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This question already has an answer here:

I think this is a basic question, but it's hard to wrap my head around.

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marked as duplicate by Daniel W. Farlow, Asaf Karagila elementary-set-theory Feb 6 '16 at 21:57

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  • $\begingroup$ By the definition, complements of open sets are closed. $\endgroup$ – Marko Feb 6 '16 at 19:22
  • $\begingroup$ Then why can't I pick (say, $X$ is open, $\emptyset$ is closed) and stick with it? Then the definition is satisfied, no? $\endgroup$ – bsm Feb 6 '16 at 19:23
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    $\begingroup$ By definition. A topology on $X$ contains $\varnothing$ and $X$ by definition of a topology, so both are open. And by definition of a closed set - $A$ is closed if and only if $X\setminus A$ is open - we see that $\varnothing = X \setminus X$ and $X = X \setminus \varnothing$ are also closed. $\endgroup$ – Daniel Fischer Feb 6 '16 at 19:23
  • $\begingroup$ Maybe you find it helpful to consider why this is true in metric spaces. Since topological spaces generalize them, the definition is just natural. $\endgroup$ – user42761 Feb 6 '16 at 19:25
  • $\begingroup$ Is your difficulty based on the fact that "open" and "closed" are opposites in normal English usage? Try to go back to the definitions, and ignore intuitions about what "open" and "closed" mean to non-mathematicians. Or are you asking why "open" and "closed" are useful ways to categorize sets? The "why" question is difficult to answer without knowing what precisely is bothering you. $\endgroup$ – Eric Lippert Feb 6 '16 at 19:31
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Both are open sets because they are neighborhoods of all their points. That's why you always want to add them in your topology, also their complementaries are open so they are indeed open and closed.

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If you're asking about why the definition is the way it is, I think it is (at least partially) because it makes the most sense for generalizing the notion of continuous functions.

If $f \colon \mathbb{R} \to \mathbb{R}$ is continuous and the image of $f$ misses an interval, you still want the preimage of that interval to be open, so the empty set should be an element of the topology. Similarly, if the image of $\mathbb{R}$ is contained in an open (or closed) interval, you want the preimage to be open (or closed), so the whole space should be an element of the topology.

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Look at a the definition of a topological space. And look at the definition of a closed set. A topological space is by definition a set $X$, together with a collection of subsets of $X$, called open sets, such that:

(i) $\emptyset$ and $X$ are open sets.

(ii) The union of any collection of open sets is open.

(iii) The intersection of a finite collection of open sets is open.

Now, by definition, a subset $F$ of $X$ is called closed if $X \setminus F$ is an open set. Check that $X \setminus \emptyset$ and $X \setminus X$ are open sets, and you're done.

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