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I just want to confirm if my proof is correct for the statement: "The empty set is closed". Although I know that the empty set is also open, the proof is similar.

Proof. Suppose ∅ is open, then for all x that is an element of ∅ there exist a ball B(x,r) that is centered at x with radius r such that B(x,r) is a subset of ∅. But this implies that ∅ is not the empty set since it contains some elements. Thus, if ∅ is open then ∅ is not the empty set, but arguing contrapositively, if ∅ is the empty set then ∅ is closed. QED

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    $\begingroup$ Something is wrong with your proof. You start out by writing that you know that the empty set is open, but then it looks like that your argue that if $\emptyset$ is open, then it cannot be empty? $\endgroup$ – Mankind Aug 8 '15 at 14:12
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    $\begingroup$ "The empty set is closed". Proof: by definition. P.S. Okay, $\emptyset=X\setminus X$ and $X$ is open by definition. $\endgroup$ – A.Γ. Aug 8 '15 at 14:13
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    $\begingroup$ Also, the contrapositive of the statement "$X$ is open implies that $X$ is not empty" is not "$X$ is empty implies that $X$ is closed". Not being open is not the same as being closed. $\endgroup$ – Mankind Aug 8 '15 at 14:14
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    $\begingroup$ Think carefully about the statement "then for all $x$ there exists a ball $B(x, r)$ ...", if there is no $x \in \emptyset$, this statement it vacuously true. It doesn't force $\emptyset$ to contain an $x$. $\endgroup$ – Seven Aug 8 '15 at 14:15
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    $\begingroup$ Any set $X$, we have $X=X \setminus \emptyset$. $\endgroup$ – Ramiro Aug 8 '15 at 14:43
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Using your method to 'prove' that $\varnothing$ is open and closed you should reason like this:

  • Is it true that for every element $x$ in $\varnothing$ there is an open ball $B(x,r)$ with $x\in B(x,r)\subseteq\varnothing$? Yes! Because there is no element in $\varnothing$ for wich this is not true. Even stronger: there is no element in $\varnothing$ at all! Conclusion: $\varnothing$ is open.

  • Is it true that for every element $x\in X$ there is an open ball $B(x,r)$ with $x\in B(x,r)\subseteq X$? Yes! That is obvious. We conclude that $X$ is open or equivalent that its complement $\varnothing$ is closed.

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  • $\begingroup$ How do you write ∅ as the complement of X? ∅=X\X? Also how about X as the complement of ∅? X=∅\∅? $\endgroup$ – shinobi20 Aug 8 '15 at 14:45
  • $\begingroup$ In your second statement, you said "for every element x∈X in ∅" how is that? $\endgroup$ – shinobi20 Aug 8 '15 at 14:57
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    $\begingroup$ Usual notations are $\varnothing=X^c$ and $X=\varnothing^c$. The part "in $\varnothing$" in the second statement was wrong (product of copying and pasting) is removed now. Thanks for attending me. $\endgroup$ – drhab Aug 8 '15 at 15:51
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No, your argument is flawed.

When you say that it contains some elements, then that's where you get wrong. If you said for all elements $x$ then $P$ where $P$ is some property, then that doesn't imply that it has an element. It's an example of what is called a vacuously true statement. If you want to argue it, you can prove that its negation is false. This is how it would go. Assume there is $x$ such that $\not\ P$ , but there is no $x$, so the negation is false. Thus, it is true.

Now, you can argue in the same manner to prove that it's closed.

Assume it contains a limit point that doesn't belong to it. But it has no limit points, then it doesn't contain any limit point which doesn't belong to it. Then, its all limit points (which are non-existent basically) belong to it, and we are done.

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  • $\begingroup$ So you think this is okay? Assume there exist an x in ∅ such that there is no ball B(x,r) whose center is x and B(x,r) is a subset of ∅. But there is no element in ∅, so the negation is false. Thus it is true that ∅ is closed $\endgroup$ – shinobi20 Aug 8 '15 at 14:26
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    $\begingroup$ Yes, this is how it goes. The thing is, any property that relates to elements of $∅$ is true. $\endgroup$ – Hasan Saad Aug 8 '15 at 14:34
  • $\begingroup$ Sorry but I think it should be "there is a ball B(x,r) whose center is x and B(x,r) is a subset of ∅" so that if the negation is flase, it should be that there is NO ball etc. Because by what I said, I proved that ∅ is open by saying that the negation is false hence there is a ball. $\endgroup$ – shinobi20 Aug 8 '15 at 14:56
  • $\begingroup$ Wait, I misread what you said, sorry. But what do balls have to do with its being closed? The definition of closed is either that its complement is open (which is straight forward) or it contains all its limit points. $\endgroup$ – Hasan Saad Aug 8 '15 at 14:58
  • $\begingroup$ The definition of an open set from Apostol's book is "A set S in R^n is called open if all its points are interior points", while the definition of interior points is "Suppose a is an element of S. Then a is called an interior point of S if there is an open ball with center at a all of whose points belong to S. $\endgroup$ – shinobi20 Aug 8 '15 at 15:13
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The empty set is both open and closed. By definition of a topology both the whole space and the empty set are open. Since the empty set is the complement of the whole set it is also closed. Your proof does not work since the condition for being an open set you want to use states that a set $A$ is open if any $x$ in $A$ is contained in a basic open set contained in $A$. But since the empty set has no elements, there is nothing to check and the condition is trivially satisfied.

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  • $\begingroup$ So you mean by arguing that there is nothing to check whether there is something that satisfies the definition or falsify it, gives already the proof? $\endgroup$ – shinobi20 Aug 8 '15 at 14:19
  • $\begingroup$ Yes, if you define a set $A$ to be open if each of its point is contained in an open ball contained in $A$, then since there is no $x\in \emptyset$, $\emptyset$ is open as the condition is automatically satisfied. But to prove that it is closed you have to do something! $\endgroup$ – GFR Aug 8 '15 at 14:23
  • $\begingroup$ Can I assume that there is an x in ∅ such that there is a ball that is a subset of ∅. But since there is no element in ∅, the statement "there is a ball that is a subset of ∅" is false. Hence there is no ball that is a subset of ∅. Thus ∅ is closed. $\endgroup$ – shinobi20 Aug 8 '15 at 14:32

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