1. ## empty set

I presented a professor with the following proof:

Prove that the empty set is closed.

Proof :

By definition : $\emptyset$ is closed <=> cl $\emptyset\subseteq\emptyset$ <=> ( $x\in$ cl $\emptyset$=> $x\in\emptyset$)

cl $\emptyset$ is the closure of the empty set,and
B(x,r) is a ball of radius r round x

But ,by definition again $x\in$cl $\emptyset$ <=> for all r>0 ,B(x,r) $\cap\emptyset\neq\emptyset$.................................................. ..................1

But , B(x,r) $\cap\emptyset =\emptyset$ => (B(x,r) $\cap\emptyset =\emptyset$ or $x\in\emptyset$) <=>

(B(x,r) $\cap\emptyset\neq\emptyset$ => $x\in\emptyset$)

And using (1) we get : $x\in\emptyset$

Thus ,we have proved:

( $x\in$ cl $\emptyset$=> $x\in\emptyset$)

And the empty set is closed.

The professor did not accept the proof as correct .

Do you agree with him??

2. Originally Posted by alexandros
I presented a professor with the following proof:

Prove that the empty set is closed.

Proof :

By definition : $\emptyset$ is closed <=> cl $\emptyset\subseteq\emptyset$ <=> ( $x\in$ cl $\emptyset$=> $x\in\emptyset$)

cl $\emptyset$ is the closure of the empty set,and
B(x,r) is a ball of radius r round x

But ,by definition again $x\in$cl $\emptyset$ <=> for all r>0 ,B(x,r) $\cap\emptyset\neq\emptyset$.................................................. ..................1

But , B(x,r) $\cap\emptyset =\emptyset$ => (B(x,r) $\cap\emptyset =\emptyset$ or $x\in\emptyset$) <=>

(B(x,r) $\cap\emptyset\neq\emptyset$ => $x\in\emptyset$)

And using (1) we get : $x\in\emptyset$

Thus ,we have proved:

( $x\in$ cl $\emptyset$=> $x\in\emptyset$)

And the empty set is closed.

The professor did not accept the proof as correct .

Do you agree with him??
Yes, I agree this isn't correct. You have said (without comment) that $x\in\varnothing$

3. Well of course I do agree with the instructor.
What you have written is hard to follow at best and most likely nonsense.
A closed set contains all of its limit points.
So the emptyset is by definition closed.

4. Originally Posted by Drexel28
Yes, I agree this isn't correct. You have said (without comment) that $x\in\varnothing$
You mean in a proof we are not allowed to write :

$x\in\emptyset$

5. Originally Posted by alexandros
You mean in a proof we are not allowed to write :

$x\in\emptyset$
Not unless you immediately put "Contradiction!", "Aha!", or "B.S.!" afterwards.

6. Originally Posted by Plato
Well of course I do agree with the instructor.
What you have written is hard to follow at best and most likely nonsense.
A closed set contains all of its limit points.
So the emptyset is by definition closed.
.

Is nonsenses because is hard to follow ?

Is not correct that:

A is closed iff closure(A) $\subset A$ ??

7. Originally Posted by alexandros

A is closed iff closure(A) $\subset A$ ??
Well, more conventionally $E\text{ is closed}\Leftrightarrow E=\overline{E}$ but considering that it is always true that $E\subseteq\overline{E}$ your definition is correct.

8. Originally Posted by Drexel28
Not unless you immediately put "Contradiction!", "Aha!", or "B.S.!" afterwards.
For example when we want to prove that: for all sets ,A : $\emptyset\subset A$, we cannot start by assuming :

Let , $x\in\emptyset$ and then try to prove ,that $x\in A$??

9. Originally Posted by alexandros
For example when we want to prove that: for all sets ,A : $\emptyset\subset A$, we cannot start by assuming :

Let , $x\in\emptyset$ and then try to prove ,that $x\in A$??
You are missing the point. The point is that there is vacuous truth here.

Theorem: Let $A$ be any set in a universal set $U$ then, $\varnothing\subseteq A$

Proof: Suppose not, then there exists some $x\in\varnothing$ such that $x\notin A$ but there is no $x\in\varnothing$. This is clearly a contradiction.

10. Originally Posted by Drexel28
You are missing the point. The point is that there is vacuous truth here.

.
I am sorry i do not get any meaning out of it ,can you explain?

You mean that the assumption:

Let , $x\in\emptyset$ is not acceptable in a proof ??

11. Originally Posted by alexandros
I presented a professor with the following proof:

Prove that the empty set is closed.

