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$\emptyset$ is called both open and closed. Trying to understand this. It's complement is $R$ for instance. Does that mean R is both open and closed? R is unbounded for sure, but every cluster point of R is contained in R (which satisfies the definition of being closed). R is closed looked from this point of view. This would make $$\emptyset$$ open.

But R could also be considered as open, since every point $p_o$ has a neighbourhood which is interior to $R$. Is this why $\emptyset$ is considered both open and closed, therefore $R$ is considered both open and closed?

Another example :

Considered the set $$S={x^2+y^2≥0}$$ Can this be considered both open and closed then?. This condition is satisfied for all $x,y$ therefore this set would be the $R$ itself. Complement of this set is the empty set, and since empty set is considered both open and closed as proved above. This set is also considered both open and closed.

Andrés E. Caicedo
  • 79,201
  • In every topological space, the empty set and the full space are both open and closed. – Ian Oct 15 '16 at 14:35
  • Please use the notation $\varnothing$ for the empty set. The notation $\emptyset$ comes from computer science, not mathematics. – Bernard Oct 15 '16 at 14:43
  • @Bernard Care to give a citation? That is the output from \emptyset under normal LaTeX settings... – Ian Oct 15 '16 at 17:47
  • I don't have one in mind at the moment, but just look at any math bookwritten before 1960, e.g. any volume of Bourbaki's treatise of mathematics. – Bernard Oct 15 '16 at 17:51
  • Could be please treat the question as "Could this proof Xenidia suggested be an answer?" instead of "What is the answer of the theorem"

    I wrote theorem since answer of the question is not given in the duplicate. Since the question is whether this proof is wrong or not, but I couldn't find an answer which would suggest that this proof is correct, in the corresponding duplicate.

     –  Oct 15 '16 at 18:25

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