
Jul 14, 2021 at 17:35
• @Alan I changed it now
– user895986
Jul 14, 2021 at 17:38

Let $$x\in A$$. Clearly, $$x\in T\cap A=A$$, so $$x\in\overline{A}$$.
If $$x\in T\setminus A$$, once again the only open set that contains $$x$$ is $$T$$ itself which has nonempty intersection with $$A$$. So $$x\in\overline{A}$$, and we should have $$\overline{A}=T$$.
In the new topology given in the current version of this question, note that every nonempty open set intersects $$A$$. So in this case, we also have $$\overline{A}=T$$.
• Hi Andrew. Sorry, I meant to put a topology where the closure would be $[1,2]$. Do you mind addressing that topology or shall I create a new question?
• In the new topology, every nonempty open set intersects $A$. So $\overline{A}=T$ even in this new topology, since for any $x\in T$, all the open sets containing $x$ will intersect $A$. Jul 14, 2021 at 17:46