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?

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

1 Answer 1


Assuming the trivial topology as in a previous version of this question, we reason as follows.

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 $\overline{A}=T$.

  • $\begingroup$ 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? $\endgroup$
    – user895986
    Jul 14, 2021 at 17:37
  • $\begingroup$ 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 $A$. $\endgroup$ Jul 14, 2021 at 17:46

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