I have to compute the interior and closure of $\mathbb{Q}$ in $\mathbb{R}$, but I know that in the Euclidean topology, the interior is the empty set and the closure is whole $\mathbb{R}$. I have to compute these when $\mathbb{R}$ is endowed with the following topologies:

  1. The discrete topology: the topology whose opens sets are all subsets of $X$.
  2. The co-finite topology: the topology whose open sets are the empty set and complements of finite subsets of $X$
  3. The co-countable topology: the topology whose open sets are the empty set and complements of subsets of $X$ which are at most countable.

I think this lemma would be very useful: $x$ is in the interior of $A$ if and only if there exists $U\in \mathcal{B}_x$ (a basis of neighborhoods around $X$) such that $U \subset A$. And $x$ in the closure of $A$ if and only if, for all $U \in\mathcal{B}_x, U \cap A \neq \emptyset$

Who can help me? Thank you!


The following is also useful: the interior of $A$ is the largest open subset of $A$, and the closure of $A$ is the smallest closed subset that contains $A$.

In the discrete topology, all subsets are open and closed. By the above the interior of $\mathbb{Q}$ is $\mathbb{Q}$ and so is its closure.

In the cofinite topology, the only closed sets are the finite ones, and $\mathbb{R}$ itself. So the closure of any infinite set is $\mathbb{R}$. The only open sets are the empty set and all sets with finite complements, none of which fit inside $\mathbb{Q}$. So the interior is empty here.

In the co-countable topology $\mathbb{Q}$ is closed, so equal to its own closure. No non-empty open subset fits inside the rationals, as in the cofinite case, so again the interior is empty.

  • $\begingroup$ Thank you very much! With this I understand interior and closure much more than before :) $\endgroup$ – jbuser430 Jan 9 '16 at 13:05
  • $\begingroup$ @JBIBB Glad to help. Your lemma can also be used, but it's better to know more methods. $\endgroup$ – Henno Brandsma Jan 9 '16 at 13:06

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