Closed/open sets

sorry! am not clear with these questions
1.why an empty set is open as well as closed?
2.why the set of all real numbers is open as well as closed?

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Why would you put that title?!? –  user641 Mar 27 '11 at 8:47

Well the definition of a topological space $X$ specifies that both $X$ and the empty set must be open sets (if the topology is defined in terms of closed sets rather than open sets, it will stipulate that they are closed). But then it is just by definition that it must be open (or closed).

Then a set $A$ is said to be closed if and only if its complement $X - A$ is open. So if you look at the empty set its complement is $X - \emptyset = X$ and $X$ is open by definition. Therefore the empty set is closed.

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By definition, a set $A$ of real numbers is open when the following condition is met: $$\hbox{\forall x\in A, \exists\epsilon>0 such that (x-\epsilon,x+\epsilon)\subset A,}$$ where $(a,b)$ denotes the open interval $\{x\in{\Bbb R}\,|\,a<x<b\}$. It should be not hard to convince yourself that the subsets $A=\emptyset$ and $A={\Bbb R}$ satisfy this condition.

Then, remember that $$\hbox{A is open \iff {\Bbb R}\setminus A is closed}$$ again by definition. You conclude since $\emptyset={\Bbb R}\setminus{\Bbb R}$ and ${\Bbb R}={\Bbb R}\setminus\emptyset$.

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I would not say by definition as the definition given above is tied to metric spaces while there already exists a definition for topological spaces. –  Wok Mar 31 '11 at 10:47
+1 Since leopard considers the real numbers only. –  Wok Mar 31 '11 at 10:52
Well, clearly there are more general definitions of topological and metric spaces (and their opens sets) of which this is but a special case, but my intention was to discuss as much concretely as possible the case put forward by the questioner, assuming implicitly that the topology under consideration is the standard one. –  Andrea Mori Mar 31 '11 at 14:03
$\emptyset$ is open. By definition, $X$ is open if $\forall x\in X$, there is a open set $U\subset X$ such that $x\in U$. So there is not any point in $\emptyset$, the condition of the definition is automatically satisfied (a logical convention).
$\mathbb{R}$ is open (check Andrea's answer), so its complement, $\emptyset$ is closed.
Therefore, $\emptyset$ is both open and closed.
OK, in topology, $\emptyset$ and $\mathbb{R}$ are open sets, by the axioms. By definition of closed set, $\emptyset=\mathbb{R}\setminus\mathbb{R}$ is closed. This simply restates Adrian's answer. –  sihong xie Oct 19 '12 at 17:15