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This question is from "topology without tears" book, page 30. Example 1.2.3

$\{1\}$ has as complement $\mathbb{N}\setminus \{1\}$ , so it is not finite . ($\mathbb{N}$ is set of natural numbers with the cofinite topology) Then the book concludes that $\{1\}$ is not open!?

My question is : where this conclusion comes from? Does it mean that if complement of a set is infinite then that set is not open?

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  • $\begingroup$ well, what topology has $\mathbb{N}$ equipped on it? $\endgroup$ – Riccardo Oct 31 '15 at 21:35
  • $\begingroup$ For example, if $\mathbb{N}$ has the COFINITE topology on it (which seems the case) then by definition of cofinite topology, a subset whose complementary is NOT finite, then it is NOT open. But still, it depends on your hypothesis $\endgroup$ – Riccardo Oct 31 '15 at 21:37
  • $\begingroup$ Perfect! You are right. This is topology: t =$ \ emptyset$ union S. S is $\subset {N} such that N - is finite $\endgroup$ – Omid Oct 31 '15 at 21:43
  • $\begingroup$ hope it's clear now, cofinite topology describe precisely what are opens and what are NOT. you just need to check wether your set satisfies the condition or not. $\endgroup$ – Riccardo Oct 31 '15 at 21:46
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It would be helpfull if you stated the complete example, especially the topology would be helpfull, just like Riccardo asked.

Now I looked up the book and it says: the topology consists of each subset $S$ of $\mathbb{N}$ such that the complement of $S$ is a finite set. Clearly $\mathbb{N}\backslash \{1\}$ is an infinite set, hence $\{1\}$ cannot be open.

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