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So We need to show that our topology is power set of $X$. how can I proceed?

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closed as off-topic by Arnaud Mortier, Shailesh, Feng Shao, The Count, 0XLR Aug 18 at 0:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Arnaud Mortier, Shailesh, Feng Shao, The Count, 0XLR
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ It's wrong in general. Otherwise there wouldn't be two distinct names for it. This question lacks context as such. $\endgroup$ – Arnaud Mortier Aug 17 at 20:04
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    $\begingroup$ What is $X$? This is not true unless you have a specific $X$ in mind. $\endgroup$ – 0XLR Aug 17 at 20:05
  • $\begingroup$ I don't know the exact details of the question, just remember my professor saying something like this. But i think the new edit gives sufficient information. $\endgroup$ – Cosmic Aug 17 at 20:08
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    $\begingroup$ With the new edit, the result is now true, but it's also really elementary, and still lacks context. Please read this post first. $\endgroup$ – Arnaud Mortier Aug 17 at 20:14
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The cofinite topology on $X$ consists of the empty set and all cofinite sets, meaning all sets with finite complement. Since every subset of $X$ is finite, so is their complement. Therefore, the cofinite topology is equal to the power set.

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Given any set $X$, we can define a topology $\tau$ on it, called the cofinite topology, by declaring the empty set $\phi$ and all the cofinite (look here for the definition of cofinite subset) subsets of $X$ to be the open sets (why is this collection form a topology on $X$?).

Now if $X$ is finite then the complement of any subset of $X$ is open in $\tau$ (why?).

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