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Let $\mathbb R$ be given the cofinite topology. Find ${\mathbb Q}^{\circ},\bar {\mathbb Q}$ and frontier of $\mathbb Q$.

The definition of and interior point $x$ of given set $N$ is that if $\exists$ open set $U$ s.t. $x\in U \subset N$. But if $\exists$ such $U$, the cardinality of $U$ must be larger than $\mathbb Q$ since $\mathbb R \backslash \mathbb Q$ is infinite. So ${\mathbb Q}^{\circ}= \emptyset$

Closure is the smallest closed set. I first compute the closed set $\{ X \} \cup \{ V \subset \mathbb R : \mathbb R \backslash V $ is infinite$ \}$. So I guess the closure of $\mathbb Q$ is $\mathbb Q$ itself.

The frontier, or the boundary, is defined as $∂A=\bar{A}-A^{\circ}$. With this, I conclude frontier of $\mathbb Q$ is again $\mathbb Q$ itself.

Please tell me if I am wrong or not, because I find it very hard in studying topology. The intuitive picture is quite clear but when I use definition to solve questions, I find it very difficult.

Thank you!

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  • $\begingroup$ $\mathbb{Q}$ is neither open nor closed in cofinite topology. So, it can't be closure. $\endgroup$ – HumbleStudent Jan 22 '17 at 7:20
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$\mathbb{Q}$ is neither open nor closed in cofinite topology. So, it can't be closure.

Closure should be a closed set; however, only closed sets are either finite or the whole thing. Therefore the closure of $\mathbb{Q}$ is $\mathbb{R}$.

Another way to see this is that every point is the limit point in the cofinite topology and hence the closure is the whole set $\mathbb{R}$.

Therfore the frontier would be the irrationals $\mathbb{Q}$ - $\mathbb{R}$

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Your reasoning for the first part is correct, although I'd word it a bit better. Something like "if $U\subseteq \Bbb{Q}$ is open, then $(\Bbb{R}\setminus\Bbb{Q})\subseteq(\Bbb{R}\setminus U)$, and the first is infinite but the second is finite, which is a contradiction".

Your reasoning for the second is wrong; note that in the cofinite topology, "closed set" is synonymous with "finite set (or all of $\Bbb R$)", because closed sets are complements of open sets.

So in fact, since $\bar{\Bbb Q}$ is the intersection of all closed sets containing $\Bbb Q$, and by what we just said, the only closed set containing $\Bbb Q$ is $\Bbb R$, then we see $\bar{\Bbb Q}=\Bbb R$.

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