Proof :

By definition : $\emptyset$ is closed <=> cl $\emptyset\subseteq\emptyset$ <=> ( $x\in$ cl $\emptyset$=> $x\in\emptyset$)

cl $\emptyset$ is the closure of the empty set,and
B(x,r) is a ball of radius r round x

But ,by definition again $x\in$cl $\emptyset$ <=> for all r>0 ,B(x,r) $\cap\emptyset\neq\emptyset$.................................................. ..................1

But , B(x,r) $\cap\emptyset =\emptyset$ => (B(x,r) $\cap\emptyset =\emptyset$ or $x\in\emptyset$) <=>

(B(x,r) $\cap\emptyset\neq\emptyset$ => $x\in\emptyset$)

And using (1) we get : $x\in\emptyset$

Thus ,we have proved:

( $x\in$ cl $\emptyset$=> $x\in\emptyset$)

And the empty set is closed.

The professor did not accept the proof as correct .

Do you agree with him??

Certainly your proof is 100% correct.

Usually professors are confused with this kind of proof

12. Originally Posted by xalk
Certainly your proof is 100% correct.
I know you're into descriptive set theory or something and technically maybe what the OP wrote is correct, but a big part of becoming mathematically mature is learning what the mathematical community accepts as a decent proof.

Usually professors are confused with this kind of proof
And that's what makes it a bad proof. I know I'll get flack for this but I'm telling you that a proof is not just composed of it's content, it's about its efficiency and coherency.

13. Originally Posted by alexandros
Prove that the empty set is closed.

Proof :

By definition : $\emptyset$ is closed <=> cl $\emptyset\subseteq\emptyset$ <=> ( $x\in$ cl $\emptyset$=> $x\in\emptyset$)

cl $\emptyset$ is the closure of the empty set,and
B(x,r) is a ball of radius r round x

But ,by definition again $x\in$cl $\emptyset$ <=> for all r>0 ,B(x,r) $\cap\emptyset\neq\emptyset$.................................................. ..................1

But , B(x,r) $\cap\emptyset =\emptyset$ => (B(x,r) $\cap\emptyset =\emptyset$ or $x\in\emptyset$) <=>

(B(x,r) $\cap\emptyset\neq\emptyset$ => $x\in\emptyset$)

And using (1) we get : $x\in\emptyset$

Thus ,we have proved:

( $x\in$ cl $\emptyset$=> $x\in\emptyset$)

And the empty set is closed.

The professor did not accept the proof as correct .

Do you agree with him??
Yes, I do. First, because the proof indeed isn't correct. Second, because it would be obfuscatingly complicated.

The mistake is when you say "And using (1), we get $x\in\emptyset$". Not because this conclusion is false or nonsense but just because your apparent logic is flawed.

Indeed, you write correctly (for $A=\emptyset$, but works in general):
a) $x\in\overline{A}\Leftrightarrow(\forall r>0, B(x,r)\cap A\neq \emptyset)$;
b) $\big(B(x,r)\cap A=\emptyset\big)\Rightarrow\big((B(x,r)\cap A\neq\emptyset)\Rightarrow x\in\emptyset\big)$.
(if you look at it right in the eye, b) is somewhat funny)
From there, you deduce $x\in\emptyset$, don't you? Then you've just proved that $\overline{A}=\emptyset$ for any set $A$.

This was an illustration of my second point: obfuscation. You are allowed to write very formal proofs, but that requires skill and insight (both from you and your reader); if you perform blind logical manipulation, you are very likely to end up writing junk, to say the least. Furthermore, maths is not just about writing correct statements that follow logically from one another; it is mainly about ideas, constructions, insights. And writing a proof is usually about wanting it to be understood, not just checked for validity. It is important that proofs are correct, but a good mathematician is also one who knows when an argument is missing without having to write the proof up to every single logic axiom involved.

A correct proof along your lines could be (according to me): if there exists $x\in\overline{\emptyset}$, then in particular $B(x,1)\cap\emptyset\neq\emptyset$, which rewrites as $\emptyset\neq\emptyset$ (the left-hand side being a subset of the empty set). This conclusion is false, hence the assumption was false as well. Thus there is no element in $\overline{\emptyset}$ and we have $\overline{\emptyset}=\emptyset$.

If you see the closure of a set (in a metric space as well) as the set of the limits of convergent sequences with values inside this set then, as Plato said, the conclusion is also straightforward: the set of the sequences inside the empty set is empty hence the set of their limits is empty as well.

14. All in favor of the instructor: Unanimous.

All in favor of the OP: Bupkis.