The problem statement, all variables and given/known data.

Let $X$ be a set and let $\tau=\{U \in \mathcal P(X) : X \setminus U \text{is finite}\} \cup \{\emptyset\}$. This is called the cofinite topology on $X$. Describe the interior, the closure and the boundary of the subsets of $X$ with respect to $\tau$.

The attempt at a solution.

If $A \subset X$, then $A^{\circ}=\{x \in A: \exists U \in \tau : x \in U \subset A\}$. Now, as $U \in \tau$, then $X \setminus U$ is finite. I don't know what this is telling me about $A^{\circ}$.

The definition of closure of a set $A$ is $\overline{A}=\cap_{F \text{closed}, A \subset F} F$. But $F^c=U$, an open set which belongs to $\tau$. Again, I don't know how to use this to describe the closure of $A$.

Now, the boundary of $A$ is $∂A=\overline{A}-A^{\circ}$.

I am pretty lost, I would appreciate if someone could help me to characterize these three in terms of the given $\tau$.

  • $\begingroup$ If $X\backslash U$ is finite then $X\backslash A$ is also finite. it follow that $A^{\circ}=A$ if $A$ is open and $A^{\circ}=\emptyset$ if $A$ is not open $\endgroup$
    – Hamou
    Aug 12, 2014 at 5:30

1 Answer 1


The interior of $A$ is defined as the largest open set containing $A$ (this is equivalent to your definition since union of open sets are open). If $A$ is open (i.e. if $X \backslash A$ is finite), it is a general fact that open sets are equal to their interior, so you should have $A^{\circ} = A$. Otherwise $X \backslash A$ is infinite, and if $B \subseteq A$, then $X \backslash B$ has greater cardinality than $X \backslash A$, so it is also infinite. This shows that $A^{\circ} = $?

Closed sets are finite sets (and $X$). The closure of $A$ is the smallest closed set containing $A$ (this is equivalent to your definition, since intersections of closed sets are closed). If $A$ is finite, then it is closed, and again closed sets are always equal to their closure in a topology. If $A$ is not finite, then what are the closed sets containing $A$? Take their intersection (it is easily computed).

If you computed the two previous steps using my hints, the boundary should be no problem ; split in cases whether $A$ and/or its complement are finite or not.

Hope that helps,

  • $\begingroup$ I want to check if I could follow your answer: if $A$ is not open, as you've said, it follows that $X \setminus B$ is infinite for every $B \subset A$, so there is no open set contained in $A$. But then $A^{\circ}=\emptyset$. If $A$ is not finite, then all $F$ such that $A \subset F$ can't be finite, from here it follows $\overline A=\emptyset$ in that particular case. $\endgroup$
    – user100106
    Aug 12, 2014 at 5:54
  • $\begingroup$ or $\overline{A}=X$. As for the boundary, we have four possible combinations $A,A^c$ finite, which means $A=\overline{A}$ and $X-A$ is finite, so $A$ is clopen, it follows $∂A=A-A=\emptyset$. Then $A,A^c$ are infinite, but then $\overline{A}=\emptyset$ and $A^{\circ}=\emptyset$, so $∂A=\emptyset$. If $A$ is finite and $A^c$ is infinite, then $A$ is closed and $A^{\circ}=\emptyset$, so $∂A=A$. Finally, if $A$ is infinite and $A^c$ is finite, then $A$ is open and $\overline{A}=\emptyset$ or $\overline{A}=X$. But as $A^{\circ} \subset \overline{A}$, then $∂A=\emptyset$ or $∂A=X-A$. $\endgroup$
    – user100106
    Aug 12, 2014 at 6:14
  • $\begingroup$ I would appreciate if you could tell me if what I wrote is ok and correct me if it isnot. $\endgroup$
    – user100106
    Aug 12, 2014 at 6:17
  • $\begingroup$ @user100106 : You always have $A \subseteq \overline A$, so $\overline A = \varnothing$ only holds if $A = \varnothing$. You should remove the case $\overline A = \varnothing$ from most of your arguments ; note that the boundary is entirely determined by the (in)/finiteness of $A$/$A^c$. If $A$ and $\overline A^c$ are finite, then $X$ is finite, so in particular it has the discrete topology and all sets have empty boundary (what you said was correct, I was just commenting). If $A$ and $A^c$ are infinite, then $\overline A = X$ and $A^{\circ} = \varnothing$. You should correct your answer. $\endgroup$ Aug 12, 2014 at 17:14

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